Talk:Monty Hall problem/Archive 12

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Archive 5

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Archive 10

Variants - Other Host Behaviours

http://en.wikipedia.org/wiki/Monty_Hall_problem#Other_host_behaviors

The term 'Host Behaviour' is used earlier in the article to dis-prove the Popular solutions. And the concept of a host's bias, without the word 'behaviour' is used numerous times to dis-prove the Popular solution. In this section, the same issue is used for some unstated (fun?) purpose. That's probably confusing to readers.

This section is inconsistent with the Monty Hall Problem in that the results shown are not from the contestant's SoK. It seems to me that except for the 'forgetful host variant' which vos Savant described as 'random' (aka Deal or No Deal) the contestant must still rely on the 2/3 average. This table is likely to confuse people, rather than enlighten them.

Does the contestant get a do-over if Monty reveals the car in the 'forgetful host (random, aka Deal or No Deal) variant'? If so, then I guess the contestant could infer that Monty is acting randomly, and determine the odds are 50/50. **Wait a minute, if it's a do-over, and he shows a goat this time, wouldn't it be 2/3 if I switch?** Still no harm in switching, though.

The paragraph under the table relating to 'game theory' also uses the term 'host's behaviour' differently than the earlier occurrence in the article. And the conclusion may not be correct. With no knowledge of the hosts behaviour (motivations?), it's understood that it's best to switch with a 2/3 probability. Glkanter (talk) 17:25, 29 November 2009 (UTC)

I'm not seeing the term used in any way other than to mean constraints on the host. In particular, it is not used at all in the section titled "Probabilistic Solution". The first three variants in the "Other host behaviors" section are all referenced, and are all notable. Sources should be added for the ones that are not sourced. In the paragraph below the table, "host behavior" also means constraints on the host's actions included in the problem statement. The "for example" sentence means if the problem does not constrain the host to always make the offer to switch, the host could be offering the switch only in the case the player has initially picked the car (per the NY Times interview with Monty Hall - see the Tierney 1991 reference). With no constraints on the host's behavior, it's understood that the probability of winning by switching can be anything. That is sort of the point of this section. -- Rick Block (talk) 05:18, 30 November 2009 (UTC)

The most meaningful point is that the contestant, whose State of Knowledge the problem uses, never knows about these biases. When shown a goat, he always sees himself as the 'average contestant' with a 2/3 likelihood of winning by switching. As I mention elsewhere, once a host behaviour is introduced, the problem statement becomes the opposite of the Monty Hall problem statement mentioned twice in the article. I mentioned a few reasons I think a reader, who knows how game shows must work, would be confused by the irrelevant tables, narrative and the illogical change in whose State of Knowledge is being discussed. Glkanter (talk) 05:59, 30 November 2009 (UTC)

The "host behaviors" we're talking about here are constraints that are either taken to be implied by the problem statement or explicitly given as part of the problem statement. The Parade version of the problem statement is widely regarded as ambiguous since these constraints are not explicitly listed, in particular whether the host must make the offer to switch and exactly how the host chooses what door to open. There are countless references that consider the impact of slightly different constraints on the host, that would presumably be the well known rules (i.e. explicit and within the SoK of the contestant). This is a subsection of a major section called "Variants". You're perhaps arguing to delete this section, but can you instead suggest a way to make this more clear? -- Rick Block (talk) 15:18, 30 November 2009 (UTC)

However you describe these things, they all rely on the contestant having knowledge of the host's behaviour. At that point it's no longer the Monty Hall problem about a game show, which is the subject of the article. They're confusing, and do not further the reader's understanding of the Monty Hall problem paradox. Glkanter (talk) 15:33, 30 November 2009 (UTC)

Again, what is being called the "host's behavior" are the rules under which the game show would be run, which would be known to the contestant. It could still be a game show if the host is not required to make the offer to switch (in fact, that might be an interpretation that would be more consistent with the actual Let's Make a Deal show on which the problem is obviously based). The fact that the Parade version doesn't fully specify the host's behavior is in all likelihood one of the reasons the "Monty Hall problem" is so very contentious. Different people read it as different problems, and then insist their interpretation is the "right" one. Discussing variants leads most people (perhaps not you) to a better understanding of the issues that are involved in the problem description. Many, many sources do this. If the article didn't do this, it would not be complete. -- Rick Block (talk) 19:42, 30 November 2009 (UTC)

(1) And again, we must assume that any knowledge that is available to the contestant is also avaiable to the person who is trying to determine what the contestant should do - the puzzle solver. The fact that it is not mentioned, especially in a non-rigorous forum like a sunday supplement (see Seymann), means that it isn't necessary to answer the puzzle. For the elements of policy, like if a switch is always offered, that means what is implied in the one example is the full policy. MvS EXPLICITLY says this is her intent in one of the sources you like to cite when it suits your purpose, but apparently not when it contradicts it. For the elements of random choice, like car placement and host choice, it means it is uniform, something even your sources assume when to do otherwise would not make an "interesting" (see Morgan, and Gilman's intentional misquote of MvS) problem.

(2) Read K&W, and don't give us your unsupported (and unsupportable) opinion. They conclude that the reason for contention has nothing to do with any ambiguity, it is an inability to assess how information based on part of the random nature affects the results. It isn't "different problems," it is not understanding what is essentially the Principle of restricted choice. The two "unconditional camps" - the 1/2'ers and the 2/3'ers are conceptualizing the same problem. And it isn;t the conditioanl problem.

(3) The only thing that discussing variants helps is understanding the variants. Again, not one of your references applies their results to the contestant. Morgan even reduces it to what they call the unconditional problem (but really isn't, it just has the same answers under the assumptions they do make that are equivalent to what they try not to make) when they try to apply it to the contestant. And there is a good reason: they were addressing a variant, not the MHP itself, and they did it only because it was interesting. Not topical. JeffJor (talk) 22:27, 30 November 2009 (UTC)

Wikipedia Editing Works Best With A Consensus

I probably know the least about this topic. But it looks like there is a lot of agreement amongst active editors. Not unanimous, of course, but that's not a requirement, is it? Glkanter (talk) 21:07, 29 November 2009 (UTC)

What specific changes are you thinking there's consensus for? -- Rick Block (talk) 01:23, 30 November 2009 (UTC)

Hi Rick. Thank you for your response. As you see, I've started 4 new sections. I'm especially interested in your responses to the various questions I ask and thought-provoking statements I made, as you appear to not be part of the consensus. We may have differing opinions, but we can work together within the parameters of Wikipedia. I'm laying all my cards out. That's what collaboration's all about, right? So, what's the Wikipedia definition of a 'proper' consensus? Glkanter (talk) 01:38, 30 November 2009 (UTC)

Again, consensus for what? Assuming you're talking about major changes to the article, I would think any reasonable definition of consensus would have to include at least most of the folks who commented at the last featured article review, archived here. There were a flurry of changes since then - running any major changes by the folks involved in those would be a good idea as well (e.g. Dicklyon, Nijdam, Henning Makholm, Glopk). -- Rick Block (talk) 04:54, 30 November 2009 (UTC)

Yes, of course, major changes, de-emphasisng the reliance on Morgan. I think that's consistent with the published views of JeffJor, Martin and myself. With all the documented problems with Morgan, it's time to get to work. I mean, the filibustering has to end sometime, right? I'm sure all those other guys you mentioned are anxious to join the discussion. Because, as I read on the subject of Wikipedia Consensus:

Not hypothetical

While everyone on Wikipedia has the right to be heard, this does not mean that discussions remain open indefinitely until we hear from them. Nor does it mean that a consensus should be overridden by an appeal to "Wikipedians out there" who silently disagree. In essence, silence implies consent. If you believe that the current discussion does not represent real opinion, either prove it by referring to an existing discussion, or suggest starting a new discussion with a wider audience.

Glkanter (talk) 05:10, 30 November 2009 (UTC)

You are misinterpreting what you've quoted from Wikipedia:What is consensus?. We're talking here about things that have already been discussed, not a new topic. We know a specific set of users who have previously expressed opinions on these topics. It's a simple matter to ask them. Do you want to, or should I? -- Rick Block (talk) 15:35, 30 November 2009 (UTC)

You, please. Don't forget my buddy. Glkanter (talk) 16:07, 30 November 2009 (UTC)

Which buddy would that be? -- Rick Block (talk) 16:16, 30 November 2009 (UTC)

Morgan's Paper Doesn't Specifically Mention The Combined Doors Solution

Is it meaningful for our purposes then, to assume, infer, or offer opinions on why or why not?

Can we infer anything about Morgan's thoughts on The Combined Door Solution if it's not specifically mentioned?

I think any such discussion would be out of order. Glkanter (talk) 22:08, 29 November 2009 (UTC)

Are you specifically talking about the sentence that says "Morgan et al. (1991) state that the popular solutions are incomplete, because they all make assumptions about the probabilities after the host has opened a door, without proof"? How about:

Morgan et al. (1991) state that many popular solutions are incomplete, because they do not explicitly address the specific case of a player who has picked Door 1 and has then seen the host open Door 3.

If this is the sentence you're talking about, would you find this wording more acceptable? -- Rick Block (talk) 15:53, 30 November 2009 (UTC)

No, a bigger picture than that. The article is edited as if Morgan's claim trumps all others. I'm pointing out that Morgan didn't state that 'The Combined Doors solution' is false/incomplete/etc'. That's an interpretation of Morgan's paper that you've made, but it's not supported by anything that's been published. So, it's wrong for the article to say, or have a POV, that the Combining Doors solution is in any way inferior/incomplete/etc., because that has not been explicitly published. Glkanter (talk) 16:05, 30 November 2009 (UTC)

Well, then, where in the article are you seeing this criticism of the Combining Doors solution? The only occurrence of the letters "combin" in the entire article is in the paragraph describing this solution. If you're not talking about the sentence I've suggested, what are you talking about? The fourth paragraph in "Sources of confusion"? We can work on that paragraph, too, but if you're not going to say specifically what you don't like it's rather hard to fix it. -- Rick Block (talk) 01:29, 1 December 2009 (UTC)

BTW, if nobody objects in the next day or so, I'll make the change I suggested above. -- Rick Block (talk) 01:32, 1 December 2009 (UTC)

Excessive Reliance On Morgan

Each instance in the article of a host behaviour, or host bias, or host prejudice is indicative of a reliance on Morgan's paper. Regardless of what is being illustrated, this topic only exists among Morgan and a few others.

Since the problem statement of both vos Savant (Whitaker) and Krauss & Wang begin with: "Suppose you are on a game show", we know that this host behaviour will not be shared with the contestant, whose State of Knowledge is the only one asked for in The Monty Hall problem.

So, while Morgan is published, his argument is flawed. It reminds me of what Nijdam once said about Devlin, 'His name is Morgan, not God'. The moment the problem is restated to rely on a host behaviour, it's no longer the Monty Hall problem. The problem statement becomes: 'Suppose you are not on a game show'. Which is the exact opposite of how both Monty Hall problem statements in the article begin.

Yes, Morgan should be referenced, but with such an obviously erroneous argument, it hardly deserves the great emphasis it currently enjoys. Glkanter (talk) 05:50, 30 November 2009 (UTC)

You are completely and utterly wrong about any mention of "host behavior" being indicative of a reliance on Morgan. The Parade description of the problem is nearly universally agreed to be under qualified. What most people intend by the problem is consistent with the Krauss and Wang version, but the difference between this version and the Parade version is precisely the topic of "host behavior". As Seymann says (the quote Martin keeps bringing up): "Without a clear understanding of the precise intent of the questioner, there can be no single correct solution to any problem." It is by specifying the "host behavior" that the precise problem can be understood. If aspects of the host's behavior are not specified you may be able to make reasonable assumptions about them, but your assumptions may not match what whoever is posing the question meant. -- Rick Block (talk) 16:14, 30 November 2009 (UTC)

Rick Block wrote above: "You are completely and utterly wrong about any mention of "host behavior" being indicative of a reliance on Morgan." Rick, I will take abnormal and tremendous joy when you are forced to eat those words. I'm only human. Glkanter (talk) 19:56, 1 December 2009 (UTC)

Rick, I am not sure what you mean when you say that the difference between the K & W version and the parade version is host behaviour. In the K & W statement quoted in this article the host is stated to choose randomly, just like the car is stated to be placed randomly. Martin Hogbin (talk) 10:02, 3 December 2009 (UTC)

"Host behaviour" means several things. Just because you can model all the different aspects with probabilities, that does not make them all fall under one umbrella. When "most people" (and I will supply as much support for that as you just did, Rick: none) use "host behavior," they interpret it to mean "does he always open a door," "can he reveal the car," and even "can he open the chosen door" which Morgan and Gillman ignore. Not "is the host biased toward one functionally equivalent choice over another, based on door numbers?" These are diffferent issues, and only Morgan (plus those that follow using the conditional problem) consider that possibility. But it is not something that can be used to directly address the question "should the contestant switch?" It can only be used by a third party who knows every factor that determines what p₁₃ is on that particular day, to address the question "given this additional information, can we second-guess the contestant?" The answer is "no, given all of these other assumptions about behaviour that are as unfounded as the one about the choice of doors." That answer is useless to the contestant herself unless the problem statement says she has knowledge of p₁₃. And nobody - not Morgan, not Gilman, not G&S, not K&W - says she does, or contradicts what I am saying here. JeffJor (talk) 19:35, 30 November 2009 (UTC)

What the Morgan paper does do.

I have added this section to try to find find some common ground regarding the Morgan paper. Basically I can agree that the Morgan paper was a response to vos Savant and her solution. As I understand it, vos Savant, proposed the set of rules that were not stated in the Whitaker question, namely that the host always offers the choice and always opens a door to reveal a goat. These rules have become the standard rules for the problem and were not criticized by Morgan.

Vos Savant also states that she takes the initial car placement to be random, this assumption is tacitly accepted by Morgan. She does not state that she takes the players initial choice to be random, and from the player's perspective I guess it is not. As it happens, this turns out not to be important and Morgan do not criticize her for not stating her assumptions in this respect.

The only mistake that vos Savant makes is that she fails to state that she takes the host's legal door choice to be random, even though it is clear that she does in fact do this and later stated that this was what she assumed. What the Morgan paper does do is to point out this omission by vos Savant.

Thus we can agree that the Morgan paper is a valid criticism of the vos Savant solution.

From my personal point of view it is a grossly excessive response to a simple failure to point out one assumption and a response that attempts to make this simple omission into a major problem requiring an 'elegant solution'.

The Morgan paper is not, in my opinion, a well-written, detailed, and comprehensive analysis of the Monty Hall problem which makes clear the exact question that it is answering or all the assumptions that it is making in its calculations. Martin Hogbin (talk) 17:40, 25 November 2009 (UTC)

The entire issue, as you propose it, boils down to whether the "Whitiker question" is about one specific instance of the game show, in which case every fact in it is to be taken literally and apply to that day only; or if it was supposed to be an illustrative example of the game in general, in which case such facts are circumstantial. They illustrtate a point of variability between different days if the game allows that variability, otherwise they describe the game itself. The problem is, that the question (as asked, not as misquoted by Morgan or Gillman) is unanswerable in the former case. We can't know if the host always offers a switch, among other things. We would only know that he did on that day.

The only way to justify the assumptions needed to make it solvable, is to treat it as an illustrative example that defines the game itself. And you can't pick-and-choose which interpretation you want for different facets of the problem. There are always three doors, because nobody in the illustrative example choose a number of doors. A door is always opened, because nobody choose whether to do it in the example. A goat is alwasy revealed, because it didn't say Monty choose between a car and a goat. However, any specific circumstances about those choices vary with each show, and can only do so randomly. That includes car placement, contestant choices (it says "you choose a door," not "you always hoose Door #1"), and host choices. Most people who read the problem do not even consider that it is a specific instance. This is effectively admitted by Morgan, who say few even considered the "conditional" problem of Choose 1 Open 3. Then they fail to recognize that those whom they claim did, addressed the solution based on the host's choice being between indistinguishable doors. So theirs was not the same "conditional" problem Morgan addressed, where bias is important.

Rick, I see little evidence that what you attribute to Marilyn vos Savant is correct. She explained no assumption in the original column, she just implied her assumptions by using them in her solution. Specifically, she never said the placement is random. In followups, she only explained one of the assumptions from that interpretation. She said "So let's look at it again, remembering that the original answer defines certain conditions, the most significant of which is that the host always opens a losing door on purpose. (There's no way he can always open a losing door by chance!) Anything else is a different question." She never even said "always opens a door." She never identified what the other conditions were, she just kept using "1/3" as the probability that the car was behind any door, and "1/2" for the probability when the host chooses between two goat doors. There was no discussion of why, contrary to what you think. But it is a clear implication of what she thought the other "certain conditions" were. All of the conditions were based on the illustrative example I described above.

Thus, what we can see is that the Morgan critisism is about a question MVS never thought she asked. It is not valid. She did not "omit that the host's choice was random," she never considered it to be possible for the user to use such information even if it was true, which makes it random in the contestant's eyes. All the paper was, in my opinion, was an attempt to apply what tools could be applied to the problem in a real-worl setting. By scholars who were too familiar with applying their science to real-world cases, to see that they were requiring a specific game, and a specific host, in what can only be a hypothetical problem. The very suggestion that Bayesian Inference could be used to refine the host's strategy demonstrates how that could not separate the real-world from the hypothetical.

But it does make its conclusions clearer than you allow for, once you read it with an educated mind. The primary conclusion is "The fact that P(W_s|D3)≥1/2, regardless of the host's strategy, is key to the solution." There is nothing unclear about that. They aren't saying what the probabiliy is, they are saying how its range affects the choice to switch. The fact that they never discuss the contestant using information about any of their parameters shows that they are keeping the two SoK's separate, even if they don't explain that to the reader. JeffJor (talk) 20:58, 25 November 2009 (UTC)

Jeff - You do realize this section was started by Martin, not me (Rick) - right? I'll respond to both of you.

Martin: the Morgan et al. criticism of MvS is not that she didn't specify that the host's choice between two goats is random, but that her solution is not the solution to the problem that is asked. We've been over this so many times I'm surprised you apparently continue to fail to understand this.

I fail to understand this because there is no 'problem that is asked', but do not take my word for it, read the comment on the Morgan paper by Seymann. In order to get a clearly defined problem certain details must be added. Whitaker actually says, 'opens another door, say #3', Morgan take this to mean, 'opens door #3', which interpretation they support by misquoting the original question. Vos Savant takes the question to mean, 'opens another door (to reveal a goat)'. This is an equally valid interpretation of Whitaker's question, in fact it is much more likely to be what Whitaker actually wanted to know. Do you really believe that he was only interested in the case where the host opens door 3?

Look at her enumeration of the cases that "exhaust all the possibilities" from her second column (reproduced here). In what she labels as games 1-3, the car location is placed behind door 1, door 2, and door 3 respectively. The listed result in these cases is the result of switching, assuming the player has initially picked door 1. Games 4-6 are the same but show the result of staying with the initial choice (of door 1). From the structure of this table we can clearly see she is assuming the doors are distinguishable (as implied by "Suppose you're on a game show"), she's assuming the car is uniformly placed initially, she's assuming the host must show a goat and make the offer to switch, and that she's following part, but not all, of the suggested case from the problem statement where the player picks door 1 and the host opens door 3. What this table literally addresses is:

P(win by switching|player picked door 1)

The probability is 2/3. She doesn't do it, but the same table can also be used to show the results given any other initial door choice, which is obviously also 2/3. No argument. But, this is not the probability that is asked about in the problem statement. What is asked about is

P(win by switching|player picked door 1 and host opened door 3)

This probability is not obvious from her table. In fact, if you casually try to infer it from her table you end up with 1/2 (the only excluded case is the one where the car is behind door 3). This is the error most people make when they initially think about this problem, i.e. they see three equally likely cases and one of them is made impossible by the problem statement leaving only two equally likely cases. And, to be clear, no one is claiming MvS is making this error, only that her approach is not addressing the problem that is asked. I suspect you've been following this page long enough to realize that there are people who "understand" the overall probability of winning by switching is 2/3 but who still think the conditional probability given the player has picked door 1 and the host has opened door 3 is 1/2.

No not at all. I have never seen that particular misconception expressed here. Most people think the probability in both cases is 1/2, see K & W, for example. Martin Hogbin (talk) 18:39, 27 November 2009 (UTC)

But you have missed my main point which is that the Morgan paper is criticism of vos Savant's explanation, it is not a well-written, detailed, and comprehensive analysis of the Monty Hall problem which makes clear the exact question that it is answering or all the assumptions that it is making in its calculations. It is thus not appropriate to base this article on the Morgan paper or its conclusions, despite its being published in a peer reviewed journal. Martin Hogbin (talk) 18:43, 27 November 2009 (UTC)

Jeff: I agree with nearly everything you say here. I think the critical difference is whether you treat the doors as distinguishable or not. You treat them as indistinguishable. Morgan et al. don't. Assume for the moment the doors are distinguishable. Given this assumption, you must agree that

P(win by switch|player picked door 1 and host opened door 2)

P(win by switch|player picked door 1 and host opened door 3)

P(win by switch|player picked door 2 and host opened door 1)

P(win by switch|player picked door 2 and host opened door 3)

P(win by switch|player picked door 3 and host opened door 1)

P(win by switch|player picked door 3 and host opened door 2)

are not all identically 2/3 unless q is 1/2 in each case, since if we're free to assign whatever q values we'd like in these cases we can make them each individually anything we want from 1/2 to 1. Getting back to where this thread started (two or three headings ago) I think what you're saying is that in a simulation of, say, 3000 iterations of the "problem" we can count ALL of them, if necessary renaming the door the player picks as door 1 and the door the host opens as door 3, and with this simulation the number of wins by switching will approach 2/3. I agree with this. However, if the doors ARE distinguishable, we can also keep track of the number of wins by switching separately, in each of the 6 combinations of initial player pick and host door, meaning we can ask about any of the following probabilities (which might be different):

P(win by switching)

P(win by switching|player picked door 1)

P(win by switching|player picked door 1 and host opened door 3)

Although it's a little harder to phrase unambiguously, I think the last of these probabilities is what most people understand as the question raised by the MHP, and is the one that most conflicts with people's intuition. -- Rick Block (talk) 18:05, 27 November 2009 (UTC)

Rick - yes, I mistook who started it. I don't see that it matters much. Your statement, "The Morgan et al. criticism of MvS is not that she didn't specify that the host's choice between two goats is random" (emphasis added) is correct. Their criticism is that she didn't answer "the conditional problem," and they point out that it matters if the host's choice is not random. G&S are clearer when they say it only matters if the host's choice is not random. Morgan, et al, never say the contestant needs to know the host's bias to answer the question, they only say it isn't necessary to assume a uniform bias to answer the question. But no source suggests the contestant should consider that possibility, but the article suggests the contestant does.

Morgan, et al, also says that MvS didn't consider a non-uniform placement of the cars. But they can't make the same conclusion here - that it isn't necessary to assume it is random - so they didn't placement bias at all. That is as much of an admission they give that it is a different problem. But it is such an admission, because they can't ignore it as they did if it is necessary. Had they provided their implied argument to justify ignoring car-placement bias, it would apply to host bias as well. I can't help it thay they omitted these arguments, but they did because even they consider host bias to be a different problem. And none of the sources you have quoted say it is necessary to consider in order to answer the MHP. They just say you don't need to assume it is uniform, either. But it is necessary to assume placement is.

And in fact, I gave you the example (OR, I know, so I am not suggesting it go in the article - only that it be used to understand what Morgan is trying to say) where the car placement is non-random. If there is a 50% chance of being behind Door #1, and 25% for #2 and #3, then Morgan's approach does not work. The probability can be less than 1/2 that switching will increase the chances for a specific host stragegy favoring a specific door. But from the SoK of a contestant who does not know which door is favored and so must treat a bias either way with equal probility, it does always go up. Just not as much as with a random car placement. The point is that Morgan did not need to address why a contestant should assume a specific host bias, because it was not necessary. In their problem, it always went up.

Yes, I will agree that those six probabilities you listed can be different, under one condition. That you agree that the contestant has no way to know which is different, and so must average the six probabilities before she can decide if switching is beneficial. That she cannot use any one of the those answers, which is EXCATLY' what Morgan means when they say the problem they are addessing can't be answered in general without knowing the host's strategy, but that she can use the non-informed combination of the three. And that if she does, the answer is 2/3m regardless of host strategy.

The numbers MvS uses in her second column are only to distinguish the three doors "Chosen," "Opened," and "Unchosen, Unopened." She does it using the numbers in the example, rather than the more complicated names, because it is far more transparent to the reader. She assigns equal probability to them not because it is impossible for a bias to exist, but because the doors are truly interchangeable so they must appear that way to the contestant. If you read the text that goes with that table you refer to, she refers to the host openeinmg "a losing door," not "Door #3," and that she compares it to a shell game with indistinguishable shells. Then she uses face-down cards, which are also indistinguishable. She did not intend to ask, and in fact did not ask, "the conditional problem." Again, the numbers used in the original problem are examples only.

And I disagree emphatically that "[P(win by switching|player picked door 1 and host opened door 3)] is what most people understand as the question raised by the MHP." Almost none do. Read the pages and pages of responses to that forum post you referenced. None see that problem, and I have only seen mathematicians who cannot see the forest for the trees do. When they do so, they never claim any typical readers to. So you need to stop inserting your own opinion, and OR, into the problem. JeffJor (talk) 18:55, 28 November 2009 (UTC)

It's nice to see the conversation directed at the emphasis on Morgan, and not some of the old red herrings. Credible people were publishing solutions along the lines of the 'Combining Doors' solution before, during and after Morgan and the others. The idea that this problem can only be solved with 'conditional' or 'unconditional' probability formulas is bogus. Symbolic logic works perfectly well, and eliminates the 'hosts behaviour' canard. That the original door choice never changes from 1/3 based on Monty's actions is all that's required. Glkanter (talk) 00:43, 29 November 2009 (UTC)

Jeff - again, I think we agree on many things. Morgan et al.'s criticism of MvS is that she didn't answer the conditional problem, and they say in this problem the host's protocol for choosing between two goats matters (to the extent that if it is not specified the we cannot answer what the probability is of winning by switching, only that it certainly doesn't go down).

They don't exactly say why they assume uniform car placement, although they do say that non-uniform car placement could be considered. I'm not sure what your point is about this. The car placement can either be uniform or not. If you assume non-uniform the problem is really not very interesting (unless, perhaps, the player knows the distribution). Whether this distribution is uniform or not, it's still a conditional probability problem. If the distribution is as in your example Morgan's approach DOES work, they just didn't bother showing this generalization. Treating the doors as distinguishable, and the problem as a conditional probability problem, allows any variant to be easily analyzed.

Regarding the six probabilities - I'll agree the contestant has no way to know which is different, but only if the doors are treated as indistinguishable. If "Door 1" means the door with the number 1 written on it (which I think it does - I understand you think it doesn't), then the contestant could know which is different. I think this boils down to what we take the introductory "Suppose you're on a game show" to mean. I (some WP:OR here) take it to mean the setting is a game show meaning real doors, on a stage, with identifying markings (numbers 1,2, and 3) so the player and host (and audience) can know which door is which. I think you're taking it to mean far less than this, perhaps only to help explain what we mean by "host" and "player" - but not "door". The assumption that the doors are distinguishable seems to me to be just as much implied as the roles of the host and player. They don't exactly say this either, but Morgan et al. are clearly treating the doors as distinguishable.

The fact that MvS uses unconditional analogies, and addresses the MHP as an unconditional problem, is precisely the criticism Morgan et al. make. In their view, the problem is conditional (almost certainly because they view the doors to be distinguishable). It might have been nice if the problem used something that really is indistinguishable or explicitly said the doors are to be treated as indistinguishable, actually making it an urn problem, but that's not how the MHP is stated. On a game show, doors are distinguishable - the player knows which one she chose and can see which one the host opens. Modeling this as a math problem where the doors are indistinguishable addresses a different problem (which is what I understand you think IS the Monty Hall problem). This is exactly the point. Perhaps the MHP is BOTH of these problems. Certainly many solutions treat it as an urn problem, but this interpretation is what Morgan et al. criticize. I don't think I've seen anything published that explicitly makes the claim that in the MHP the doors should be treated as indistinguishable. From your POV they're simply solving the problem. What Morgan is saying is that they've oversimplified their model of the problem. -- Rick Block (talk) 16:19, 29 November 2009 (UTC)

Rick, once again, nobody except the people who "solve" the "conditional" problem think that the MHP is about the "conditional problem." MvS did not. None of the lay people who read it thought it was. Very few of the mathematicians who read it though it was. Gardner did not. The literal interpretation of the MvS problem is not. The EXPLICIT statement of the MHP that defines what this article is about does not. I don't know how to more clear on this point, yet you refuse to acknowledge that it means the MHP is not "the conditional problem." Regardless of whether Morgan thinks so.

Yes, the host's strategy in the "conditional problem" matters. But whether or not it matters is completely irrelevant, since the "conditional problem" is not the MHP. No matter how many times you ignore this, it still has to define how the article is written. The "conditional problem" is a different problem than the MHP, whether or not Morgan and Gillman think it is due to their misquoting it.

And I can also repeat why the "car placement" distribution is important, but you will call it OR. It isn't. It is a trivial argument used universally in probability that is being applied inconsistently by Morgan as you read it. Where no explicit reason is given in the problem statement for why one option is more probable than the others, all equivalent options have to be treated equally in the solution of that problem. It is so trivial, that few people reference the reason when they use it. It is the Principle of Indifferece, it is well known, it is not OR, and it applies to both the host's choice and the car placement. Even if they are named, and so lok different. It is a principle that is NECESSARY for the solution to apply to both the host's strategy and the car placement. And that does not mean that anybody thinks thoses chocies are, in fact, uniform, which is where you keep making your mistake. It means that even if there is a bias, the contestant does not know what it is. So from the contestant's point of view, it LOOKS uniform. This is part of the trivial point of probability. It is part of the Principle of Indifference. It is not OR.

When Morgan and Gillman take their TWO liberties, (1) addressing the conditional problem instead of the unconditional one that is intended and explict, and (2) letting the host have an undefined bias, they are addressing a different problem, and addressing it in a special way. They are saying that that one particular choice is unimportant EVEN IF IT IS CONSIDERED TO BE NONUNIFORM. The "interest" it has is not that it defines "the solution" as you seem to think, it is that it removes a question some might have about the correct solution. No refernce ever says it should be considered for the general MHP, they are just applying an interesting twist that says it is unimportant to that question. Morgan, et al, ignore the fact that there could be similar questions about car placement until the end, where they leave it as an exersize to the reader to apply it and find that there is no similar "interesting twist" there.

About the "six probabilities." How do you suggest the contestant in this problem treat them differently? If you can suggest a way the contestant can do it, that is supported by the problem statement, I will withdraw everything I have said. But if you can't, you have to admit that the contestant has no way to do it and therefore can't. That means the contestant can't apply "the conditional solution," becasue teh contestant must treat the chocie as unbiased. And I will (again) reiterate that Morgan and Gillman never suggest a way, either. They don't say that the "conditional solution" is a solution to the MHP in general. They never say the doors can be treated as distinguishable. They only say that IF THEY ARE so treaed, it doesn't affect the answer to "should you switch?" You are confusing their hypothetical "if they are distinguishable" from a statement that "they are distinguishable." JeffJor (talk) 18:08, 29 November 2009 (UTC)

Jeff - I'm sorry, but Morgan et al., Gillman, and Grinstead and Snell are very, very clear. All the following are quotes:

Morgan et al.: Ms. vos Savant went on to defend her original claim with a false proof and also suggested a false simulation ...

Morgan et al.: Solution F1: If, regardless of the host's action, the player's strategy is to never switch, she will obviously will the car 1/3 of the time. Hence, the probability that she wins if she does switch is 2/3. ... F1's beauty as a false solution is that it is a true statement! It just does not solve the problem at hand.

Morgan et al.: Solution F2: The sample space is {AGG, GAG, GGA}, each point having probability 1/3, where the triple AGG, for instance, means the auto behind door 1, goat behind door 2, and goat behind door 3. The player choosing door 1 will win in two of these cases if she switches, hence the probability that she wins by switching is 2/3. ... That it [F2] is not a solution to the stated conditional problem is apparent in that the outcome GGA is not in the conditional sample space, since door 3 has been revealed as hiding a goat.

Gillman: Marilyn's solution goes like this. The chance is 1/3 that the car is actually at #1, and in that case you lose when you switch. The chance is 2/3 that the car is either at #2 (in which case the host perforce opens #3) or at #3 (in which case he perforce opens #2)-and in these cases, the host's revelation of a goat shows you how to switch and win. This is an elegant proof, but it does not address the problem posed, in which the host has shown you a goat at #3.

Grinstead and Snell: This very simple analysis [as a preselected strategy, staying wins with probability 1/3 while switching wins with probability 2/3], though correct, does not quite solve the problem that Craig posed. Craig asked for the conditional probability that you win if you switch, given that you have chosen door 1 and that Monty has chosen door 3. To solve this problem, we set up the problem before getting this information and then compute the conditional probability given this information.

Your argument about what what people think the problem is would be much more convincing with references. I've been here before with Martin. The problem is, as far as I can tell, no one who treats the problem unconditionally bothers to make it clear exactly what they're talking about, but those who treat it conditionally do make it clear and go on to say that the unconditional interpretation is not what the problem says. So, who are we to believe - reliable sources that implicitly say X, or other reliable sources who explicitly say both not X and that the sources that say X are incorrect? As a matter of Wikipedia policy (WP:NPOV), the article needs to neutrally present both POVs here (which is what I think the current version of the article attempts to do). Whether you or I personally agree with any particular POV is irrelevant.

Rather than continue this seemingly unproductive argument, can I ask what specific changes you think should be made to the article? It might be helpful to start a new section for this. -- Rick Block (talk) 01:11, 30 November 2009 (UTC)

Rick, about your quotes: Morgan, et al, misquoted MvS. They thought she was trying to solve a problem where specific doors were included, and rewrote the problem so that it looked like she did. With her wording, to quote K&W, "semantically, Door 3 in the standard version is named merely as an example." That invalidates your quotes about it. Semantically, the actual MvS problem is not about Door #3. Semantically, the MHP is the so-called unconditional problem, not the conditional problem. Any reference that addresses the conditional problem has semantically changed the problem into a different one than was intended by MvS; whether they do it explicitly by changing the wording, like Morgan and Gillman did, or implicitly by allowing that the interpretation might be different, like G&S did.

K&W also say that the solution to the conditional problem "focuses on the behavior of the host rather than on that of the contestant. Consequently, the change from the contestant’s perspective to Monty Hall’s perspective corresponds to a change from non-Bayesian to Bayesian thinking." Not one reference you have quoted ever justifies how a specific host perspective can be used BY THE CONTESTANT to answer the question "should I switch?" Or how a specific Bayesian prior could be chosen. They only establish the set of results that could exist, never a specific result that the contestant can use. They only establish that it is unnecessary to assume a specific host strategy w.r.t. opening a door, when the car is behind the chosen door, in order to answer that question. They all continue to make the standard assumptions for every other aspect - car placement, always opening a door, always revealing a goat - because those assumptions are necessary to answer the question. K&W tried to write those into the problem, to avoid having to assume them.

The quote you gave from G&S is the one that is most (I would claim it is the only one that is) useful. You just skipped the parts that you don't want to include. "[The conditional problem] means only two paths [of the twelve in the unconditional problem) through the tree are possible (see Figure 4.4). For one of these paths, the car is behind door 1 and for the other it is behind door 2. The path with the car behind door 2 is twice as likely as the one with the car behind door 1. Thus the conditional probability is 2/3 that the car is behind door 2 and 1/3 that it is behind door 1, so if you switch you have a 2/3 chance of winning the car, as Marilyn claimed." In other words, under the standard assmptions, the two approaches to the solution give the same answer through different combinations of possibilities. G&S then go on to say "Now suppose instead that in the case that he has a choice, he chooses the door with the larger number with probability 3/4. In the 'switch' vs. 'stay' problem, the probability of winning with the 'switch' strategy is still 2/3. However, in the original problem, if the contestant switches, he wins with probability 4/7." All this does is demonstrate how the different solution methods give different answers under specific assumptions. It does not justify what specific assumptions should be made, and the difference in the answers is moot without such a justification.

And while incorrect in several ways, Morgan takes the correct approach for how the conditional solution applies to the contestant when they say "The unconditional problem is of interest too, for it evaluates the proportion of winners out of all games with the player following the switch strategy. It is instructive to express this as a mixture of the two conditional cases: [derivation omitted] Pr(W_S)=(p₂₃+p₃₂." This is incorrect because it is a still the conditional problem, it just uses four of G&S's twelve paths instead of two. Where they say "unconditional," they mean "not conditioned on knowing the host's strategy when Door 1 is chosen and has the car." They leave it in terms of two parameters that are a part of the host's strategy, but the usual assumption (explicit in the K&W statement) is that both are equal to one. In this conclusion, Morgan says the answer from the contestant's point of view is 2/3.

I've given you references for people who treat it unconditionally. MvS. Gardner. Add Tierney. Mlodinow. I haven't read the others to find more such references, but I'm sure they are there. But all you need for proof is Morgan themselves, in their list of so-called false solutions. They all are employing the unconditional problem in one way or another. Your problem with seeing that, is that you only recognize one of the two ways of employing it. You can do it as G&S do, by making twelve paths through the set of possibilities; or you can do it by employing either two or four paths, and recognizing that the numbered doors are interchangable. THAT IS A CONCLUSION THAT CAN BE DRAWN DIRECTLY FROM THE UNCONDITIONAL INTERPRETATION, BUT ALLOWS USING DOOR NUMBERS IN THE SOLUTION. That is, assume the doors are lettered A, B, and C, and so are distinguishable in your approach (as opposed to your assumption, and it is an assumption, that the doors are numbered). Where the problem says "Door #1," that represents a uniform distribution of Doors A, B, and C. The twelve paths represented by G&S using A, B, and C reduce to four using 1, 2, and 3. Where the problem says "Door #3," that represents a uniform distribution of the two possible remaining doors (a different set in each case fo Door #1). That reduces four paths to two. The two-path solution does indeed represent the case where Door #1 is chosen, and Door #3 is opened. It is just form the contestant's SoK, which is the only SoK that is useful in answering the question.

There is a good reason you have not seen more scholarly work on the unconditional problem. It is a simple problem that from a mathematical standpoint was completely understood and presented in citable works before MvS asked it. Nothing in any of the references you cite invalidates anything about it, and much supports it. The answer (for the probability) is 2/3, quite simply, with nothing to be learned about it.

To "neutrally present" the conditional problem is what Martin and I are trying to accomplish. That isn't what is currently done. The current article's approach to it does not address the question in the MHP, "Should the contestant switch?" It addresses "given additional information, what can be said by a third party about the choice?" If you disagree, please show me a reference that specifically mentions how the contestant can use the conditional solution to decide. Not what some mathematician thinks the answer would be if additional information were available. This is about the third time I've asked, and you haven't done it. The conditional solution isn't an alternate POV, which Wikipedia policy says must be neutrally presented, it is a different problem that assumes there is a way to apply it that is unstated. A variant, to be sure, but a different problem. And I don't just mean conditional vs. unconditional. To use it requires additional information that is not in the problem statement, and so is not assumable.

As we have said before, if that section is kept at all, it should be moved below the discussion of "Other Host behaviors" in the "Variants" section. It is not a "Probabilistic solution" to the general MHP, it is a solution to an alternate problem called "the conditional problem." It should say that some readers think specific doors are meant, but that the original fomulator did not (because she never said she did, and in every word she wrote she clearly thought the specific door numbers did not matter). As Seymann points out in the response to Morgan, she can't be held to rigorous standards in a sunday supplement. To prove that she emant the conditioanl problem, you need to provide positive proof that she intended those doors to be different. And there is none.

The article should make clear that the Moragn's conclusions require two assumptions: (1) the conditional problem is considered, AND (2) the game-show strategy includes certain biases (but not others). Only then you can still conclude that switching is better. But this applies only to those specific biases, not to the general case. You can even cite Morgan for that last statement. The "Bayseian analysis" section can't be included as is without a discussion of the two different meanings of "Baysian." People too often forget that "Bayesian" can mean "Using conditional probabilities," or "Starting with a specific prior, using conditional probabilities to get answsers that apply in just that specific case." The latter meaning cannot be used in a hypothetical problem, since there is no basis for assuming a informed prior (a noninformative prior is not really assuming a prior, it is assuming that nothing can be distinguished even if there is nonuniformity). As such, it is really just the solution to the conditional problem, not a separate analysis that repeats the same conclusions. Regardless, it needs to be more closely associated with the conditioanl problem, since all it is, is that variant's solution.

Finally, it is not a FAQ that people ask about the conditional problem. It is only mentioned by people who solve the conditional problem, or are reading an article about it so their quesiotn can be asnwered there. It needs to be de-emphasized anywhere it is mentioned. Oh, and most of this discussion needs to be archived. [Signature added later - I never get it right :)] JeffJor (talk) 19:11, 30 November 2009 (UTC)

I can get on board with the changes to the article described here. Glkanter (talk) 16:27, 30 November 2009 (UTC)

Claiming that Morgan is solving a "different" problem instead of the "real" one is at least as misguided as Morgan claiming, vos Savant was solving not the "real" problem. This is an ambiguous problem that can be modelled in several ways depending on which specifics you chose to replace the ambiguity and from which perspective you argue. Moreover it is not up to us to pick between vos Savant, Morgan or others, but we simply report their published approaches and criticism.--Kmhkmh (talk) 17:46, 5 December 2009 (UTC)

Observation and suggestion

I noticed that request for comments and recent posting related to it already follow the same pattern as the old discussion. The same old points get reiterated by the mostly by same people over and over (right now including me I'm afraid). Also there is tendency of explaining each's (personal) understanding of the problem rather than sticking to the sources and there is a rather endless arguing about relatively minor points ("which goes first"). This "game" is now played by the same actors (partially including me) for over a year without any productive result.
Due to the observation above I'd like to repeat my recommendation above. The article is best served, if it gets a review by a group of knowledgeable, competent and "neutral" editors. All former editors and constant participants of this endless discussion should state their point in a single comment and other than that (voluntarily) stay out of the way. They should only give further comments if they are explicitly requested by the reviewers. The former editors that should stay out of the way explicitly include Martin Hogbin, Glkanter, JeffJor, Rick Block, Nijdam and Kmhkmh.

--Kmhkmh (talk) 13:13, 5 December 2009 (UTC)

The MHP has been described as the world's most intractable brain teaser. It contains elements of both mathematics and philosophy. I think a suitable body of knowledgeable and neutral editors who will be able to understand to issues involved will be hard to find.

As an alternative I would suggest some form of mediation, possible by someone who makes no attempt to address the problem itself but who mediates on general content principles and policies might help. Martin Hogbin (talk) 14:15, 5 December 2009 (UTC)

Finding a group of people being better or at least as qualified as he current long members of the discussion is not that hard. The reviewers don't have to be the perfect mix but just good enough for the job. There real issue here is, if people are willing to step back and accept the sound judgement of others, even if it does not support their favoured solution. As far as mediation i have no objection if the other editors are fine with it, but i'm quite skeptical regarding the success.--Kmhkmh (talk) 15:06, 5 December 2009 (UTC)

Where are they now then? Are you thinking of a bunch of mathematicians? Most of them will probably not be that interested, the maths is pretty simple, all the problems lie elsewhere. Martin Hogbin (talk) 16:42, 5 December 2009 (UTC)

I think most third part expert (mathematicians or otherwise) are put off by the endless (and often pointless) discussion and they are definitely not interested in participating in such a thing over months or even years (neither am I). They might however be willing to help out with a review/3rd opinion provided this can be handled in timely fashion and that the results will be heeded. This means the first step would be that the "warring parties" explicitly agree beforehand to accept the result and refrain from further edits.--Kmhkmh (talk) 16:59, 5 December 2009 (UTC)

Whichever side lost in that case would feel that they had not had their case taken seriously. I think mediation, where the current proponents can present their cases to a mediator, is a better bet. Martin Hogbin (talk) 17:02, 5 December 2009 (UTC)

I have no objections against a mediator, but again the current proponents present their cases for years now.--Kmhkmh (talk) 17:31, 5 December 2009 (UTC)

One time, in English please...

Would one of the Morgan supporters, using whichever published source you prefer, please finish this sentence:

Solving the unconditional K & W Monty Hall problem statement is not sufficient, and a conditional problem approach must be utilized due to the instance where the host chooses between 2 goats. Because...[Please continue here].

This is essentially your argument against the three similar proposals, right? Thank you. Glkanter (talk) 19:05, 5 December 2009 (UTC)

Are you asking why this view should be in the article, or are you asking for an explanation of this view? If the former, see WP:NPOV. If the latter, see Talk:Monty Hall problem/FAQ or the Morgan et al., or Gillman, or Grinstead and Snell references. -- Rick Block (talk) 22:15, 5 December 2009 (UTC)

6 Years and 9 Archives Comes Down To The Definition of a Game Show.

"Suppose you are on a game show..."

That's how both Monty Hall problem statements begin in the article.

Call it whatever you want in Probability Speak, symmetry, indifference, random, unknown, or equal.

In game shows, it is understood that nothing will be communicated about this to the contestant. So he remains blissfully unaware of anything but symmetry, indifference, random, unknown, or equal.

This is not trivial, to be argued or interpreted away, it is the 5th and 6th words of the problem statement. GAME SHOW. Literate people the world over are expected to know and understand what they are watching on TV. It is not this article's responsibility to consider otherwise.

As part of the definition of Game Show, it was unnecessary to state outright as a premise: 'Host will not share location of car with contestant'. Because that's what you claim he 'could', bizarrely, illogically, somehow do.

So, go ahead and argue. Tell me, as always, that I am mis-interpreting, or simply wrong, or whatever. Do me one favor, though. Just tell the truth. Glkanter (talk) 22:27, 5 December 2009 (UTC)

The explanation is given further up and btw i explained you that already almost half a year ago for the first time.--Kmhkmh (talk) 22:50, 5 December 2009 (UTC)

Mediation?

Mathematics is the one area where there will be general agreement amongst those who understand the problem. The arguments are all about questions like: What is the MHP? What is the most notable aspect of the MHP? How should an encyclopedia article be presented? How should we use reliable sources? Which sources are the most reliable?

This list is not intended in any way to be exhaustive it just gives examples of the kind of thing that we are arguing over and both sides claim to have WP policies on their side. Does anyone think that we might benefit from mediation on the policies and principles involved here? Martin Hogbin (talk) 16:53, 5 December 2009 (UTC)

Yes, I think a good mediator could be helpful in getting this long-festering discussion to converge sensibly. Dicklyon (talk) 17:23, 5 December 2009 (UTC)

Sounds like a reasonable idea to me. -- Rick Block (talk) 18:23, 5 December 2009 (UTC)

As a professional mathematician (indeed, a probabilist and statistician) I am excited there is a new serious discussion about what the MHP is all about. The problem with problems which appear mathematical problems, posed by non-mathematicians, is that the mathematician may well discover there are several ways to interpret what the original problem-poser meant. Mathematicians often work backwards. Someone asks a problem. The problem is in fact ambiguous. One discovers a way to disambiguate, which allows a pretty solution. Often, the disambiguation involves making some of the "hints" of the original poser much harder, and ignoring others. I was delighted to discover their are two ways to disambiguate the problem: one is the conditional version, one is the unconditional version. I was also delighted to discover that in order to get "the good solution" to the conditional problem, one needs to demand different conditions, than to get "the good solution" to the unconditional problem. I also discovered a game-theoretic resolution of the problem, and I discovered some more interesting literature, I'll add citations later. Since my own work (on my web page) is not yet converted into a regular mathematical paper, and not yet submitted to, let alone passed by, peer review, I don't suppose wikipedia can or should use my insights, for the time being! But I continue to follow the discussion with great interest and I'll certainly let you all know my five cents worth if I think I have few cents to chuck in. Gill110951 (talk) 15:40, 6 December 2009 (UTC)

Just looked at the Morgan et al paper. It is published work in a well-known journal (but not a very high status one) and it is 20 years old, and there have been plenty more insights since then. Mathematics is never finished. Mathematicians will go on finding new variants of the problem and hopefully getting new insights about it. Also discussions between mathematicians and non-mathematicians about this particular problem will continue to expose new variants. Great!!! As a mathematician I am not particularly interested in what vos Savant really mean, what she thought was the solution, and whether or not it was correct. There are many nice problems here and many nice solutions. A mathematician's job is to simplify and unify and understand. Often, as I said, working backwards: to which questions is the answer "change doors" the (a) right answer? What I am saying is that the mathematician has somewhat different interests to the encyclopedia compiler. Also since mathematics is in principle more or less verifiable independent of knowing all kinds of facts about the real world, the idea that an encyclopeadia may not contain new results, is crazy. Finding new mathematics results is a creative process, but checking that they are correct is an exercise in logic, not a matter of citing authorities. Especially in this case, since understanding most versions of the MHP will presumably only involve elementary mathematics. I don't care too much about what "authorities" have written about this problem in the past. A mathematician is more convinced by a beautiful and transparent proof, with some new twists, of an a priori surprising result, than by a citation to some dusty printed matter. Gill110951 (talk) 16:27, 6 December 2009 (UTC)

Finally for today, I actually edited the MHP page including the insights from my own researches from last March. Of course a wikipedia editor is not supposed to write about his or her own research. So you can all edit it all out again, but please think about what I am trying to say first. Maybe you can be convinced by the logic. If so you can decide whether notes for his students and for the many journalists who keep asking again about this problem, on the home page of a well known authority on mathematical statistics, who has been telling his students about MHP for years, is an authoritative enough source for wikipedia. If you like I can post the notes on arXiv.org so they will still be there after I'm retired... And maybe Boris Tsirelson can say if he agrees with my maths. I'll send him an email. Gill110951 (talk) 17:21, 6 December 2009 (UTC)

How about this as a start

Those in favour of change start a separate development article, no doubt based on the current one, that concentrates on the 'simple' problem and convincing explanations. Much of this is already in the current article. This would make clear exactly what changes are being sought. If the pro-changers cannot agree on what changes are wanted then the article stays as it is by default. On the other hand once we have an agreed alternative version, we can seek mediation on how to proceed from there. This will involve questions like: should we have one or two articles? What should they be called? How could the two versions be combined? What order should the sections be in? etc. The advantage of this approach is that it lets the mediator understand what the two sides are after. Martin Hogbin (talk) 12:41, 6 December 2009 (UTC)

I'm sorry Martin, I think this suggestion just continues the horrid legacy of the 6 years and 9 archives (plus countless other argument and talk pages scattered throughout Wikipedia's servers, plus this HUGE talk page) filibuster engineered by Rick. You can spend all the time and effort you want, then in my opinion, Rick and Nijdam will ultimately exercise some sort of veto power, despite any majority opinion (including respected Mathematics professors) that would otherwise, by Wikipedia standards, be indicative of a clear consensus.

For example, here's Rick's comment, before he decided to "reword to avoid potential misinterpretation as ownership".

"(as if by convincing me that their POV is "correct" I would then agree to change the article as they wish)."

http://en.wikipedia.org/w/index.php?title=Talk:Monty_Hall_problem&diff=329396402&oldid=329394782

Nijdam seems to think it's his way or the highway. So far, anyways, he's been right.

No, you've got the attention of the whole Wikipedia Mathematics group. Follow through with gaining the consensus before wasting any more time on sandbox-style editing. Heck, if I recall correctly, there's already a 'new' page prototype somewhere under your own user name. I've contributed to it. And the only issue of contention is 'Is the contestant aware?'. Most everything else looks resolved, to me. Glkanter (talk) 13:10, 6 December 2009 (UTC)

If not a development version the how about some kind of outline for the proposed article? I am trying to check that we are all after the same thing. Martin Hogbin (talk) 13:36, 6 December 2009 (UTC)

Martin, there's nothing ambiguous here:

Changes suggested by JeffJor, Martin Hogbin, and Glkanter

If you're here because you've been invited to comment, there are ,~~two~~,. three (related) suggestions.

Glkanter's suggestion: Eliminate all 'host behaviour, etc' influenced discussion, save for the Wikipedia minimum necessary references to Morgan and his ilk, as the 'conditional' problem is the converse of "Suppose you are on a game show."

JeffJor's suggestion: The so-called conditional problem needs to be a separate article, with "conditional" in its title.

Martin Hogbin's suggestion: This article should concentrate on the unconditional solution with the Morgan's conditional solution in a variations section.

http://en.wikipedia.org/wiki/Talk:Monty_Hall_problem#Changes_suggested_by_JeffJor.2C_Martin_Hogbin.2C_and_Glkanter

We've been clear and consistent throughout. And the consensus is really already here, save for whatever veto powers exist within Rick and Nijdam. Glkanter (talk) 13:46, 6 December 2009 (UTC)

8 to 4 (Henning Makholm's comments clearly indicate he's opposed to this change) with one side arguing a fundamental Wikipedia policy and the other arguing mostly WP:OR is not a consensus, and even if it were it cannot be used as a blank check to violate NPOV. -- Rick Block (talk) 15:30, 6 December 2009 (UTC)

I wholeheartedly disagree with your characterization of this good faith discussion by many editors. Essentially, you're describing it as 'Good vs Evil', and 8 of us, so far are on the side of Evil. I'm not buying it. I reject your argument prima facie. Just more filibustering on your part. Glkanter (talk) 15:36, 6 December 2009 (UTC)

Rick, perhaps you could ask Henning to put his name in the appropriate section and sign it.

Outline for an alternative article

This is where I would start, the current contents are given for reference. Martin Hogbin (talk) 16:12, 6 December 2009 (UTC)

Current

Problem
Popular solution
Probabilistic solution
Sources of confusion
Aids to understanding
1. Why the probability is not 1/2
2. Increasing the number of doors
3. Simulation
Variants
Other host behaviors
N doors
Quantum version
History of the problem
Bayesian analysis
See also
1. Similar problems
References
External links

Proposed

Problem - Whitaker statement - Sourced unconditional restatement
Solution - The unconditional solution, from one of many sources - clear diagrams.
Sources of confusion - Much as it is now
Aids to understanding
1. Why the probability is not 1/2
2. Increasing the number of doors
3. Simulation
Conditional Problem - Reasons the problem is strictly conditional.
Variants
1. Morgan Scenario
2. N doors
3. Quantum version
History of the problem
Bayesian analysis
See also
1. Similar problems
References
External links

I'd disagree with that. Morgan is not just variant but he explicitly deals with the original problem as well, so do other sources with a conditional problem. The conditional approach is not a different problem as the TOC seems to suggest now(?) but a different perspective or approach to the same (ambiguous) problem. The way you attempt to reframe, that in your table of content is clearly in contradiction to the treatment in various reputable sources and hence in this form nonnegationable. Not to mention it also contracts your own comments on Nijdam's recent posting ("No, your calculation is fine and as I have made clear before I accept that the MHP problem (as defined by the K & W statement in which specific doors are mentioned) is a problem of conditional probability."). If it this is just a case of ambiguous wording (i.e. I'm simply misreading you ) then please correct that before further discussion.--Kmhkmh (talk) 16:28, 6 December 2009 (UTC)

I did not really expect you to agree as you are one of those who are against change here. This section was for those who want to change the article to check that they are all on the same wavelength. I will respond to your point about Morgan in a new section. Martin Hogbin (talk) 17:03, 6 December 2009 (UTC)

Why Is Mediation Even Being Considered?

I think a consensus already exists for the proposed changes. Or, it will exist soon enough, when everyone has had the opportunity to comment. Using the 80/20 rule, I figure we've already heard from most everybody that's going to weigh in.

I count:

3 editors for maintaining the status quo

8 editors who are in agreement with one of the 3 similar suggestions

1 editor who's preference is undetermined

10 'comments requested' editors who have not commented to date. I understand from my Wikipedia readings that ultimately 'silence implies consent'.

We know unanimity isn't required, or Wikipedia would call it 'Gaining Unanimity' instead of 'Gaining Consensus'. So unanimity isn't our goal. And certainly not going to happen.

And, while voting is improper, *counting* is necessary.

So, which is it:

'All editors are equal'

or

'All editors are equal, but some editors are more equal than others'?

Glkanter (talk) 14:34, 6 December 2009 (UTC)

From WP:NPOV: "Neutral point of view" is one of Wikipedia's three core content policies, along with "Verifiability" and "No original research." Jointly, these policies determine the type and quality of material that is acceptable in Wikipedia articles. They should not be interpreted in isolation from one another, and editors should therefore familiarize themselves with all three. The principles upon which these policies are based cannot be superseded by other policies or guidelines, or by editors' consensus. [emphasis added]

All 4 editors in the minority (Kmhkmh, Nijdam, Henning Makholm, and myself) have directly or indirectly cited this policy as the reason against the changes that have been suggested. I have made this point clear in my comments above and over and over and over and over again in response to your tendentious attempts to introduce this change. Although you know I am an administrator on this site (which means not that my opinion on content issues matters more than any other editor's, but that I am VERY familiar with Wikipedia policies), you apparently do not believe me. The point of mediation would be to bring in an outsider, also familiar with Wikipedia policy, who might help see some middle ground between what we 4 see as a direct violation of Wikipedia policy and the change that you're seeking to make.

Is that clear? -- Rick Block (talk) 15:19, 6 December 2009 (UTC)

+1--Kmhkmh (talk) 15:43, 6 December 2009 (UTC)

I would point out first, that you, of all people, invoking Wikipedia's NPOV as a reason to maintain the MHP article is ironic and comedic. There's a consensus that doesn't agree with your interpretation of how to apply WP:NPOV. I see it as just more filibustering by you.

Kapish? Glkanter (talk) 15:27, 6 December 2009 (UTC

Are you agreeing the article must be NPOV? If so, please explain how the changes you're seeking comply with this policy. I'm hearing what you're suggesting as "make the article take the POV that Morgan is wrong, or is addressing a different problem". If this is NOT what you mean, please explain what you DO mean. -- Rick Block (talk) 17:52, 6 December 2009 (UTC)

Of course, NPOV is a requirement. I strongly and vocally for 14 months now have disagree with your assessment that the article currently satisfies WP:NPOV, especially for a Featured Article. I think it's drastically skewed to the conditional solutions, in emphasis, textual amounts, FAQs, additional optional narratives, etc. Doing the reader no good whatsoever. All I've said is give Morgan and his ilk the Wikipedia required reference for having been published. Not elimination, as you've mis-quoted me more than once. How much emphasis they receive remains to be determined by consensus. Glkanter (talk) 20:56, 6 December 2009 (UTC)

The Uber Reliably Published Source

Rick, you've used this argument before, and are likely to use it, again, soon.

So, to save us all a lot of time and typing, would you mind explaining why it's Morgan? Glkanter (talk) 00:39, 7 December 2009 (UTC)

Deal or No Deal

Asked whether the contestant should switch, vos Savant correctly replied, "If the host is clueless, it makes no difference whether you stay or switch. If he knows, switch" (vos Savant, 2006).

I read this, and then the following section on "increasing the number of doors" and had the instant comparisson in my mind to "Deal or No Deal". I came back here to see the debate over that game show, but to me it's quite obvious. If there are 22 cases, and you select one, your initial probability is 1/22 of having the mil.

The two situations are clearly fundamentally different as demonstrable:

We've all see contestants painfully open a case that has the Mil$. This is because the contestant does NOT know the contents and is randomly selecting. The probabilities say that if there are 22 cases, this should happen 20/22 times with only 2/22 probability of getting down to your case and one other (the "switch" situation). This is why it is so uncommon for the Mil to still be in play at the switch point of that show [In reality it's even less common than two in every 22 contestants getting to the switch with a Mil$ due to the irrelevant fact that those who still have a Mil$ in play often take a deal rather than risk eliminating the Mil$ with three or four cases left]. But assuming the true math problem, If you happen to get down to the switch for a Mil$, your odds are equal. God knows which case you picked. You got lucky enough to leave the Mil in play. Now you're down to 50/50 odds.
The Monty Hall situation would be if you select your case (1/22 odds) and then the HOST, KNOWING which case has the mil randomly selects 20 other cases to eliminate that DO NOT have the mil. It doesn't matter if he eliminates them one at a time or all at once because it's random and he can't eliminate the Mil$. He ends up with case 13 up there, and you have your case. What do you think the odds are that you picked right in the first place and he HAD to leave 13 because it has the Mil$, vs the odds that you picked right and he randomly left 13 up there? Well it's clearly 1/22 odds that you picked right. If you picked ANY of the 21 of the 22 cases that didn't have the Mil, he would have left the same case up there: The one with the mil. Only in that 1/22 case where you have the Mil does he need to be random. So the odds when the revealer KNOWS and AVOIDS revealing the million are the odds of your original choice being wrong.

The point I was going for here in this lengthly analysis is, perhaps the now-common format of Deal or No Deal would make for a good example of the contrast between Monty Hall Problem (revealer knows and avoids opening the prize door) and Deal or No Deal (revealer doesn't know and got lucky to leave two random cases). It's a very good everyday life scenario that many readers will understand and relate to in terms of getting the idea across. I guess the problem is it needs citation, but there must certainly be published works out there analysing the math of Deal or No Deal that can be cited... TheHYPO (talk) 06:29, 4 December 2009 (UTC)

Yes, I also consider this aspect as the most instructive part of MHP. See above, in my comments: "I compare two cases: (a) the given case: the host knows what's behind the doors, and (b) the alternative case: he does not know, and it is his good luck that he opens a door which has a goat." Boris Tsirelson (talk) 06:41, 4 December 2009 (UTC)

I agree, that is an interesting, and indeed surprising for most people, case to consider. Most people are amazed to discover that whether the host chooses a goat door by chance or because they know which door is which makes a difference. Of course, you need to decide what happens if the host chooses randomly and happens to choose the car.

My explanation would be along the lines that when the host knows what is behind the doors you consider all games but when the host chooses randomly you consider a non-random sample of games.

This is the kind of thing that this article should be concentrating on rather than the rather bizarre scenario where the player knows the hosts door opening strategy. Martin Hogbin (talk) 09:57, 4 December 2009 (UTC)

Yes. And here is an analogy. The (elementary) geometry of a circle is a special case of the (differential) geometry of an arbitrary curve. However it would be terrible to treat the circle (in an encyclopedia) only inside an article on curves in general. The symmetric case is more notable, more important, this is the point. And let me repeat: we surely have our POV about importance (rather than content). Boris Tsirelson (talk) 12:19, 4 December 2009 (UTC)

Exactly. The whole point about the symmetric MHP is that is is a very simple problem that nearly everybody gets wrong. The Morgan scenario (player knows the host door opening policy) obfuscates the central problem and leaves us with a more complicated and rather boring problem that most people are not interested in. Martin Hogbin (talk) 12:51, 4 December 2009 (UTC)

Yes. And let me add: I have nothing against conditional probabilities. I only note that in the symmetric case there is no difference between conditional and unconditional probabilities. And it is not a coincidence. Not at all. An unconditional probability is an average of conditional probabilities (the total probability formula); these conditional probabilities are mutually equal (by symmetry); the equality follows. Boris Tsirelson (talk) 14:16, 4 December 2009 (UTC)

No difference?? Even in the symmetric case there is a difference between the unconditional and the conditional probability. They are different probability measures. The only point is that for certain events they share the same value. Nijdam (talk) 15:10, 4 December 2009 (UTC)

(back from the weekend) Of course, different measures. Of course, for certain event. But Martin Hogbin understands what I mean, see below "Words, words, words". And again, I do not object to conditioning. I only say that the symmetric case is more important for an encyclopedia.

Boris, when you teach this problem, do you or do you not introduce conditional probabilities? If not, how to you distinguish the "host forgets" case from the "host knows" case? Martin is still misinterpreting what Morgan et al. says, which is not that the player necessarily knows the host's door opening policy (in the "host knows" case), but that to solve for a specific conditional probability, e.g. P(car is behind Door 2|player picks Door 1 and host opens Door 3), you have to be given or make an assumption about how often the host opens Door 3 if the car is behind Door 1. Symmetry says this is 1/2 is a perfectly valid way to make this assumption. The player has no knowledge of a potential host preference between Door 2 or Door 3 so must treat these doors as indistinguishable is another perfectly valid way to make this assumption. If your solution says nothing about this, your 2/3 answer is not the answer to P(car is behind Door 2|player picks Door 1 and host opens Door 3) but something different, like perhaps P(win|player switches). Which of these conditional probabilities more accurately represents the conditions described by the MHP which specifically allows the player to choose whether to switch after seeing the host open a door?

(back from the weekend) It seems, everything is already answered in the "Words, words, words" section below. Except for the "personal" question... Yes, I do introduce conditional probabilities. Indeed, I do not teach MHP; I just illustrate conditional probabilities by MHP. So what? First, Wikipedia is not a textbook. Second, I do not object to conditioning. I only say that the symmetric case is more important for an encyclopedia.Boris Tsirelson (talk) 15:18, 5 December 2009 (UTC)

BTW - the paragraph just added to the article has no references. Citation standards for feature articles are quite demanding. Essentially every independent thought should be referenced (this ensures that what is said is verifiable). Unless a reference can be provided that uses the 100,000 door example to contrast MHP with Deal or No Deal, this paragraph should be deleted. -- Rick Block (talk) 15:20, 4 December 2009 (UTC)

On that note: If the standards as a result of this endless discussion and edit conflict cannot be maintained, then the article should lose its featured status, which by the way is exactly what happened to the German version.--Kmhkmh (talk) 10:12, 5 December 2009 (UTC)

Rick, you said "Morgan et al. says ... that to solve for a specific conditional probability ... you have to be given or make an assumption about how often the host opens Door 3 if the car is behind Door 1." That isn't really what they say. They say that it is an interesting generalization of the MHP to consider one specific host strategy that avoids such an assumption. They also say that the fully gerneralized host strategy is not interesting in this way. They never say that the assumption q=1/2 is an invalid assumption, that it is not implied by the problem statement, or offer any way for a contestant to obtain information about q. They only say it isn't necessary to make it the assumption in order to answer the question. They do say that for other generalizations, such as placement bias, you are required to assume something to answer the question (because to not do so is "unlikely to correspond to a real playing" of the game). Their only point is to treat that one option as a vehicle for teaching conditional probability, not to say that any formula gives "the probability" for the MHP. So you are injecting your own POV into the article. The POV that a contestant can obtain this information and use it, even if how is unstated. That is an uncited, and uncitable, original thought that should not be included in the article by the very standards you insist others must adhere to. Why do you think you can avoid it?

So it isn't Martin that is misinterpreting Morgan, it is you. Martin understands that not only are "symmetric assumptions" valid, but they are required for the very reason that no bias values are given. Martin undersatnd that mot published treatmens of the problem make that assumption, and yes I have told you some of them. He understands that what you call the "symmertic assumptions" can be implemented in two ways: either by assuming q=1/2, or by assuming the doors are symmetric and therefore not distinguished. And that the two are isomorphic problems that have different solution methods, but the same answer (of course). That the former symmetric assumption requires Morgan's analysis, but the latter can be solved with most of the methods Moragn calls "incomplete." And that the point of the MHP is to show that that simpler soluiton produces an unintuitive result, not to get an actual probability.

How can you argue, that the doors cannot be distinguished? When obviously anybody can distinguish them (comletely independent on which solution approach you pursue)? Even more the original problem formulation itsself distinguishes them explicitly? You may argue a "symmetry assumption" for solving (or simplifying) the problem, but you cannot argue indistinguishable doors, unless you blindfold the candidate and rob him of any orientation.--Kmhkmh (talk) 12:48, 5 December 2009 (UTC)

I don't argue that the doors cannot be distinguished somehow, each one from the others. (But please note, since you are trying to argue for a very strict reading of the Whitiker statement: it does not say the doors have numbers, or are distinguishable. That is an assumption you make. It does mention numbers, but it doesn't say whether the numbers are a manifestation you created in your mind, or are seen by everybody. And don't, as others have, tell me there were numbers on the Let's Make a Deal show - they never played a game remotely like this and it was also not mentioned in the Whitiker statement.) I argue that no factors relating to the uncertainty in the outcome can be applied to the doors, as they are distinguished that way, when solving the puzzle. Suppose I were to roll a pair of dice, one red and one green. If I ask you for the probability one die will roll higher than the other, would you give different answers based on color? How? JeffJor (talk) 13:55, 7 December 2009 (UTC)

The doors are there to keep the contestant from seeing behind them. Which they accomplish 100% effectively. What if the contestant was sequestered away, and couldn't see the front of the doors or the relative locations of the doors? He relies on Monty's description, as verified by the studio audience. The contestant still has no more knowledge than that switching 2/3 of the time will win the car, before or after Monty has opened a goat door. How has this extra level of 'identicalness of doors', that you claim is required for the unconditional solution, helped or hurt the contestant in any way? How does it apply to the Combined Doors solution, in which Monty reveals 'which' door has a goat, but that has no effect on his offer to switch, or the response by the contestant?Glkanter (talk) 13:28, 5 December 2009 (UTC)

All of what you've stated here does not affect the distinguishability of the doors, I was talking above. I was making no statement regading the solution and I'm definitely not going to rehash the seemingly endless (and imho rather boring) discussion on that. I just pointed out, that claiming the doors he doors are indistuingishable, is factually false as far as reality or the perception of the contestant concerned. The contestant always see a left, middle and right door, i.e. he can distinguish them. Whether it might make sense to declare the doors indistinguishable in a model to solve the problem is another matter I did not address.--Kmhkmh (talk) 13:50, 5 December 2009 (UTC)

The point of separating the articles is not to eliminate any POVs. It is to emphasize them. To not let one facet of the MHP (simple solution w nonintuitive result) become overpowered by the other (good teaching tool for conditional probabilites). If we don't physically separate them, we need to more clearluy divide the article. The first half should be about the classic (unconditional) MHP, as stated by MvS (not K&W), and listing the set of assumptions she has said (and 99.9% of readers agree) are implied: interchangable doors, and any kind bias becoems irrelevant because of interchangeable doors. Then a section about game protocals (part of what you call host stratgies) such as always opening a door or revealing a goat, WITHOUT mention of bias or conditional problems. Finally, you can cite Gillman (not Morgan) as a reference that introduces the possibility that the conditional problem is intended, but matters only if there is a bias. Use the K&W sattemetn here, not Gillman's misquote. Gillman is better than Morgan because it is clearer, includes placement bias, and does not launch into possibilities that we are never told how to use. I think this is pretty consistent with Martin's suggestion. JeffJor (talk) 17:41, 4 December 2009 (UTC)

Rick you keep saying that I am misinterpreting what Morgan says without proof or justification. Every mathematician here agrees that, if the player does not know the host door opening policy and we are to address the problem from the player's state of knowledge, we can only take it that the host is equally likely to open either goat door when he has a choice. No parameter q is required, it is fixed in value at 1/2.

Whether the problem has to be considered from the players ("contestant") perspective or from an informed 3rd person's perspective ("problem solver") is entirely unclear and depends on the perspective you take. Meaning the original question could stand "How would you decide as a player?" or "what would you recomend the player?" Various publications on the topic have examined those different perspectives. Also if the player doesn't know the host's policy, he still has 2 options for his decision making. Simply pick the most reasonable assumption (fix q at 1/2) and base his conclusion on that. Or instead analyze a large variety of possible host policies to see whether all (or at least most) yield to the same conclusion anyway (q varies).--Kmhkmh (talk) 13:31, 5 December 2009 (UTC)

Both problem statements begin: "Suppose you are on a game show..." I see no ambiguity in that we are taking only the contestant's perspective (State of Knowledge) into our analysis. Glkanter (talk) 13:37, 5 December 2009 (UTC)

That is correct if you just look at those statements in a literal sense. However it is not so clear if look at some "interpretations" or variation of the problem in literature (see Mark Steinbach in German article as a source). But no matter how you regard that, it does not change the second part, where even from the player's perspective the bayesian analysis (variation of q) is still an option for his reasoning.--Kmhkmh (talk) 13:59, 5 December 2009 (UTC)

Regarding Morgan's attempt to answer a more general problem, they make a very poor job of this. For the most part is is hard to determine exactly what problem they are trying to answer. But that is irrelevant here, we know the problem that we are trying to deal with and it is not the same one that Morgan address.Martin Hogbin (talk) 20:06, 4 December 2009 (UTC)

Why the Morgan scenario is a variant.

The problem statement in the current article is the unambiguous K&W version. In this, the host is stated to choose randomly when he has a choice of goat doors.

In the Morgan paper the authors assign a parameter q to the probability that the host will choose a specific door if the player has initially chosen the car. It is clearly envisaged that this parameter might have a value other than 1/2, thus Moragn are clearly considering a different scenario from that addressed by this article.

To be specific, Morgan are addressing the case that the car is initially placed randomly but the host is known to choose a goat door non-randomly when the player has initially chosen the car. They do not say this, in fact they say nothing about the initial car placement or the host's policy at the start of the paper. We are left to deduce the problem that they are addressing from the mathematics that they use to solve the problem. The fact that Morgan take the probability of the car initially being behind any given door as 1/3 tells us that they take the car to have been randomly placed at the start. This agrees with the K&W problem statement. The fact that Morgan take the probability that the host will open a specific door when the host has initially chosen a car tells us that the host is assumed to choose non-randomly in this case. This is not in agreement with our K&W problem statement.

The Morgan scenario is therefor a rather bizarre and unrealistic one in which we know that the car was placed randomly by the producer (or his agent) but we know that the host will choose a legal goat door non-randomly. Martin Hogbin (talk) 17:19, 6 December 2009 (UTC)

Obviously Morgan paper (and other similar treatments) are a generalization of the non ambiguous case and as such can be seen as variant as well. However that does by no means change, that they use their generalization to deal with the non ambiguous case (K&W) too and explicitly state so themselves.

That aside the attempt to frame the (K&W) formulation as "the" MHP problem is not appropriate either and not in line with sources. (K&W) is the most common approach to remove the ambiguity - not more, not less.--Kmhkmh (talk) 17:35, 6 December 2009 (UTC)

As I have said before, I am not attempting to frame the K&W formulation as the MHP, others have done this for me, it is the formulation that a consensus of editors decided should represent the problem in this article. I do, in fact, support the K&W statement because it is based on their published and verified view of how most people interpret the problem.

Morgan's problem statement, on the other hand, is criticised in the very journal in which it is published. Read Seymann's comment. It is about as critical as one can get whilst conforming to the strict protocol required for publication in peer reviewed journal. Why do you think it is there? It is quite unusual to have such a commentary immediately following a published paper

K&W clearly state the problem that they are addressing, the only way to determine the problem that Morgan address is to work backwards from their calculations.

Finally, to address a generalisation of a problem is to address a different problem. In Morgan's case it is an incomplete and unrealistic generalisation. Martin Hogbin (talk) 17:59, 6 December 2009 (UTC)

I don't think the Morgan scenario is bizarre and unrealistic. I don't think the K&W version is a good version. There are different ways to disambiguate the original problem as posed by vos Savant; some are interesting in that they have interesting solutions. Why must we assume that the host chooses his door at uniformly at random, and so on? I think it is interesting that the unconditional problem has got the "paradoxical" (good) solution "always change" under rather weak assumptions: namely that the *player* chooses his initial door at random. He doesn't need to assume anything about the behaviour of the quiz team (where the car is located) nor of the quiz-master. Whatever probabilities are used for the quiz-teams and the hosts choices, the player will definitely win on 2/3 of the time. This is called an equalizer strategy in game-theory - the minimax solution is actually such that it doesn't matter a damn what the other player does, as long as you choose your equalizer strategy. The minimax solution or Nash equilibrium is the solution (which exists if you allow all choices of all parties at all stages to be random) such that for each player, if they use that strategy, they cannot do worse than the "value" of the game. ie if the player uses the solution "initial choice uniformly random, thereafter always switch" he is guaranteed at least a 2/3 (unconditional) chance of winning the car. If the team always locates the car uniformly at random and if the host always uniformly randomly opens a door revealing a goat and not chosen initially by the player, then the TV show is guaranteed at least a 1/3 (unconditional) chance of keeping the car. If either moves away from the minimax strategy, the other could in principle do better (by guessing the strategy of their opponent). So one might suppose that "reasonable" opponents will settle at Nash equilibrium. But why should anyone be reasonable? In practice, the players certainly weren't reasonable. Gill110951 (talk) 17:37, 6 December 2009 (UTC)

Whether the K&W is a 'good' version is a matter of opinion but it is, according to the most reliable published source on this matter (K&W), the way most people actually interpret the problem.

The MHP is notable because it is a very simple mathematical puzzle that most people get wrong. That is why there was such a furore about vos Savant's original and correct answer, most people did not believe it. It is the simple non-conditional problem that is notable and it is this that this article should initially address.

Beyond this there are many ways to complicate the problem and I have no objections to these being discussed after the essential paradox has been dealt with. We can assume that the car is placed non-randomly, the player chooses non-randomly and that the host chooses non-randomly. With no further information, this makes the problem very uninteresting. There is no logical reason to take the host action to be non-random any more than there is to take the producer's initial car placement to be non-random, in reality they were probably both approximately random. We could envisage that the player has studied the history of the show. This would give information equally about the initial car placement and the host policy. There is no special reason to assume the car is placed randomly but host acts non-randomly except that it produces and 'elegant solution'. Martin Hogbin (talk) 18:18, 6 December 2009 (UTC)

From the rejoinder to vos Savant's reply to the Morgan et al. article: "One of the ideas put forth in our article, and one of the few that directly concerns her responses, is that even if one accepts the restrictions she places on the reader's question, it is still a conditional probability problem. One may argue that the information necessary to use the conditional solution is not available to the player, or that given natural symmetry conditions, the unconditional approach necessarily leads to the same result, but this does not change the aforementioned fact."

What I'm hearing you suggest is that the article exclude this POV, i.e. that the article should explicitly say (or implicitly imply, by omission) the "basic" MHP is NOT a conditional probability problem. Is that what you mean? -- Rick Block (talk) 19:54, 6 December 2009 (UTC)

I accept that even with the K&W formulation, the problem is strictly one that requires conditional probability but that is not what makes the problem notable. Despite that fact, there are plenty of sources that treat the problem unconditionally and there are many reasons and justifications for us to do that here which we have been through many times.

By the Morgan scenario, I mean the (rather unrealistic and somewhat contrived) formulation in which q can be other than 1/2. This clearly is not the K&W formulation that is stated at the start of this article. Thus Morgan address a different and (bizarrely) more general problem. This is what I call the Morgan scenario. Regrettable Morgan do not make clear the problem that they are addressing in their paper. Martin Hogbin (talk) 20:05, 6 December 2009 (UTC)

Yes, there are plenty of sources that treat the problem unconditionally. But you're ALSO suggesting removing the conditional solution section and treating this as a variant (aren't you?). The article ALREADY leads with an unconditional solution. Excluding the conditional solution, or suggesting it applies only to a "variant" is what I'm objecting to. -- Rick Block (talk) 20:35, 6 December 2009 (UTC)

Not quite. I suggest that after the unconditional solution we have 'Aids to understanding' and 'Sources of confusion' for the unconditional solution, then a bit to explain that in some formulations even the symmetric version should, strictly speaking, be treated conditionally but this makes no difference to the answer and there are good reasons not to do this. After that I would mention the Morgan scenario in which the host chooses non-randomly making a conditional approach essential. Martin Hogbin (talk) 21:48, 6 December 2009 (UTC)

And, by doing this, aren't you making the article take the POV that the unconditional solution is complete and correct? YOU think that this is simply the truth, which is your prerogative. However, multiple sources (it's not just Morgan et al.) dispute this and say that the MHP is fundamentally a conditional probability problem, and that the unconditional solutions address a slightly different problem (that being the problem you think IS the MHP, but again this is your view). You want to elevate the unconditional approach to be the primary one, making the article NOT neutral on this issue. -- Rick Block (talk) 22:11, 6 December 2009 (UTC)

I am making the point that the unconditional approach is the the notable one. If have accepted many times, even today, that for some problem formulations (such as the current K&W) the problem is, strictly speaking, conditional. The article is not currently neutral, it places far too much emphasis on the Morgan paper.

'Sources of confusion' and Aids to understanding' clearly belong to the unconditional section. Generally when people get to the stage that they realise that some problem formulations are conditional they are not confused and already understand the problem. Little of the the text in these sections relates to the conditional nature of some problem formulations. Martin Hogbin (talk) 22:36, 6 December 2009 (UTC)

You say above that K&W is the "most reliable published source" on the matter of how people interpret the problem. From p.5 of the PDF version: "Although, semantically, Door 3 in the standard version is named merely as an example ("Monty Hall opens another door, say, number 3"), most participants take the opening of Door 3 for granted and base their reasoning on this fact. In a pretest we gave participants (N = 40) the standard version [the original Parade version], asking them to illustrate their view of the situation described by drawing a sketch. After excluding four uninterpretable drawings, we saw that 35 out of the remaining 36 (97%) indeed drew an open Door 3, and only a single participant (3%) indicated other constellations also remain possible according to the wording of the standard version. The assumption that only Door 3 will open is further reinforced by the question that follows: "Do you want to switch to Door Number 2?" Note that once formed, this assumption prevents the problem solver from gaining access to the intuitive solution illustrated in Figure 1."

97% drew an open Door 3. Are they thinking of the unconditional problem, or the conditional one? What, exactly, is confusing them?

The unconditional solution is what is generally presented as the answer. But 35 out of 36 of K&W's test subjects try to solve the conditional problem. -- Rick Block (talk) 23:00, 6 December 2009 (UTC)

What the unconditional solution says is something like: ha ha, the problem tricked you into looking at the problem wrong - what is really being asked isn't the conditional case you're trying to solve but the general chance of winning by switching which any idiot can see is 2/3. What the conditional solution says is: your approach was fine, you just didn't execute it quite right because you forgot that the host opens Door 2 sometimes when the car is behind Door 1. Both approaches are equally valid. That's what it means to be NPOV. -- Rick Block (talk) 00:31, 7 December 2009 (UTC)

No, you are absolutely wrong about one thing, which is that any idiot can see that the chance of winning by switching for the unconditional case is 2/3. I have never seen that view expressed anywhere else, either here of in the literature. The average punter does not know what conditional and unconditional mean. I agree that the problem that they in fact address, when presented with the Whitaker question, is the problem that is strictly speaking conditional but there is no evidence that they see and understand and attempt to solve it this way. As you say above the unconditional solution is the most common.

What we agree on is that the unconditional solution is the most common and notable one. I also agree that, if the problem is presented in a conditional form, the solutions should strictly speaking be conditional, we agree that this is one of the valid points made by the Morgan paper, however the unconditional solution is by far the most notable one and it must be thoroughly addressed and explained in this article, including 'Aids to understanding' and 'Sources of confusion'. That is all that I am asking for. This may not be easy, whilst basing our article on reliable sources, but it is worth a try.Martin Hogbin (talk) 12:29, 7 December 2009 (UTC)

Sorry - I was specifically talking about popular sources being the ones that are generally presented. The "any idiot" bit was meant to be part of what they say, not something I'm saying (they say any idiot can see that your initial chance of selecting the car is 1/3 so if you don't switch that must be your overall chance of winning). IMO, why people resist this solution is precisely because it violates their (conditional) mental model of what the problem is asking. To thoroughly address and explain this solution, you HAVE to talk about this.

And, again, by presenting a more or less complete article (many sections) about the "unconditional" solution - in particular presenting a "solution" section that does not include a conditional solution - would make the article take the POV that the unconditional solution is the preferred (or most correct) solution. If you survey the breadth of the literature (not just the popular literature), this is a highly distorted view. -- Rick Block (talk) 14:24, 7 December 2009 (UTC)

Another five cents worth by a mathematician

Right now I agree that the conditional version of the problem is a "minor variant" and that the article should focus on the unconditional version. Here's some explanation for this opinion.

I think that the "nicest problem" is the unconditional problem which is solved by always changing doors, and in order to justify that, all you need is that the initial chance of choosing the good door is 1/3. This could be the case because the player chose a door uniformly at random, or because the quiz-team hid the door uniformly at random. A wise player will not allow the quizteam to fool him so he will choose his door uniformly at random. It's the minimax strategy (guarantees a win probability of 2/3) and even an equalizer strategy (win probability is 2/3 independently of strategy of the quiz team. [all probability statements here are unconditional ones]. I guess that Marilyn meant us to think about the unconditional problem because it is so beautiful, clean, paradoxical. But I haven't looked at all her writings (the obvious thing is to look at her answers, and see what problem she actually solves). This is a question of historical research and proper documentation, nothing wrong with that.

When the [unconditional] problem was first posed to me by my mathematician friend Adrian Baddeley I gave the wrong answer but afterwards could easily be convinced what was the right answer. My mother who is not a mathematician, indeed hardly had any schooling at all, instantly got the right answer by imagining the situation with 100 doors; you have chosen one, the quizmaster opens 98; do you switch? I think only a fool wouldn't.

But people (even Marilyn) feel it necessary to add to the question, "eg if you chose door 1 and the quizmaster opened door 2". This suggests to people who have heard about conditional probability that you ought to look at the conditional problem. Of course, nothing forbids you to study whatever problem you like. Then there is not a unique answer. I can tell you that if the player uses his minimax strategy (choose door at random, always switch) then there are strategies of the quizteam such that for some outcome of the first three moves (quizteam's car location, player's choice, host's choice) the car is certainly behind the player's initial door, for other outcomes of the first three moves the car is certainly behind the other closed door. So the conditional win chances, if your strategy is "switch, irrespective" vary from 0 through to 1. But on average you'll win 2/3 of the time. It's necessary to assume that the quizteam does not hide the car with uniform random chances for this situation to arise. EG the car is always behind door 1. You choose door 1. Quizmaster opens door 2. You switch. You don't get the car.

Things are not quite so bad if we insist that the car is hidden uniformly at random. Then, however you choose your door, and however the host makes his choice, your conditional probability of finding the car behind the other closed door is somewhere between 1/2 and 1. Again (of necessity) it averages out at 2/3.

At least we can say, in this situation it is never (ie, condionally) to your disadvantage to change.

But I repeat that in my opinion a wikipedia article on MHP ought to concentrate on a succinct expose of the unconditional problem, written in terms which will convince the man in the street (by its logic!) as well as satisfy the trained mathematician (who will know how to translate the english sentences into logical statements and mathematical expressions). Make some space for discussing variants.

The beauty of that is that one does not have to look at conditional probabilities, which as I mentioned could be anything, and in order to somehow force them to the nice answer you have to make assumptions about the host's behaviour. Why should you? All we know is that he always asks this question, every show, again and again. Such an article can be short and sweet and no maths elaboration is needed at all.

A section on variants of the problem could discuss the conditional formulation, and various answers to that under various assumptions.

The beauty of the simple unconditional pure MHP is that it is a true good mathematical paradox in the sense that everyone is first fooled but afterwards easily agrees what is the good answer (unless you are lawyers: in a survey at Nijmegen university, it was not possible to convince a lawyer that their initial answer was wrong). A paradox for talking about in the pub or at a party is not good if it requires tables of calculations and mathematical notation. The unconditional version is the version which you can solve by simulation, repeating over and over again (a splendid exercise, which forces one to fix each of the stages of the game). Gill110951 (talk) 04:43, 7 December 2009 (UTC)

Yes; but also the conditional version can be "solved" by simulation. I always emphasize to students: you cannot enforce a condition to hold, but you can form a sub-sample by postselection! Boris Tsirelson (talk) 06:54, 7 December 2009 (UTC)

There seems to be a bit of misconception though, if i read Gill's comment correctly. So to be sure here, vos Savant did not pose an (unconditional) problem, but she offered an (correct/reasonable) unconditional solution to a (conditional) problem posed by Whitaker. Moreover the original question goes back to Selvin, who suggested it for a statistical magazine and afaik as a conditional problem or with a conditional solution in mind. So while we can feature the unconditional solution first for its beauty/simplicity/accessibility and given the ambiguity as a perfectly correct solution, we cannot simply declare the problem itself unconditionally per se, since this mathematically speaking not quite true and grossly misrepresenting the bulk of publications/literature on the subject. So while you can argue that conditional perspective is not required, you cannot argue the conditional approach does not address the original problem (i.e. is mere variant). My 2 cents as a Wikipedian and fellow mathematician.--Kmhkmh (talk) 13:10, 7 December 2009 (UTC)

You have been swayed in your opinion here by Morgan, who tried to take a question from an amateur in a popular general interest magazine to be a formal probability problem statement. What vos Savant did was to use her judgment to answer the question based on her interpretation of the intent of the questioner (as later suggested by Seymann). This is surely the better approach Martin Hogbin (talk) 13:41, 7 December 2009 (UTC)

No i have not and i have told you so repeatedly in the past. In fact I haven't even read Morgan's paper. However I have read many other publications on the subject in German and English and as I've pointed out several times already almost all of them treat it at least as a conditional problem as well and sometimes as a conditional problem only.--Kmhkmh (talk) 15:00, 7 December 2009 (UTC)

Or perhaps, Martin has been swayed in his opinion here by vos Savant. The relevant question here is not one of opinions, but about what the literature says. I hear Kmhkmh saying the literature as a whole does not give any preference to the unconditional solution. NPOV says then the article shouldn't either. -- Rick Block (talk) 14:39, 7 December 2009 (UTC)

+1--Kmhkmh (talk) 15:25, 7 December 2009 (UTC)

I guess his is where the mediation will come in. My opinion, stated many time before, is that Morgan et al. are experts on statistics, thus a paper published by them should be treated as a reliable source in matters of statistical calculation (after all they get most of this right). They have no more right to pontificate as to what 'The Monty Hall Problem' is than the average man in the street and considerably less credibility in this respect than vos Savant, whose job it was to regularly interpret vague questions from the general public. We say at the start of this article that the Monty Hall problem is a probability puzzle, and that is how it should be primarily treated here - as a simple puzzle that most people get wrong. Martin Hogbin (talk) 17:21, 7 December 2009 (UTC)

Article removed from my watchlist.

This article keeps disappearing from my watchlist. Is anyone else having this problem? Martin Hogbin (talk) 12:48, 7 December 2009 (UTC)

This Is All I Really Care About

If, after the Popular solution, these words were given great prominence before all that conditional stuff, etc., my main goals would be satisfied.

"The Monty Hall problem is unconditional. That is the whole paradox; the rest is the explanation; go and learn." Glkanter (talk) 23:13, 7 December 2009 (UTC)

Is This Chronology Correct?

1975 - Selvin poses the problem.

It is solved unconditionally. It is hailed as a great Paradox.

Devlin and many others write articles and text books as reliably sourced references using only the unconditional solution. They make no mention of Morgan or conditionality.

This might be a good time to explain why you put this here (and other places, multiple times, after being hinted that no Devlin paper earlier than 2003 is in evidence; and Morgan hasn't published his bit yet, so not mentioning it could probably go without saying at this date.). Dicklyon (talk) 07:10, 8 December 2009 (UTC)

1990 - Whitaker poses the problem to vos Savant.

It is solved unconditionally. It is hailed as a great Paradox.

Devlin and many others continue to write articles and text books as reliably sourced references using only the unconditional solution. They make no mention of Morgan or conditionality.

1991 - Morgan 'restates and formalizes' the problem.

It is solved conditionally, but only if the 'two remaining goat doors' constraint is fixed at 1/2. Morgan criticizes the unconditional solutions as being deficient in some way. Otherwise, Morgan offers no solution.

You tell me. 16 years after Selvin, is the MHP still hailed as a great Paradox because of the 'old' unconditional solution? Or is it now considered a great Paradox because of the 'new' conditional solution, which requires working all the way down to the 'equal goat door constraint'? Or, is it no longer considered a great Paradox because of the 'new' conditional solution, which requires working all the way down to the 'equal goat door constraint'? My position? Selvin and the unconditional solution stand unbowed.

Devlin and many others continue...

Wikipedia editors decide that the conditional problem is why the MHP is a great paradox.

The Wikipedia article is written to emphasize the lack of understanding of Probability by those we are satisfied with the unconditional solution.

Devlin and many others continue...

There are extensive arguments at Wikipedia, nearly always on unconditional vs conditional.

Devlin and many others continue...

Glkanter points out that the universal understanding of a TV game show denies any possibility of the conditional problem statement.

There are extensive arguments at Wikipedia, nearly always on unconditional vs conditional.

Is this roughly how we got here, today? Glkanter (talk) 14:05, 7 December 2009 (UTC) Glkanter (talk) 23:46, 7 December 2009 (UTC)

Roughly. But if you try to keep it more accurate, you will do better making your point. Dicklyon (talk) 16:37, 7 December 2009 (UTC)

I think you're way out of line with those edits. But if you support the proposed changes, I won't mention it again. Glkanter (talk) 16:42, 7 December 2009 (UTC)

Really? No unconditional papers by reliably sources were published between Selvin and Morgan? I find that unlikely. Glkanter (talk) 16:47, 7 December 2009 (UTC)

I see Dick Lyon discusses his numerous edit wars on his user page. I cannot compete with that. What is my best approach to take to return my interpretation of how events happened, which I want to share on this talk page, to it's original form? Thank you. Glkanter (talk) 17:24, 7 December 2009 (UTC)

Look at this edit summary from Dicklyon:

(cur) (prev) 11:34, 7 December 2009 Dicklyon (talk | contribs) (792,974 bytes) (Reverted 1 edit by Glkanter; Obviously incorrect, pointy addition, bordering on vandalism. (TW)) (undo)

I don't deserve that. What, I vandalized a section I just created? Why would someone write that? Glkanter (talk) 17:30, 7 December 2009 (UTC)

Dicklyon, I don't want any trouble. I just want my interpretation of the chronology to be on this talk page. Unedited, but certainly commented on below. So, do you want this section, and I'll start a new one? You want to start a new one, for your interpretation, and I'll fix this one back to my vision? Just let me know. Thanks. Glkanter (talk) 17:47, 7 December 2009 (UTC)

You can just write a sensible summary instead of interspersing repeated stuff as a roundabout way to make a point. And my user pages doesn't discuss most of my edit wars, just the "dramatic" ones. Dicklyon (talk) 17:50, 7 December 2009 (UTC)

Dick, deleting material from a talk page is rarely justified and not so in this case. Maybe glkanter could have formatted his contribution better but that is no reason to delete it. It certainly was not vandalism. Martin Hogbin (talk) 18:44, 7 December 2009 (UTC)

I don't think I removed any actual material, just interspersed repetitions of a statement that made the chronology hard to follow. If he or someone wants to put it back I won't bother with it. Dicklyon (talk) 19:14, 7 December 2009 (UTC)

Thanks for that. Despite very strong opinions this discussion has remained completely civil, let us keep it that way. Martin Hogbin (talk) 19:18, 7 December 2009 (UTC)

Of course; I hope I didn't do anything that hinted otherwise. Dicklyon (talk) 19:22, 7 December 2009 (UTC)

Well, that's quite a Wikipedia education I was administered today. And I want to give thanks to the man who taught me this lesson, Dicklyon.

Thank you Dicklyon for:

Vandalizing my edit

Then, accusing me of being the vandal (That was outstanding. Really. Not many can pull this one off successfully! Kudos!)

Calling my personal opinion on a talk page 'obviously incorrect'

Lecturing me on the proper 'tone' to use on Wikipedia talk pages

Denying in carefully chosen words that you did any of the above

Making it necessary for another Wikipedian to defend me. That does wonders for my self esteem.

Teaching me who the baddest edit warrior on Wikipedia is.

Giving me permission to return the section I created to it's original status.

Assuring me that you won't further violate Wikipedia rules regarding my edits in talk pages

I'm sure there is much more I'm overlooking. I'm only human.

But anyways, an apology is in order to all you Wikipedia editors who were forced to sit through this.

In the spirit of the guy who got shot in the face by VP Dick Cheney, I apologize to Dicklyon for being the recipient of your unprovoked savage violations of my good faith edits to a Wikipedia talk page. Glkanter (talk) 21:49, 7 December 2009 (UTC)

You really are a piece of work! Your current rant is no less disruptive than this edit that I reverted thinking it looked like vandalism. If it was not intended as disruptive, I misinterpreted your intention; is that my fault? Dicklyon (talk) 04:32, 8 December 2009 (UTC)

Is it your fault? No, of course not. I read your edit warring exploits. It's never your fault. This time, it's my fault, like I said above. You just out of the blue decide to edit some guy's stuff on a talk page? How could you be to blame? I see you as the victim in this situation. In fact, I would speak on your behalf about how, despite your benevolence, Glkanter has wronged you. Glkanter (talk) 05:05, 8 December 2009 (UTC)

Good, then we agree: my misinterpretation of your intentions was your fault. Case closed. Dicklyon (talk) 07:06, 8 December 2009 (UTC)

To avoid further drama, I'll take this article off my watch list again, as I had the good sense to do last April when it was clear that progress would not be possible. I hope nobody invites me to comment again, as I might repeat my failing and come back for more. Dicklyon (talk) 07:34, 8 December 2009 (UTC)

Gentlemen. We do not need this. Despite strong feelings on both sides the discussion has remained civil. As I said above, I think Dick went a bit too far when he deleted text from the talk page, but he has said the he will not object if it is restored. Glkanter, why not just restore the text and leave it at that. Martin Hogbin (talk) 10:27, 8 December 2009 (UTC)

That's too bad he's stopped watching this page. He's missing something extraordinary. Somebody should give him a head's up. Glkanter (talk) 22:45, 8 December 2009 (UTC)

Relevance of History

I found the history of the MHP enlightening. Now here's my point of view as to what that means for the article. The MHP became famous because of Marilyn vos Savant, and she has made clear that she was thinking of the unconditional problem, just as most people interpreted her words; that problem has a number of beautiful short and simultaneously rigorous solutions. It's a beautiful problem with a beautiful solution, that's why it has become folklore. (In my humble opinion) she mentions door 1 and door 2 merely as a pedagogic device, to make things concrete so you can imagine the situation better; IMHO I think she wants an answer which doesn't depend on which door is called door 1 and which door is called door 2. Door 1 *is* the door you first chose, whatever door that actually is; door 2 *is* the door the host chooses, whatever door that actually is. She wants a universal answer independent of the numbers of the doors; the answer "always switch" is the good unconditional answer as long as the door the player has chosen had 1/3 chance of the car being behind it, which is moreover what most people (IMHO) immediately imagine.

This unconditional game with its hidden assumption and the corresponding unconditional solution, I think, *is* what most people out there, whether mathematicians or plumbers, think of as *the* MHP.

Conditional versions of the MHP are interesting and it is enlightening to consider them. The "original" MHP might have been conditional. Who knows, maybe the Aztecs had an even earlier version which was unconditional. So what.

So today at 10:26 am CET I think the article should, for historical and cultural, and pedagogical reasons, start by elucidating and solving the unconditional problem, and then go on a) to describe the (pre)history of MHP (neutrally, objectively) and b) to discuss important variants, such as the conditional version of the problem (neutrally, objectively). Gill110951 (talk) 09:29, 8 December 2009 (UTC)

I think the primitive chronology I created shows something similar. From Selvin, 1975 through vos Savant, 1990 it was consider a great paradox and had only been considered unconditional. In 1991 with Morgan, does this change to great patadox - conditional, stay as great paradox unconditional, or meh, if it's got to be conditional with the equal goat door constraint, it's not much of a great paradox anymore. I have my opinion, I'd like to hear the other side's response. Glkanter (talk) 11:20, 8 December 2009 (UTC)

That is pretty much what I am suggesting and to some degree how the article is now. The change I would like is to link the 'Aids to understanding' and 'Sources of confusion' sections to the 'Popular solutions' (unconditional) section. If you read the 'Aids to understanding' and 'Sources of confusion' you will see that they relate far better to the unconditional problem than the conditional one.

Following the unconditional solution I would like to have a section explaing why the problem, in certain formulations, is, strictly speaking, a conditional one but why this is not important for the symmetric (host chooses randomly) case. The Morgan scenario (host is known to choose non-randomly but car is placed randomly) I would like to relegate to the variants section. I am not going to try to hide my distaste for the Morgan paper. I think it did the whole MHP a major disservice by turning a simple problem that everybody gets wrong into a complicated problem that nobody cares about. Martin Hogbin (talk) 10:20, 8 December 2009 (UTC)

Is The Contestant Aware?

Has it been agreed by the editors of this article that regardless of how Monty handles the 'two goats remaining' situation, the contestant has no knowledge of the method?

It seems to me that this is a (unstated) premise of the problem, as both vos Savant (Whitaker) and Krauss and Wang begin the problem statement with: 'Suppose you're on a game show'. I read this as clearly stating it is only the contestant's point of view we are concerned about. And, being a game show, the host is prohibited from divulging to the contestant either where the car is, or where the car is not.

Is there agreement on this, or is this in dispute? Glkanter (talk) 11:28, 29 November 2009 (UTC)

As far as I know, this is the only known instance where a contestant at home was able to determine a game show's strategy. It was an aberration, an unexpected outcome, and steps were immediately taken to prevent it from happening again.

http://en.wikipedia.org/wiki/Press_Your_Luck#Michael_Larson

In my humble opinion, a particular contestant gaining usable information from the hosts actions, which would make that contestant's SoK something different than the 'average' contestant's SoK of '2/3 likelihood of a car if I switch' is inconsistent with the published problem statement: 'Suppose you're on a game show'. The problem statement would become: 'Suppose you are not on a game show'. This is not merely a 'variant', or a new or changed premise. It is a completely different problem. And why this completely different problem should be referenced as often and prominently as it is in the Wikipedia article does not make sense to me.

There are countless reliable published sources which use the combining doors solution to derive the '2/3 likelihood of a car if I switch'. The contestant uses this method to determine the 2/3, then properly says to himself, 'Monty's actions haven't given me any new information, so I'll go with the 2/3 for the average contestant'. Glkanter (talk) 14:49, 29 November 2009 (UTC)

Is there a specific change you're suggesting? If not, I'd suggest moving this thread to the /Arguments page. -- Rick Block (talk) 01:15, 30 November 2009 (UTC)

Rick, this is the central element in my criticism of the over-reliance on Morgan's paper in the article. In order to develop a consensus that includes you, I need to know your position on it, so I would greatly appreciate your response to the original question:

"Has it been agreed by the editors of this article that regardless of how Monty handles the 'two goats remaining' situation, the contestant has no knowledge of the method?."

I see this as an 'editing' question more than a 'mathematics' question, so I'd prefer to leave it here. Thank you. Glkanter (talk) 13:51, 30 November 2009 (UTC)

Is there a published source that takes the stance that "Suppose you're on a game show" means what you're suggesting? If not, then what you're talking about is WP:OR which makes it a moot point as far as editing is concerned. I'm not saying it's a bad or invalid argument, just that if it's not published it's worthless for editing purposes. -- Rick Block (talk) 14:52, 30 November 2009 (UTC)

Yes, I have previously linked to the Wikipedia articles showing the only 2 instances where individual contestants had a different State of Knowledge than would the average contestant. The 1950s Quiz Show Scandal and Whammy/Press Your Luck. Both were considered as extraordinary events.

It wouldn't matter, even if it was OR. I'm not going to put this critical mistake of Morgan's in the article (unless other editors want to build a consensus for it). I'm only using it to decide, using facts rather than my personal opinion, how much emphasis Morgan should get in the article.

Really? We're going to argue over the common understanding of a game show? I'd rather not. Glkanter (talk) 15:22, 30 November 2009 (UTC)

Rick, I have directly asked you this question many times, and have never seen a direct 'yes or no' answer from you. As this is a crucial element of the consensus that has been built, it is essential that we understand your reasons if you do not agree with the paragraph above:

"It seems to me that this is a (unstated) premise of the problem, as both vos Savant (Whitaker) and Krauss and Wang begin the problem statement with: 'Suppose you're on a game show'. I read this as clearly stating it is only the contestant's point of view we are concerned about. And, being a game show, the host is prohibited from divulging to the contestant either where the car is, or where the car is not."

I look forward to your response. Glkanter (talk) 23:02, 1 December 2009 (UTC)

Rick has responded under his comments section. For clarity and closure, I'll copy them into this discussion.

"Glkanter asks why I haven't responded about his "Is The Contestant Aware?" question. Why should I? Glkanter has repeatedly demonstrated a complete lack of comprehension of nearly everything I've ever said. It's like trying to explain something to a cat. At some point you just have to give up. However, I'll give it another go. Meow, meeeow, meow, meowww. I'm not sure I have that quite right since I don't speak cat, but it's probably about as comprehensible to him as anything else I could say."

http://en.wikipedia.org/w/index.php?title=Talk:Monty_Hall_problem&diff=329587461&oldid=329562564

I've included only the relevant portion above. The response of 01:53, 4 December 2009 (UTC) is much longer. Glkanter (talk) 18:59, 4 December 2009 (UTC)

Nijdam, your response to the question at the beginning of this section is also of great interest to the other editors. In the MHP that begins with "Suppose you are on a game show", is the contestant aware of Monty's door choice behaviour when there are 2 goats remaining? Thank you. Glkanter (talk) 15:14, 4 December 2009 (UTC)

Glkanter - your questions have already been answered, see for example here or here. More explicitly

Has it been agreed by the editors of this article that regardless of how Monty handles the 'two goats remaining' situation, the contestant has no knowledge of the method? Yes, but that doesn't mean the contestant is prevented from wondering what affect this might have on her chances of winning by switching.

Rick, how am I to parse 'yes, but' in the above paragraph? Is that "yes, I agree that regardless of how Monty handles the 'two goats remaining' situation, the contestant has no knowledge of the method?" Or, "but..." Let the contestant wonder, it's a free country. But that doesn't affect the game play. Glkanter (talk) 18:51, 5 December 2009 (UTC)

Rick, I still show the above question as an open item in our on-going discussion. Thank you. Glkanter (talk) 14:12, 10 December 2009 (UTC)

And, being a game show, the host is prohibited from divulging to the contestant either where the car is, or where the car is not. Is there agreement on this, or is this in dispute? The problem specifies that a game show is the setting. How much information the host divulges to the contestant depends on how the problem is specifically phrased, not on what "must" be true of game shows.

By definition, information is given to all contestants equally in game shows. The content of that information varies by specific game rules. Glkanter (talk) 18:51, 5 December 2009 (UTC)

These are both questions that are more appropriate for the /Arguments page than here, since as editors what we believe to be the TRUTH is ultimately of no importance. What is important is what reliable sources say. Our task as editors is to accurately represent what these sources say regardless of whether we individually agree. This one of the three fundamental policies of Wikipedia, see WP:NPOV. -- Rick Block (talk) 17:44, 5 December 2009 (UTC)

What "the conditional problem" and "the unconditional problem" mean

I wasn't quite sure where to respond, so a started a section.

There are multiple ways to address any probability problem. If I ask "What is the probability a die rolled a 3, if we know it rolled odd?" we can solve it conditionally or unconditionally. We can say there are six equiprobable outcomes, and the conditional probability is found by P(3|odd)=P(3 and odd)/(P(1)+P(3)+P(5)) = (1/6)/(1/6+1/6+1/6)=1/3, or that there are three equiprobable outcomes that are odd, so P(3)=1/3. The unconditional problem is isomorphic with the conditional one.
The MHP IS a conditional problem, but depending on how you address it, you can solve an isomorphic unconditional problem in its place. Just like my trivial example. That is not what we want for the so-called "unconditional problem" that the main part of the article should address.
What we want, is for the contestant to base her decision on P(switch to car|she choose a door and host opened another door) as opposed to P(switch to car|she choose door #1 and host opened door #3). And I worded that carefully, because the first option is EXACTLY what the Whitiker problem statement says, and what K&W admit is the sematic meaning of the statement. We call it "the unconditional problem" because it is not conditioned on knowing how the host treats the doors differently. The doors cannot distinguished from each other in terms how probability applies. That does not mean that they can't be distinguished, it just means it can't be used.
The question in the classic MHP is "should the contestant switch." We want "the unconditional problem" because we are not told how the host treats the different doors in any statement of the problem that I am aware of. So there is no known source that uses the unique aspects of "the conditional problem" to DIRECTLY answer the MHP. None.
The two sources that seem to be recognized as starting the "game show" variation of this problem (as opposed to the related and much older "Three Prisoners" problem and Bertrand's Box Paradox), that is to say Steve Selvin's 1975 The American Statistician article and Marilyn vos Savant's Parade article, both used names for the boxes/doors as examples. But neither used those names to treat them differently in the solution. So they are both addressing the unconditional problem. Marilyn vos Savant, at least, has clarified that she intended the the "unconditional" problem.
There are only two aspects to "the conditional problem." (There are more that are what I call "game protocol," like whether the host always offers a switch. I disregard those, because we have to assume the game protocol is well represented in the problem statement.) There are two random chioces, not one, that are part of the uncertainty in the game and that require assumptions. Not all of the "conditional problem" sources use both (Morgan does not), but those that do universally make the assumptions about the car placement that they eschew for the host's choice. Tat is, they treat it like "the unconditional problem." No conclusions can be drawn, nor HAVE been drawn, about the answer to the full "unconditioanl problem."
There are, in general, three types of sources for the MHP: (A) Those that address problem from a common-knowledge standpoint, and that universally use "the unconditional problem" WHETHER OR NOT THEY USE CONDITIONAL PROBABILITY in their solution method. (B) Those that address the cognitive issues that cause confusion in people when first presented with the problem, and that emphasize in non-intuitive aaspect of switching to improve probability. This has nothing to do with the "condititional problem." (C) Those that explore it from a more rigorous mathematical viewpoint. Some emphasize either the unconditional, or the conditional. Call them CU and CC. Now, I haven't read all the references, but I did a quick run-down of the ones listed in the article. Of those that I could find quickly, 14 were in category A, 5 in category B, 4 in category CU, and 3 in CC. That shows that most people view the MHP as unconditional problem.
Nobody is saying to disregard "the conditional problem." It is just a variation. JeffJor (talk) 20:22, 7 December 2009 (UTC)

Well said! Are you a new editor or did you forget to log in? I removed the linefeeds to fix your numbering. Martin Hogbin (talk) 18:40, 7 December 2009 (UTC)

No, I do log in - but somehow in switching windows I keep ending up in a non-logged-in window, and I don't notice that when the edit comes up. JeffJor (talk) 20:22, 7 December 2009 (UTC)

@128.244.9.7:

I completely agree
I agree for the most part, however the bold line is a bit iffy. Who is exactly is "we"?
This is bit iffy. K&W don't admit but assess what the real meaning is according to them, other publications differ on that. Your reading of Whitakers intent is fine, but alas it is only valid option to read the problem. You can also see it as P(switch to car|she choose a door and host opened a particalur door). If you use particular instead of another you end up exactly where Morgan & bulk of mathematical treatments go and you do use the particular door, i.e. doors can be distinguished and used. Note that both version treat door 1 and door 3 as examples.
Not knowing the host behaviour per se does not mean we can't model it nor that we cannot draw conclusion from it. In fact Morgan & Co argue that no matter what the host behaviour is, you are never worse off by switching.
Your descripion is not quite correct here. Selvin the mathematician/statistician who coined the problem 15 years before the parade buzz provides a unconditional and conditional solution to it (see Rosenhouse). This btw also defeats a point Martin keeps rising for more than half a year now, i.e. that the problem is essential not a math problem, but the mathematical perspective is just a minor sideshow. If the original problem was posed, solved and named by mathematician in math journal and this is not supposed to to be a math problem, then frankly i don't what is. That vos Savant solved the problem as an unconditional one is fine, however since she did not pose the problem, this tells us nothing of the "real" intent/perspective of the question.
Here i admit, i'm not quite sure what point you're trying to make. What exactly is the "full unconditional" and to what literature/publications areyou refering for having drawn no conclusion. I assume you are aware that by now almost any recent primer into probability (in various language) contains a conditional treatment of MHP.
I disagree with your assessment of (B). One thing about conditional probilities is that they are considered "unintuitive", i.e. unintuitive aspects and in particular the handling a posteriori/additional knowledge for reasoning are all about conditional probabilities. As far as (C) goes I don't quite buy your category count at first glance but even if that turns out to be accurate it does match for overall treatment in literature to my experience. As i pointed out already almost any modern probability primer has a conditional approach to the problem.
yes and no, that depends on reading between the lines and carefully reviewing all the statement made here over time. Also see the recent suggestion by Jeffor and Glkanter, which to some degree you can consider as an attempt to remove the conditional approach from this article. Another thing is here.how you read/use the term variation. In light what else has been said here recently, I'd rather argue if you consider 2 problems to be isomorphic they are the "same".
One additional reminder: In our attempts to understand/explain to "what was really meant", we easily overlook that as far as WP is concerned (in particular for contentious issues), we are supposed to describe what various reputable sources say and not what we think is right.

--Kmhkmh (talk) 20:32, 7 December 2009 (UTC)

2. "We," I believe, are those who think the "conditional" problem is a variant. Sorry if it sounded to general.

3. K&W say that the doors numbers are examples only; that means it isn't intended literally. But AFAICT their treatment does not allow their participants to consider that the host is biased. Both are needed to make it "the conditional solution."

4. While you can model it, you can't use it to directly answer the question "Should the contestant switch?" because that question has to be answered from the contestant's SoK. It is just a coincidence that, for that one bias, it always goes up. For example, if the the host tells you the placement is biased 50%:25%:25% for D1:D2:D3, and that his opening bias (afte you choose Door #1) is either 25%:75% or 75%:25 for D2:D3, but not which? Then the probability of winning goes up in one case and down in the other, BUT IT IS STILL ADVANTAGEOUS TO SWITCH BECAUSE THE AVERAGE PROBABILITY GOES UP. The point is that it doesn't help the contestant IN GENERAL to say what the probability is based on hidden knowledge, if that knowledge is not also given.

5. Yes, Selvin did. See #2 above. Did you miss the part where Selvin said that the key to his solution was that he assumed the host chose randomly between two doors, making his solution a conditional approach to what we call the unconditional problem? Martin's point is entirely correct, and was the point of Selvin's and Savant's questions.

6. If door numbers are important, the solution has to treat the possibility of placement bias the same way it treats opening bias. If you do that, the question "Should she switch?" is unanswerable. THE CONDITIONAL PROBLEM CANNOT BE ANSWERED. So all the treatments eventually get around to assuming car placement is 1/3 to each door. The question they actually answer is halfway in between being the conditional one and the unconditional one.

7. Show me a survey where respondants took host bias into account.

8. Both approaches are to be included, even if separated into parts or even different articles. But any problem that does not address (1) The formal statement for the article or (2) The problem intended by either of the two articles that started the controversy, does not belong on an equal footing.

9. Agreed. In what way is the suggestion to keep them separate not? JeffJor (talk) 22:16, 7 December 2009 (UTC)

Jeff - You've quoted the K&W statement about the semantic meaning of the problem before. I must assume you've read the rest of the paragraph. If not I've quoted the entire paragraph above (this edit). What they're actually saying is that although Door 3 is semantically used as an example most people (97% of their sample) don't treat it this way. You (and they) are taking this to mean that the semantically precise interpretation is the "correct" one and thus that most people are more or less tricked by this problem statement into trying to solve the problem conditionally (and when they do this they come up with the 1/2 answer). Given this source says most (nearly all) people interpret the problem conditionally (meaning they're thinking about the specific case where Door 1 is the initial pick and Door 3 is the door the host has opened), and that this conditional problem is completely isomorphic to what K&W claim is the actual problem, and that other sources say that this conditional problem is what is meant, wouldn't it be rather helpful to our readers to address this issue head-on by providing a correct conditional solution? This is totally aside from the issue of whether addressing the problem only unconditionally satisfies NPOV. I don't know if anyone exactly puts it like this (I have most of the sources cited in the article - I'll look at some point), but I very strongly suspect that MOST of the refusal to believe the unconditional 2/3 answers comes from people using an incorrect conditional solution (not from failing to understand that the problem is "supposed" to be interpreted unconditionally).

As I put it above, the fully unconditional solution is essentially saying "Ha ha, the problem tricked you into looking at the problem wrong - what is really being asked isn't the conditional case you're trying to solve but the general chance of winning by switching. You can easily see this must be 2/3 like this: <your favorite unconditional solution here>." What the conditional solution says is "Your approach was fine, but it's not quite right because you forgot that the host opens Door 2 sometimes when the car is behind Door 1. You see the host open Door 3 all the time if the car is behind Door 2, but only 1/2 of the time if the car is behind Door 1. So, if you see the host open Door 3 the car is twice as likely to be behind Door 2." The segue into the "half conditional" problem happens quite naturally because of the "1/2" in this solution. Where exactly does it come from and what does it mean? It is of course the much maligned host preference. I would be fine deferring consideration of this "1/2" to a variant, but insisting that this "conditional approach" be treated as a variant or as addressing a "different question" seems absurd.

If you reply to this can you please try to separate your opinions from what you think the preponderance of reliable sources might say? For example, do you agree with Kmhkmh's assessment that "almost any modern probability primer has a conditional approach to the problem" (I'm taking this to mean either only a conditional approach or both conditional and unconditional approaches)? If you agree with this, I think NPOV says we MUST treat the conditional and unconditional approaches equally without favoring either one. -- Rick Block (talk) 03:46, 8 December 2009 (UTC)

1. I do not agree. The conditional solution calculates the conditional probability in the unconditional probability space, but the so-called unconditional solution calculates an unconditional probability in a conditioned probability space. There always is a condition. Nijdam (talk) 14:09, 8 December 2009 (UTC)

Yes, Rick, I've read the whole paragraph. The only point I ever made from that quote is that it is valid (and in fact proper) to interpret the Whitaker/MvS problem statement as not requiring specific door numbers; that they were examples. I didn't try to address whether K&W were using the conditional or unconditional problem, since they solve the full conditional one later (more below), and addressing it could be seen as POV. The "clarification" you provided, which you interpreted according to your POV, only shows that K&W recognized that respondants treated the door numbers the same way MvS did; as examples. Not as implying any dependency on the actual door numbers. And it was, at least in part, because the non-required use of examples was reinforced by K&W (not Whitaker/MvS) in the form their next question took: "Do you want to switch to Door #2?" That can't be answered without using the examples. So K&W followed in Morgan's and Gillman's footsteps, in rewording the MvS problem statement to make it become the conditional problem. But that paragraph, and my excerpt of it, had nothing to do with how the problem is addressed anywhere.

Once again, since I am apparently still speaking in cat when I say this (or your POV gets in the way of comprehension): The MHP is a conditional problem, but depending on what you see in it, it can be solved unconditioanlly. It is valid to do this: see G&S, p138, where they choose not to make an assumption which they say could be made without loss of generality. Their Figure 4.3 illustrates what Nijdam described using his own POV, as "an unconditional probability in a conditioned probability space" (emphasis added). Assuming it is a conditioned space means you are assuming there is a bias in either car placement and/or host choice. But that treatment is not what I call the unconditional problem. The "unconditional solution" is represented on p139 in Figure 4.4, and is a conditional solution that is isomorphic with the fully unconditional one (the isomorphism comes from the generalization they declined).

The point we (well, at least I) am trying to make here is that this pair of solutions represents what most people see in the MHP. K&W bear this out. People can a take what appears to be an unconditional approach (Figure 4.3 in G&S), but that is more complicated and provides no more generality than the alternative. With the generlization that a single door used in the solution represents the uninformed (to the contestant) distribution of all possibilities, and so its probability distribution must be the average of those possibilities, they can use conditional probability to get the same result. (It is possible that biases exist in this system: I am not assuming them away. But they all get eliminated in either solution. And I have no reference for this, but it is the approach most people, who can answer simple probability questions, actually take. I've seen it refered to as "symmetry." The paradox in the MHP is that they overextend this symmetry, not realizing how the revealed information makes things assymetric.) What we want the article to separate out, is the solution G&S add after Figure 4.4, where suddenly a bias is assumed without valid reason. That is what I have called "the conditional problem," because it requires the use of conditional probability, and ignores the symmetry which K&W always apply to their formulation of the probability. They don't use the conditional problem - they allow for it but always put in parametric values representative of symmerty, and thus the unconditional solution. They even provide arguments for why people assume symmetry. And I can't speak for "most primers," but I would think that what Kmhkmh has noticed is a conditional probability not based on biases; or if it is, wheere biases get eliminated by symmetry. 128.244.9.7 (talk) 16:34, 8 December 2009 (UTC)

As far as most primers are concerned, you get some impression via Google Books as well (or alternatively sample recent lectures on university websites or the classic libraries/bookstore approach). Also reading the beginning of your comment, there seems to be some misunderstanding. I'm not aware of anybody here (aside Nijdam maybe) claiming the unconditional would not be valid - certainly not me. As often in science there a various valid perspectives on a particular problem and various ways to model it accurately. There is also imho no issue with featuring the unconditional solution first and separate from the conditional one. There is however an issue with (subtle) attempts to redefine the problem as such in a manner, which is not in line with the bulk of the sources. This concerns in particular formulations insinuating, that the conditional approach is not solving the "real" MHP problem, that the MHP is not a math problem (but a layman's creation), that an "question-answering-expert" knows best what the problem "really" is and what the intent of the question being posed to her was, or that psychological study on how average people react or think of the problem, tells us what the problem really is. Because these various points are at best if not outright wrong the reflection of an individual publication and not the bulk of literature on subject. Hence the WP article cannot assume such a position. I'm not opposed to rearrangements/modification of the article as long as they don't create the critical points I've mentioned above. From my personal perspective the article could be arranged rather differently (starting with the ambiguous Parade formulation and the simple unconditional solution, after that separate chapters for mathematical analysis (conditional,conditional,criticism, subtle changes of perspective,various modelling approaches and the requirements, problem variants, etc.), psychological/cognitive analysis (K&M,Mueser, Granberg and others), occurrences of MHP in other areas/sciences and a historical overview/timeline (from Gardner until now). Having said that however in general i prefer a rather conservative editing policy for featured articles, because every major change requires a reevaluation of the features status. Also in my experience in contentious cases good articles can easily deteriorate with a lot disagreeing people editing around, that's actually one of the main reasons why I won't support a change until I see a feasible editor compromise emerge first, that a) provides a meaningful improvement rather than marginal changes only and b) is in line with WP guidelines.--Kmhkmh (talk) 17:13, 9 December 2009 (UTC)

Jeff - can you be more specific about the exact changes you're talking about? Is it:

Delete the entire section "Probabilistic solution"
Delete the 4th paragraph of "Sources of confusion"
Delete "Other host behaviors" under "Variants" (or only some)
Delete "Bayesian Analysis"

Or, are you suggesting something else?

From your comments above, it sounds like you have no fundamental issue with a conditional analysis so long as it assumes no host bias - i.e. you're only insisting that it be taken by definition that the host open one of two goat doors with equal probability. This makes me think you should be OK with the "Bayesian analysis" section, and even the "Probabilistic solution" section with fairly minor changes (like deleting the last paragraph in this section about the "host preference" variant or moving this paragraph to a "variant" section) - or even a single "Solution" section more like what was there following the last FARC (i.e. this version) which made no mention of a "host preference" variant. I think how we got to where we are is contention over the transition paragraphs in this solution section (the paragraph starting with "The reasoning above ..." and the following one). Would you be OK with reverting to this version and working on these paragraphs? -- Rick Block (talk) 20:34, 9 December 2009 (UTC)

The unconditional problem

Let us formulate the unconditional problem: A player will be offered the choice of one out of three doors, one of them hiding a car. What is the probability she wins the car? Nijdam (talk) 14:13, 8 December 2009 (UTC)

More OR? Ho hum. Let's focus on sources, and the commonly understood meaning of a game show. This is Wikipedia, and the MHP from Selvin (who may have originally called the puzzle 'Let's Make a Deal problem', the name of Monty Hall's show) through K & W begins 'Suppose you are on a game show...' Hosts don't tell contestant where the car is on a game show. Many reliable sources use logical notation, and/or words, to solve the problem. Is there a problem with that technique? Glkanter (talk) 14:27, 8 December 2009 (UTC)

That is a very interesting question and it shows why it is not just mathematics that is required to answer probability problems. For anyone to answer they need to know exactly what you mean. Do you want the question answered from the point of view (state of knowledge) of the player. If so, are we to assume that she has no knowledge of where the car was placed?

Perhaps you would prefer a more modern approach. In that case I would want to know the initial distribution of the car behind the doors and the distribution of the players door choice.

The point is that without stating exactly what it is that you want to know your question cannot be answered. Martin Hogbin (talk) 15:45, 8 December 2009 (UTC)

A problem for you in return

There is an urn containing 9 balls numbered 1-9. You have to pick ball 9 to win. We assume that your picks are at random. What is the probability that you will win in one pick? I think we can agree that it is 1/9.

Now suppose that you have two picks and the ball from the first pick is not replaced before your second pick. You do not win on your first pick. What is the probability you will win on your second pick and how would you calculate this? Martin Hogbin (talk) 16:31, 8 December 2009 (UTC)

I assume with pick you mean random pick.

Yes that is why I said, 'We assume that your picks are at random'. This is, of course, generally assumed anyway in an urn problem.

I have to calculate:

P(X_{2}=9|X_{1}\neq 9)

with

X_{1},X_{2}

the outcome of the first and second pick respectively. The calculation may be with the use of the apropriate laws, but you may arrive at the same answer by using a symmetry argument, stating that the second pick is from a conditioned urn, with 8 balls with number 9 amongst them. Note, that you always calculate the conditional probability. Nijdam (talk) 21:01, 8 December 2009 (UTC)

You can obviously see what I am getting at. Nobody would suggest that the correct calculation is to take all the possible conditions after the removal of the first ball (That is to say ball 1 missing, ball 2 missing...ball 8 missing) calculate the probability of getting a 9 separately in each case then combine these probabilities weighted according to the (equal) probability of each condition. We can just observe that, provided it is not a 9, the first ball picked makes no difference to the probability of picking a 9 on the second pick.

So it is with the MHP. We can note that, if the host opens a legal goat door randomly, it makes no difference to the probability of winning by switching which door he opens. In other words the conditional problem can obviously be treated unconditionally. Martin Hogbin (talk) 22:01, 8 December 2009 (UTC)

I think the problem here is the right wording. Of course one may use symmetry arguments, as BTW I did before, but you do this in order to calculate the conditional probability (for the car to be behind the picked door!) in an easy way, as its value is the same as the unconditional. Nevertheless the probability of interest is the conditional one!! Nijdam (talk) 11:09, 9 December 2009 (UTC)

OK, I accept what you say but my suggestion is that we ignore the issue of conditional/unconditional for the first section where we treat the problem in the simplest manner possible, showing why the answer is 2/3 and discussing the reasons why people get it wrong.

After that we could say that 'strictly speaking' the problem is one of conditional probability but for the symmetrical case we can ignore this fact. Finally, in in a separate section, we should discuss the necessarily conditional case, the Morgan scenario, where the player knows the non-random host door opening strategy (but not the initial car placement). 12:06, 9 December 2009 (UTC)

Well Martin, as you may have noticed from our lengthy former discussion, neither Rick nor me insist on mentioning the word conditional, but we somehow want to make clear that after a door has been opened new probabilities are at stake. And that a simple solution, along the lines "the car is in 1/3 of the cases behind the chosen door, also with 2/3 it is behind one of the other doors and hence (???) after opening of one of them with 2/3 behind the remaining closed door", is not complete. Nijdam (talk) 16:09, 9 December 2009 (UTC)

For the symmetrical case the probability that the car is behind the chosen door remains 1/3 after a legal random goat door has been opened. Most people assume this, and there are several-easy-to-understand reasons why it is so. In my opinion for the symmetrical case we can just state this fact in the starting section, to be discussed in more detail later. Martin Hogbin (talk) 17:55, 9 December 2009 (UTC)

As I said about wording, what does it mean if you say: "the probability that the car is behind the chosen door remains 1/3"? There is your and many other's problem. What do you mean with probability? A probability always remains the same! What actually is the case - as we have discussed over and over - the probability changes, without changing value. Nijdam (talk) 11:44, 10 December 2009 (UTC)

Most recent FARC version

Can somebody please explain specifically what they don't like about the version of the article as of the last FARC, i.e. this version? I think the article is actually much worse now than it was then and it might be easier to proceed from this version than the current one. Just a thought. -- Rick Block (talk) 20:40, 9 December 2009 (UTC)

I do not like any of the 'Solution' section including the diagram and the mention of conditional probability. We need a simple solution, followed by some convincing explanations. Martin Hogbin (talk) 21:45, 9 December 2009 (UTC)

Up to "The reasoning above" does not mention conditional probability, and says something very similar to what is in the initial paragraph of the current version (doesn't it?). Do you like the large diagram in the current version? It seems isomorphic to me (it varies the player pick rather than the car location - but both are based on a specific concrete example). Are you thinking there's something wrong with the image in the FARC version? -- Rick Block (talk) 23:06, 9 December 2009 (UTC)

Rick Block states above: "I think the article is actually much worse now than it was then..." This was how the Solution section began as of the last FARC:

Solution

The overall probability of winning by switching is determined by the location of the car.

until it was deleted with this diff: http://en.wikipedia.org/w/index.php?title=Monty_Hall_problem&diff=next&oldid=247025116

So, I think the newer version is a whole lot better.

But this new section is just another filibustering technique by Rick. Glkanter (talk) 12:29, 10 December 2009 (UTC)

The excerpted quote above is a perfect example of the muddlement of this article. I refer to the quote: "The overall probability of winning by switching is determined by the location of the car." Huhh??? What does this mean? Again, let me emphasize, I have no idea who it is that was being quoted, which quote I just repeated. I don't mean to attack anyone in particular. But I have to ask ... First, what is the difference between the "probability" and the "overall probability"? Second, what does the probability have to do with the location of the car? I don't get it. The location of the car has nothing to do with the probability. If the writer has a point to be made, then he/she should try to state the point non-cryptically. Sorry for this rant, but this kind of thing is very frustrating ... End of rant. Shalom. Worldrimroamer (talk) 04:26, 10 December 2009 (UTC)

OK. So far, we have "delete the first sentence". Fine. Let's assume this change, which makes the Solution section look more like this version. And Martin's issue with the image (which he hasn't clarified) and his issue with anything that says "conditional probability". The referenced version has one sentence that says "conditional probability" in the 2nd paragraph below the figure basically identifying what the "subtly different" question is actually called. Possibly we could delete this sentence, although I'm not sure I see the point. Anybody else have specific gripes with this version? -- Rick Block (talk) 05:25, 10 December 2009 (UTC)

Rick, I started discussing this in October, 2008. I have no interest whatsoever in comparing and contrasting to a May, 2008 version. I think it is unreasonable to request that we do so. It has the effect of negating discussions that led to all the edits to the article since the FARC version of May, 2008. Is that your intent? Glkanter (talk) 12:29, 10 December 2009 (UTC)

Well, in-spite of my preference for the popular solution to be the main focus of the article I still feel that the conditional solution should be mentioned in the Variants section. The FARC version linked above seems to have nothing in it about the conditional solution at all. Also I like the current lead much better. Personally I think that simply moving the conditional section to the Variants heading would be the best compromise as that way no information will be lost from the article but the flow will still proceed from the most common understanding of the MHP towards more in-depth analysis. I would be more than willing to make the change myself as an editor who has only been part of this discussion for a short time. How does that sound to you guys? Colincbn (talk) 07:28, 10 December 2009 (UTC)

Colincbn - You apparently mean something different by the "conditional solution" than at least Glkanter - which is why I would really prefer we talk about specific changes. I gather you are referring NOT to the entire "Probabilistic solution" section, but only the paragraph in that section starting with "Morgan et al. (1991) and Gillman (1992) both show a more general solution ..." and not the figure that follows. Is this right?

Sorry about that, I'll try to be more specific but I might get my terminology wrong as I'm no mathematician (I'm a history, religion and biology guy). Basically I figure the conditional problem statements referenced to Morgan would be best explained in a single subsection under the variants section. ie: take the first paragraph of the Probabilistic solution section, that starts with "Morgan et al. (1991) state that many popular solutions are incomplete...", as well as the third paragraph that starts with "Morgan et al. (1991) and Gillman (1992) both show a more general solution ..." (as you said), and put them together in the new Conditional problem subsection. This might require some rewording of course. I might also suggest moving the probability tree image (File:Monty tree door1.svg) to the new subsection along with giving it any other useful visual aids or expanded explanations that are available. If the section gets too big we can add a "main article" link at the top of that subsection and continue expanding in a separate article that deals with the conditional problem extensively. I figure that way we get to keep all the information currently here (which I am in favor of) as well as restructuring the layout to flow from the most common interpretation to more advanced and in-depth analysis including variations and more focus on the Parade version of the question (which I am also in favor of). I think this is a fairly good compromise as no one can claim undue weight or anything like that, but we also get to cover the most prominent variants to the popular solution. Colincbn (talk) 15:52, 10 December 2009 (UTC)

Colincbc, I think your suggestions have a lot of merit, at the appropriate time, as I posted just a few minutes ago. I would like to say however, I don't agree that the additional sections beyond the current 'Popular Solution' section are a "flow from the most common interpretation to more advanced and in-depth analysis". I consider those sections as unsupported criticisms and confusing different problems that might be of interest to some people. And they're not part of the Monty Hall problem paradox itself, but the published literature that only uses the MHP as a starting point. Glkanter (talk) 16:06, 10 December 2009 (UTC)

While I can certainly see your point, my main concern is finding a compromise. I figure there are some things that will always be disagreed upon, and both sides will need to accept some things they don't like, but its the only way to really progress. The way I see it if neither side can have an article they think is 100% perfect both sides getting one they feel is 80% perfect is the best way forward. (Now we just need to figure out where that 80% is...) Colincbn (talk) 16:33, 10 December 2009 (UTC)

Yes, a compromise is appropriate. Based on these discussions, I modified my proposal a couple of days ago. http://en.wikipedia.org/wiki/Talk:Monty_Hall_problem#Glkanter.27s_suggestion Glkanter (talk) 16:41, 10 December 2009 (UTC)

Glkanter - I think comparing to a version that passed a Featured Article review is entirely appropriate. I'm talking about considering replacing just the "Popular solution" and "Probabilistic solution" (not the entire article) from the current version with the "Solution" section from this older version. I think it's clear not many people like the structure of the two current sections under "Solution" (even me). Choosing a different starting point might make progress easier. -- Rick Block (talk) 15:06, 10 December 2009 (UTC)

Well, we'll just have to agree to disagree. Actually, I'd like to see some sort of 'freeze' put on the article until this consensus is allowed to edit the article as proposed. Glkanter (talk) 15:29, 10 December 2009 (UTC)

Colincbn - can you comment directly on the idea of using this version of the Solution section? I think it may well provide a reasonable compromise. -- Rick Block (talk) 18:04, 10 December 2009 (UTC)

Correction: "good starting point for a reasonable compromise". -- Rick Block (talk) 18:20, 10 December 2009 (UTC)

As a starting point definitely. I do like the Cecil Adams explanation and the two smaller images on either side of it in the current Popular solution section as a good "layman's" explanation though. And I imagine that we will need to move the paragraph that starts "A subtly different question is which strategy is...". Possibly replacing that paragraph with something that reiterates that the solution presented is only for the unconditional version and that the Parade version has spawned variants with a "see below" link. Colincbn (talk) 23:37, 10 December 2009 (UTC)

This whole discussion could have reached an amicable end long ago

If the Morgan supporters would please read and respond to the existing section "Is The Contestant Aware?", and it's question:

"Has it been agreed by the editors of this article that regardless of how Monty handles the 'two goats remaining' situation, the contestant has no knowledge of the method?

"It seems to me that this is a (unstated) premise of the problem, as both vos Savant (Whitaker) and Krauss and Wang begin the problem statement with: 'Suppose you're on a game show'. I read this as clearly stating it is only the contestant's point of view we are concerned about. And, being a game show, the host is prohibited from divulging to the contestant either where the car is, or where the car is not.

"Is there agreement on this, or is this in dispute? Glkanter (talk) 11:28, 29 November 2009 (UTC)"

http://en.wikipedia.org/wiki/Talk:Monty_Hall_problem#Is_The_Contestant_Aware.3F —Preceding unsigned comment added by Glkanter (talk • contribs) 16:15, 5 December 2009 (UTC)

Yes this is in dispute. The view(s) that have to be described in the article, are the views taken in publication on the Monty Hall problem and there treatmeants which assume different perspectives. What we probably can be agreed on, that those different perspectives should be handled in a variations or generalization section and that they don't belong into the article lead. Also it is probably not appropriate to assume (specific und possibly) legal gameshow regulations. Since most of our readers, much of the original audience, potentially even whitaker himself and some of the people that have published on it, are possibly not aware of specific regulation and legal restrictions and definitely do not mention them. This means while the article can and should discuss how actual regulations (or the real monty's behaviour) might influence the analysis of the problem, but that definitely doesn't belong in the lead section nor can any such regulation simply be considered as an obvious "given" for analysing the problem.--Kmhkmh (talk) 17:01, 5 December 2009 (UTC)

But it is indisputable. I don't care whether the problem starts with "Suppose you are on a game show," or what you think that implies. In every version of the MHP I have ever seen, the question is "Should the contestant switch?" It is not "What is the probability of winning the car if the contestant switches?" Or "Describe a parametric formula that allows the contestant, once given the proper data, to decide whether or not to switch." It is a yes/no question in the purest sense: only two answers address the question. "Yes, she should switch" and "No, she should not switch." If you are required to make assumptions to reduce your answer completely to one of those, there are acceptable means to do so. The point about the contestant being aware is indistutable because we have to make whatever assumptions are necessary to reduce the answer to "'yes' in all possibilities considered" or "'no' in all possibilities considered," which means that we have to apply the same knowledge we assume the contestant will use. When they address the actual MHP question, the sources that address the conditional problem make such assumptions; just not that specific assumption about how the host chooses between two goats. They do make the neutral contestant-knowledge assumptions for what I call "game protocol" (e.g., always revealing a goat) and placement bias.

Morgan's point is NEVER that the answer to the MHP is 1/(1+q), or whatever. It is that one specific assumption is not really necessary when the problem is taken literally in conditional form (which itself is disputable: the sources that claim it is conditioanl misquote their sources in such a way that makes it conditional). You don't need to make the assumption that reduces 1/(1+q) to 2/3, since 1/(1+q) is always greater than 1/2. But that conditional form loses the real beauty of the MHP, that the seemingly paradoxical answer is correct. Far more sources discuss this aspect of the problem than address the "conditional solution," which is where some people's POV is interfering with their ability to approach the article objectively. Yes, the conditional problem is addressed by some sources. It is a minority, and neutrality means it should be treated as a variant. JeffJor (talk) 13:31, 7 December 2009 (UTC)

Words, .... Yes, good question, should she switch or not? One question: How does the player reach at a decision? Nijdam (talk) 13:44, 8 December 2009 (UTC)

Where is your reaction, Jeffjor?? Nijdam (talk) 09:22, 12 December 2009 (UTC)

I imagine Selvin explained his version of the game show this way: The contestant makes a random choice of doors. The probability she did NOT choose the car is 2/3. Monty reveals a goat, giving her no new information about either remaining door. She is indifferent to which door Monty opened. Being a sentient being, she says to herself, my selection's 2/3 chance of not being the car has not changed, therefore, on average, the remaining door has a 2/3 chance of being the car. I don't think I can do better than that. I'll take the switch! Glkanter (talk) 13:59, 8 December 2009 (UTC)

These are the only 2 known English language Game Show situations (Whammy and the 1950s Scandal) in which the particular contestant had more 'information' than an 'average' contestant would have. Both situations were considered highly unexpected aberrations, and many steps were taken to prevent either from being repeated.

So, yes, there is a commonly understood expectation of all game show watchers that each contestant has no information otherwise not available to other contestants, certainly not coming from the host (or the shows producers). And there's tons of laws and lawyers watching for this in the US.

http://en.wikipedia.org/wiki/Whammy_%28Press_Your_Luck%29#Michael_Larson

http://en.wikipedia.org/wiki/Quiz_show_scandals

This is not ambiguous: 'Suppose you're on a game show' Glkanter (talk) 17:33, 5 December 2009 (UTC)

You miss the point you are making assumptions based on your knowledge and perception of game shows. There is no evidence whatsoever that most of the readers have the same legal background as you have nor do most publications on subject deal with legal restrictions or "gameshow regulations". Basically this boils down to stick to the sources and no speculations by WP-Editors.--Kmhkmh (talk) 17:52, 5 December 2009 (UTC)

Please see my response above. -- Rick Block (talk) 17:50, 5 December 2009 (UTC)

It is so pervasive, nobody bothered to mention it explicitly as an assumption. Everyone in the US understands this rule of game shows. The idea that one of the contestants on the screen has 'inside' knowledge? Folly. It's in the fine print that runs at the end of each and every episode. And I documented the only known times it improperly happened. This is not my opinion or interpretation. It's a defining characteristic of a game show, without which, it is no longer a game show. Your twisted interpretation more resembles a street hustler with a card table and a deck of cards. Or three shells and a pea, perhaps. Glkanter (talk) 18:11, 5 December 2009 (UTC)

You make a bold claim (It is so pervasive, nobody bothered to mention it explicitly as an assumption.) without providing any evidence and moreover you miss the point again. The question was not posed to actual game show contestants (being most likely aware of all regulations) but to readers of weekly column and later through various (international) publications to a broader audience. Even if your most likely false claim was true for all readers of Marylin's column, it is certainly not true for the international audiences reading the various publications. I've sampled quite a lot of material on the subject in English and in German and on top of my head i cannot recall any of them mentioning legal restrictions and national game show regulations. In particular the German articles on the subject with a readership definitely being unaware of any such regulations do not mention any legal conditions whatsoever. The 2 cases you've posted above are not in question, however they do not prove your claim. What's in question here is your (false) line of reasoning.--Kmhkmh (talk) 18:35, 5 December 2009 (UTC)

The MHP is really, as this article confirms, a probability puzzle. In such puzzles it is usual to make certain assumptions, which in the MHP would be that the cars are initially placed randomly, the player chooses randomly, and that the host chooses a legal goat door randomly. The assumptions are precisely those Glkanter claims above.

If in the other hand, you want to claim that the puzzle is based on the question by Whitaker then I suggest that you first read the real Whitaker statement rather than the inacurate version misquoted by Morgan. In Whitaker's original question it is clear that the door number given for that initially chosen by the player and the door number opened by the host are intended to be examples rather than specified doors. I might add that, as no information is given in the question, we should still take all the human choices to be random by the principle of indifference. Martin Hogbin (talk) 11:27, 12 December 2009 (UTC)

Who decides?

Consider 3 so-called 'variants':

Instead of goats, behind 2 doors are cows.

Instead of 1 car & 2 goats, it's 2 cars and 1 goat.

The host may have a 'bias', or 'behaviour' or 'method' to determine which goat to reveal when the contestant selects the car. This may or may not be somehow made aware to the contestant.

Who decides when a 'variant' is no longer the Monty Hall problem? Glkanter (talk) 12:39, 10 December 2009 (UTC)

Reliable sources decide. -- Rick Block (talk) 18:05, 10 December 2009 (UTC)

Yes, for Wikipedia purposes. How do academics or other professionals do it, and if you don't mind, how would you decide in each example, just for our own discussion purposes? Glkanter (talk) 20:10, 10 December 2009 (UTC)

In mathematics two problems are considered the same if they are isomorphic (see isomorphism). For example, the MHP and the Three Prisoners problem are isomorphic so even though they have different names and somewhat different descriptions they are actually the same problem. If you're asking for opinions you're on the wrong page (see /Arguments).

The isomorphism with the Three Prisoners problem is an interesting point that has already been raised. The Morgan style argument is never heard concerning the TPP. It is never suggested that the warden may have had a preference for one of the prisoners. This is if course because Gardner specified that he chose randomly if he had a choice. He obviously did this specifically to avoid the Morgan style argument and to make the problem simple. That is exactly what we should do with the MHP, especially as the Morgan scenario is now shown to be just a contrafactual conjecture. Martin Hogbin (talk) 10:14, 11 December 2009 (UTC)

No, what we should do is adhere to FUNDAMENTAL Wikipedia policy and neutrally represent what reliable sources say. To be clear, the specific point I"m talking about is whether a conditional probability analysis (using the assumption of equal probability of host choice between two goats) needs to be presented - more or less like it was in this version of the Solution section - not the generalization where the host preference is left as a variable. -- Rick Block (talk) 17:47, 11 December 2009 (UTC)

That is absolutely fine and it is exactly what I want to do. We have a number of sources who treat the problem in a simple manner (Selvin, vos Savant, Devlin etc) and we have a number of sources that treat it in a more complex manner (Morgan etc). It is therefore quite reasonable, and advantageous for most readers, to start with a section that treats the problem and its solution and 'Sources of confusion' and 'Aids to understanding' in a simple manner. We can then treat the problem in a more complex manner for those interested in such things. Note that a simple treatment of a subject followed by a more detailed one cannot be described as a content fork, it is standard practice. The problem is that, when this has been suggested, you tell us that Morgan say that the sources that treat the problem simply are wrong or inapplicable, that they present 'false solutions'. This is an obviously pro-Morgan POV that should not be permitted.

The fact is that we do have reliable sources that do treat the problem simply (do not distinguish between goat doors) and there is no reason not to represent these sources neutrally (not as false or incomplete or flawed) in the article. Martin Hogbin (talk) 18:33, 11 December 2009 (UTC)

I do not like the diagram you refer to because it shows the choice of two goats as separate pictures, horribly complicating the problem for a beginner. We desperately need to keep it simple; the problem is plenty complicated enough for most people. If we do not do this we fail in our fundamental purpose of informing our readers.Martin Hogbin (talk) 18:33, 11 December 2009 (UTC)

And I'm being somewhat non-responsive since I think one of the problems we've had on this page is too much focus on opinions and not enough focus on what reliable sources say. The three fundamental Wikipedia policies concerning content are WP:V, WP:OR, and WP:NPOV. If anyone hasn't read any of these they really should (how about right now?). What WP:V means is that everything an article says has to be sourced to a WP:reliable source. What WP:OR means is basically the converse of WP:V, i.e. you're not allowed to add content based on your own opinion. In the extreme, this is the case even if you know with absolute certainty that what you're saying is true but you can't find a reliable source to back you up. WP:NPOV means basically two things. First, that Wikipedia MUST fairly (neutrally) represent what reliable sources have to say. Second, if reliable sources disagree that Wikipedia articles are not allowed to take sides. In the aggregate, these policies mean that even though every editor has opinions and even though a collection of editors might have a collective opinion about content issues Wikipedia doesn't give a damn about these opinions. -- Rick Block (talk) 03:29, 11 December 2009 (UTC)

Well, for the 14 months I've been around, I haven't 'known' you to be shy. So, I'm afraid I think there is an additional reason you decline to share your opinion. It's just a talk page, not an article. Glkanter (talk) 03:51, 11 December 2009 (UTC)

Let's Say Some Huckleberry Played Repeatedly

Let's say some huckleberry played repeatedly. They play for $1 per play, rather than the car.

Huckleberry has figured things out using the Combining Doors solution. He doesn't understand, in fact, he is not even aware of the term 'conditional' as it applies in this instance, as he is less educated than others.

So, with just the brains and common sense he was born with, he wins 2/3 of the time. His method worked fine. Perfectly, really. Nobody could give Huckleberry ANY ADDITIONAL INFORMATION OF ANY CALIBER WHICH CAN PRODUCE A BETTER RESULT. And if this were repeated endlessly, it would always work out to 2/3.

There are other solution techniques which may produce the SAME result, but Huckleberry doesn't understand that the 'equal goat door constrain must = 1/2'.

It was an elegantly simple Paradox for Selvin, and vos Savant, and Huckleberry. 1/3 = 1/3. And always did. Glkanter (talk) 18:03, 11 December 2009 (UTC)

Parallel Universe Experience: Same random distribution of cars. Huckleberry makes all the same door choices. Monty picks the exact opposite equal goat door as he did above. Huckleberry always switches. There is NO CHANGE TO THE 2/3 OUTCOME, EVEN IF YOU PLAY FOREVER. And the individual play outcomes are identical. Huckleberry wins or loses the same exact instances regardless of which goat door Monty opens. Glkanter (talk) 14:50, 12 December 2009 (UTC)

I Guarantee It. Old Paradoxes Are Like New Technology.

“New technology goes through three stages:

First, it is ridiculed by those ignorant of its potential.

Next, it is subverted by those threatened by its potential.

Finally, it is considered self-evident.” –unknown —Preceding unsigned comment added by Glkanter (talk • contribs) 04:39, 12 December 2009 (UTC)

Looks like I started this section just in the nick of time: Moved conditional analysis involving host preference q to variant section.

There's an old saying: "If you see a parade, get in front of it." I think Rick must have heard this old saying, too. Maybe Rick will argue for the proposed changes when we get to mediation as well? Glkanter (talk) 05:23, 13 December 2009 (UTC)

Words, words, words

Because in words one often doesn't distinquish between probability and conditional probability, a lot of the misunderstanding arises. As I tried before, I ask anyone of the discussiant to formulate the problem and the solution in rigorous mathematical notation. I'll do the kick-off:

Three doors with numbers 1,2 and 3

C is the random variable indicating the number of the door with the car

X is the random variable indicating the number of the door chosen by the player

C and X are independent

H is the random variable indicating the number of the door opened by the host

As C and X are independent we may without loss of generality consider the case "X=1" and without explicit mentioning condition on this event. Probabilities are:

\!\,P(C=1)=P(C=2)=P(C=3)={\tfrac {1}{3}}

\!\,P(H=1)=0

\!\,P(H=2|C=1)=P(H=3|C=1)={\tfrac {1}{2}}

\!\,P(H=2|C=2)=P(H=3|C=3)=0

\!\,P(H=3|C=2)=P(H=2|C=3)=1

Did I leave something out? Is something wrong? Please comment! In my opinion the whole point of the MHP is the calculation of P(C=2|H=3). I'm looking forward to other opinions. Nijdam (talk) 12:40, 5 December 2009 (UTC)

No, your calculation is fine and as I have made clear before I accept that the MHP problem (as defined by the K & W statement in which specific doors are mentioned) is a problem of conditional probability. The opening of a door by the host reduces the sample set (clearly the player can no longer change to that door the host has opened) thus the sample set has been conditioned and the problem is one of conditional probability.

But, even with this problem formulation we can ask the question as to whether we need to use conditional probability to answer the question. We can immediately note that there is an obvious symmetry in the question. Namely that:

\!\,P(H=2|C=1)=P(H=3|C=1)

and therefore that: P(C=2|H=3)=P(C=3|H=2) Thus, as Boris has ,said the conditional solution, whichever door the host chooses, must equal the unconditional solution. There is nothing wrong with this approach, many intractable mathematical problems have been solved by noticing some symmetry in the problem.

>>>Even so, as you say, we have to calculate the conditional probability. Nijdam (talk) 15:57, 5 December 2009 (UTC)

Alternatively we might say that, because the hosts choice is random (when he has a choice) it gives no information about the initial placement of the car and thus the original

\!\,P(C=1)=P(C=2)=P(C=3)={\tfrac {1}{3}}

must hold good after the host has opened a door.

>>>??It doesn't: $P(C=3|H=3)=0\neq P(C=3)={\tfrac {1}{3}}$

You are right, of course. I should have just said that P(C=1)=1/3; Martin Hogbin (talk) 16:39, 5 December 2009 (UTC)

Indeed is P(C=1)=1/3, it is one of the assumptions, but it doesn't bring us much nearer to the solution.Nijdam (talk) 10:09, 6 December 2009 (UTC)

So to sum up, with the K & W formulation, the problem may be strictly conditional but it is clear that a simple unconditional solution will give the correct answer.

However there is more to the issue than this. The MHP is essentially a mathematical paradox and thus it is logical to formulate the problem so that the solution is as simple as possible. This is what was done before the Morgan paper was published. The door numbers were considered irrelevant. I have answered your question, now perhaps you will answer mine. Do you really believe that what Whitaker wanted to know was what the probability of winning was given only that specific doors had been chosen and opened? I would very much like to hear your answer to that question.

>>>Whitaker definitely wanted to know what for a chosen door and an opened one the (conditional) probability was for the car to be behind the remaining unopened door. Nijdam (talk) 15:57, 5 December 2009 (UTC)

I would say that the question that he actually wanted the answer to is: 'Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, and the host, who knows what's behind the doors, opens another door, which has a goat. He then says to you, "Do you want to pick the remaining door?" Is it generally to your advantage to switch your choice?'

He obviously wants to know what is the best strategy in playing the game: to switch or not. This a simple question with a simple answer that most people get wrong.

>>>Always given the situation the player is in. Nijdam (talk) 15:57, 5 December 2009 (UTC)

I can only repeat, yet again, the words of the wise Prof Seymann (with my emphasis) in the hopes that you will heed them: ' Without a clear understanding of the precise intent of the questioner, there can be no single correct solution to any problem. Thus, with respect to the three door problem, the answer is dependent on the assumptions one makes about the intent of the one who initially posed the question '. In my opinion Morgan et al. have completely misrepresented the original point of the question, resulting in all this conditional nonsense which only serves to obfuscate an interesting problem. Martin Hogbin (talk) 14:02, 5 December 2009 (UTC)

>>>There you have a point. The original problem may be interpreted in different ways. Also in the way of Morgan et al. BTW. But my main concern is that even in the precise definition (like the K&W formulation) many people, seemingly also mathematicians, probabalists and statisticians, give the wrong analysis. Nijdam (talk) 15:57, 5 December 2009 (UTC)

I don't completely disagree with your description. However making claims about Whittaker's intentions is clearly WP:OR. Either his intentions are published somewhere, then they are known and we can incorporate them or they are not. In the latter case personal speculations of WP editors do not belong in the article. As far possible or real misinterpretation are concerned, WP has to describe or summarize how Whitaker was interpreted in various publication - period.--Kmhkmh (talk) 14:12, 5 December 2009 (UTC)

I understand your point about sources; that is why I have not edited the article but continued to argue here. On the other hand may editors feel passionately that the current format of the article is wrong. There are many sources on this problem and it is, to some degree, up to us which ones we use and how we use them. See my suggestion below on mediation. Martin Hogbin (talk) 14:33, 5 December 2009 (UTC)

I agree completely with the analysis done above by Nijdam and Martin Hogbin. Of course, the symmetry assumption should be formulated more accurately in the "official" text; not only C and X are independent, but also C is uniform on the 3-point set, and H is conditionally uniform on the rest (be it 1-point or 2-point set). Boris Tsirelson (talk) 15:04, 5 December 2009 (UTC)

Actually Martin's description leaves me a bit confused since in my perception it is not quite in line with what he was pushing for earlier. However if we all essentially agree now, I don't quite get what the actual dissent is now. Are we just arguing about the order of various sections but agree about the general content (see also the outline under my comment)? I can live with more or less any order in the article as long as the (sourced) content stays (in whatever section). Or to be specific regarding the much debated Morgan, his approach (or that in similar publications) hast to be mentioned and his caveat to the vos savant's approach/simple solution as well, in which section of the article this happens however doesn't really matter.--Kmhkmh (talk) 15:22, 5 December 2009 (UTC)

As for me, I only say that the symmetric case is more important for an encyclopedia than the general case. (Likewise, a circle is more important for an encyclopedia than an arbitrary curve.) Boris Tsirelson (talk) 15:32, 5 December 2009 (UTC)

Why not formulate it in the above terminology? Nijdam (talk) 15:57, 5 December 2009 (UTC)

Indeed; but I'm afraid, Rick Block will say that a featured article must be perfectly sourced... Boris Tsirelson (talk) 16:15, 5 December 2009 (UTC)

I meant, please respond to my call. Formulate here your ideas about the symmetric case and its solution in the above terminology!! Nijdam (talk) 16:21, 5 December 2009 (UTC)

OK, wait a bit. Boris Tsirelson (talk) 16:25, 5 December 2009 (UTC)

Not just words

Notation:

Three doors with numbers 1,2 and 3
C is the random variable indicating the number of the door with the car
X is the random variable indicating the number of the door chosen by the player
H is the random variable indicating the number of the door opened by the host

Assumptions:

C is uniform on the set {1,2,3}
C and X are independent
H is conditionally uniform on the complement of the set {C,X} in the set {1,2,3} (be the complement a 1-point or 2-point set)

The "unconditional" solution:

A pure strategy of the player is a function that maps every possible pair (X,H) to either X or the "third" element of {1,2,3} (different from X and H).

A mixed strategy of the player is a function that maps every possible pair (X,H) to a probability measure on {1,2,3} that vanishes on {H}.

The winning probability is a function of a strategy. It is invariant under the rearrangement group of {1,2,3} (by the symmetry, of course). Therefore it is sufficient to consider only invariant mixed strategies. Such a strategy is nothing but a probability of switching. It is sufficient to consider only extremal values (0 and 1) of the probability. Thus, the question boils down to: to switch or not to switch. The simple unconditional calculation completes the analysis.

The "semi-conditional" solution:

The conditional winning probability without switching, P(X=C|X,H), is a function of X and H. It is invariant under the rearrangement group of {1,2,3} (by the symmetry, of course). Therefore it is the constant function. Taking into account that its expectation is equal to the unconditional probability (the total probability formula) we see that P(X=C|X,H) = P(X=C); that is, the conditional probability is equal to the unconditional probability. The simple unconditional calculation (the same as above) completes the analysis.

(You see, the conditional probability is used; but is reduced to the unconditional probability.)

There is also the "all-conditional" solution, not mentioning the unconditional probability; but probably I do not need to write it here.

What do you think about it? Boris Tsirelson (talk) 17:08, 5 December 2009 (UTC)

Maybe some words here sound a frightening math for some people; but they are just a formalization of the symmetry argument intuitively clear to most people. So much clear that they probably do not feel any need in formalization... Boris Tsirelson (talk) 17:27, 5 December 2009 (UTC)

As I said; words, words, words and more words. Give me the correct formulation of this decision theoretical approach in formulas. BTW: I hope you don't tell this story to your first grade students in your introductory course. What I am actually interested in is what you tell in this course and then in the above formalism, with no more words than needed. Think you can manage? Nijdam (talk) 10:03, 6 December 2009 (UTC)

Sorry, now I do not understand what do you mean by "words". Every mathematical paper consists of words (and formulas inside). The formulation above is correct. Yes, this is not what I teach. The reason is that I do not teach MHP; I teach probability, especially conditioning, and at that moment I use MHP as an instructive example. The goal of Wikipedia article is probably quite different. Boris Tsirelson (talk) 13:26, 6 December 2009 (UTC)

I am not used to explain math in a sarcastic environment. If someone will ask me a reasonably specific question in a reasonably polite manner, I'll be reasonably helpful.

I claim that the above texts are correct proofs (well, somewhat sketchy). I am not an anon; I am a professional mathematician.

If you want to say that this is anyway too complicated for the article, just say so. In fact, I never proposed these solutions for the article. And I did not object against conditional probability. I only object against the general case treated as no less important (for Wikipedia, not for science) than the symmetric case. Boris Tsirelson (talk) 14:48, 6 December 2009 (UTC)

I think the argument for this extensive treatment of the general case is, that is somewhat reflects the academic publications/treatment of that topic. I have hardly any objection against Martin's reaction to Nijdam's comment. Featuring the special case or unconditional solution and its benefits prominently in the first part of the article is perfectly fine. However if you review some of Martin's other or earlier comments and even more so some comments of JeffJor or Glkanter, then you can see they are pushing for things which partially factually false or unsourced and definitely not in line with bulk of reputable literature on the subject. Among these are glkanter "legal arguments", JeffJor's attempt to define the conditional approach as a "non MHP"-problem. There also seems to be an attempt to remove reputable sources from the article (via badmouthing) and ignoring that other sources more or less state same conclusions anyway. In particular there is one thing the WP-article cannot do, that is treating the MHP as a unconditional problem only, while the bulk of the literature treats it at least as a conditional problem as well. Another thing is, that the WP article cannot redefine the MHP to remove its ambiguity, since the original problem is ambiguous and much of the literature explicitly deals with the ambiguity. WP needs to report the definitions of others and not set its own.--Kmhkmh (talk) 15:38, 6 December 2009 (UTC)

I agree. Boris Tsirelson (talk) 16:02, 6 December 2009 (UTC)

I'm confused. If you agree then why are you listed here as supporting changing the article (to attempt to redefine the MHP to remove its ambiguity and to remove, badmouth, and ignore reputable sources). I suspect we don't have a common understanding of what changes are suggested. -- Rick Block (talk) 17:19, 6 December 2009 (UTC)

Here is my position (once again): I only say that the symmetric case is more important for an encyclopedia than the general case. (Likewise, a circle is more important for an encyclopedia than an arbitrary curve.) Boris Tsirelson (talk) 18:43, 6 December 2009 (UTC)

Then, I would suggest you change where your name appears in Martin's "summary" of opinions. I believe what Martin, and Glkanter, and JeffJor are actually arguing for is to banish any mention of conditional probability to a "variant" section. And they think you agree with this. -- Rick Block (talk) 20:20, 6 December 2009 (UTC)

But I do support some change. First, prepare a list with a different formulation. Boris Tsirelson (talk) 20:42, 6 December 2009 (UTC)

Rick, you are overstating my request. I do not want to banish any mention of conditional probability to a variant section but I do want to state that it is only really important if the host chooses non-randomly - the Morgan scenario. Martin Hogbin (talk) 21:55, 6 December 2009 (UTC)

Formulas, not words

Here is the "semi-conditional" solution rewritten for these that hate words and like formulas:

P(X=1,H=2) + P(X=1,H=3) + P(X=2,H=1) + P(X=2,H=3) + P(X=3,H=1) + P(X=3,H=2) = 1;

therefore

That is, the conditional probability is equal to the unconditional probability (in the symmetric case, of course). Boris Tsirelson (talk) 20:21, 7 December 2009 (UTC)

Well, I don't hate words, but I do hate them when they hide an underlying problem. And, sorry, I do not understand what "semi-conditional" means. I gave before extensive solutions using Bayes and symmetry arguments. In both ways it is necessary to calculate a conditional probability. I would very much like anyone who claims there is a solution without such a calculation, demonstrate it, in proper terminology. Till now, no-one tried. Yes, in words, words, words. That's why. BTW: The above analysis, without words, also calculates a conditional probability. Nijdam (talk) 13:39, 8 December 2009 (UTC)

"also calculates a conditional probability"? Ultimately, yes, of course. However, this way the fact that it equals to the unconditional probability becomes clear BEFORE we get the values of this or that probability. Not "numeric coincidence" but a logical necessity. Does it matter for you? Boris Tsirelson (talk) 16:53, 8 December 2009 (UTC)

As another example, imagine you have to calculate the perimeter of an ellipse with major axis j and minor axis n. You are told that j = n. Is it better to do the calculation for an ellipse and then set j = n or should you observe that the ellipse is, in fact, a circle and do the calculation on that basis? Martin Hogbin (talk) 12:20, 13 December 2009 (UTC)

Martin, this may serve as an example of the parameter q=1/2 in Morgan's general treatment, making an approach with a symmetry argument possible. But it doesn't circumvent conditioning. Nijdam (talk) 23:01, 14 December 2009 (UTC)

Theorem ( Boris Tsirelson, Dec 2009, published in the first time, all rights reserved :-) ) Let an even E and a random variable X be such that the conditional probability P(E|X) is a constant function of X. Then the conditional probability is equal to the unconditional probability: P(E|X) = P(E).

Eureka! Boris Tsirelson (talk) 17:38, 8 December 2009 (UTC)

What's your point? Nijdam (talk) 21:05, 8 December 2009 (UTC)

You puzzle me. What can I say more? Everything is said. What kind of point do you expect? Well, I repeat, but surely you will not be satisfied (and I do not understand why).

First, I find the symmetric case more interesting. Second, I am glad to see that in this case it is rather easy to see (before specific calculations, just from simple general arguments) that the conditional probability (you know, of which event) is equal to the unconditional probability. I believe that this is a (or rather, the) mathematical formalization of a fact understood by many non-mathematicians by intuition and/or arguments about "no other useful information available" and all that. Boris Tsirelson (talk) 21:27, 8 December 2009 (UTC)

I summarize my position in two points:

1. The symmetric case is more important for an encyclopedia than the general case. (Likewise, a circle is more important for an encyclopedia than an arbitrary curve.)

2. The coexistence of the conditional and the unconditional can be more peaceful. (Not just "numeric coincidence" in the symmetric case.)

Boris Tsirelson (talk) 06:38, 9 December 2009 (UTC)

I agree with you, and indeed is the conditional probability of the considered event not just coincidentally equal to the unconditional. But in my opinion it is necessary to mention the conditional probability as the probability of interest. That's why I asked you to formalize your reasoning. Nijdam (talk) 11:17, 9 December 2009 (UTC)

For a better understanding of my position, I may add, that a lot of people, amonst them students, teachers etc., reason as follows: the car is with probability 1/3 behind the chosen door. Hence the probability to find it behind one of the two remaining doors is 2/3. After the opening of one of them this (!) probability now applies to the remaining unopened door. I consider this as insufficient. One needs to add here that the opening of one of the door "does not influence" the probability of the car being behind the chosen door. And what is the meaning of "does not influence"? Well that the conditional probability and the unconditional one are the same. Agree? Nijdam (talk) 11:26, 9 December 2009 (UTC)

Agree! Boris Tsirelson (talk) 16:08, 9 December 2009 (UTC)

Well that's the only point Rick and I and later Kmhkmh defend. We do not insist on mentioning the word conditional, but we want to make clear that the above reasoning is not complete (or addresses another problem). Nijdam (talk) 16:19, 9 December 2009 (UTC)

I did not object against conditional probability.

"After the opening of one of them this (!) probability now applies to the remaining unopened door." — For me this is the central (!) point of MHP. The only (!) point making it instructive to my students (in my opinion). A quote from #Boris Tsirelson's comments: "I spend 20-30 min on the Monty Hall paradox. I compare two cases: (a) the given case: the host knows what's behind the doors, and (b) the alternative case: he does not know, and it is his good luck that he opens a door which has a goat." I do so exactly for explaining why in (a) the 2/3 jump to the remaining door, while in (b) it scatters to both doors. Boris Tsirelson (talk) 16:28, 9 December 2009 (UTC)

What makes things cumbersome? First of all, the lack of symmetry. And second, – to a much less extent, – conditioning. Boris Tsirelson (talk) 16:34, 9 December 2009 (UTC)

I don't follow you any more. To put things straight: the probability 2/3 of the event that the car is behind one of the not chosen doors, differs (in type, not in value) from the conditional probability the car is behind the remaining unopened door given the other door opened. It concerns different probability measures. That's why I asked for formulas. Nijdam (talk) 11:51, 10 December 2009 (UTC)

To make it better understandable: a correct formulation of the simple solution, would be: After the opening of door 3 by the host the (new) probability for the car to be behind the chosen door is due to the symmetry also 1/3. Nijdam (talk) 12:01, 10 December 2009 (UTC)

Nijdam, are the 2 paragraphs above consistent with Morgan's paper? Or is this OR? Glkanter (talk) 12:22, 10 December 2009 (UTC)

Nijdam, I completely agree with your words above. Why do you feel that you don't follow me any more? Boris Tsirelson (talk) 14:45, 10 December 2009 (UTC)

Where you write: >>"After the opening of one of them this (!) probability now applies to the remaining unopened door." — For me this is the central (!) point of MHP. The only (!) point making it instructive to my students (in my opinion). << Because this is not in line with my explanation above. It is not this probability, but a different probability with the same value as the former. Nijdam (talk) 22:05, 10 December 2009 (UTC)

(unindent) Ah, now I see; I was missing your point. Yes, of course. But on the other hand: in many cases people feel intuitively that some conditioning does not change some probability; and they are right often (probably, not always). There is a big difference between the approach of a mathematician (unless he speaks informally) and that of others. And I think, we should not be too insistent. I'd recommend a tone like that: ...thus the probability is 2/3. This probability is called unconditional, since ... In contrast, by the conditional probability one means ... It is worth to ask what is the conditional probability. It seems plausible that it is the same (2/3) since ...; and it is really the case. In general, conditional probability differs from unconditional probability, it is either bigger or smaller, depending on the condition; but on the average it is equal. In particular, in our case the conditional probability does not depend on the condition by symmetry. Thus it cannot be bigger or smaller, but only equal to the unconditional probability. However, without symmetry it is different; see Section "Asymmetric generalizations" below. Boris Tsirelson (talk) 07:11, 11 December 2009 (UTC)

This has been Rick's and my opinion all along. We won't put much emphasize on it. But anyhow it should be mentioned that after opening of the door by the host a new situation has risen in which becuase of the symmetry the probability for the car being behind the chosen door is the same (has the same value) as before. This is not clear to many of the discussiants and some even fight this fact. Like in what is called the "combined door solution". I hope we may find you on "our" side? Nijdam (talk) 09:39, 11 December 2009 (UTC)

It is a bit risky to say "yes" (or "no"), since we sometime misunderstand each other. Another example: let two fair coins be tossed; two independent events A,B of probability 0.5 appear. Now someone could say: P(B|A) = 0.5 = p(B), but this is a numeric coincidence only; P(.|A) is another probability measure different from P(.); never say "this 0.5", say "another copy of the number 0.5", etc. As for me, this is much too formal; definitely inappropriate when talking to non-mathematicians. It seems to me, you told this way. But maybe I again misinterpret your position. If so, then the better. Boris Tsirelson (talk) 11:39, 11 December 2009 (UTC)

Nor Rick, me or Kmkmh have the intention to be formal (except of course in our minds). What we want is to be sure the distinction between the probability before the opening of a door by the host and the probability thereafter is somehow mentioned. I.e. we want to make clear or at least mention they are not the same. Why? Well, because some people, as I wrote before, reason as follows, or in a similar way: the probability of the car behind the chosen door 1 is 1/3. After opening of door 3 the probability for the car to be there is 0, hence (???) the probability for the other unopened door is 2/3. This "simple" explanation (solution?) is wrong. It turns up in many forms, all omitting the difference between the unconditional and conditional probabilities. This way of reasoning is copied by teachers, pupils, students and alas also by some mathematicians. It is the reasoning in the "combined door solution". It also turns up in many of the simulations on the internet. I suggested a very modest phrasing month ago: After the player has chosen a door, the probability it hides the car is 1/3. This probability is not influenced by the opening of a door with a goat by Monty, hence after Monty has opened a door with a goat, the probability the original chosen door hides the car is also 1/3. Because clearly the open door does not show the car, the remaining closed door must hide the car with probability 2/3. Hence switching increases the probability of winning the car from 1/3 to 2/3, which is formally correct. In a next, more formal section we could go into some more detail about the probability not being influenced. Of course this all is not Wikipedia:OR. It is found in many texts about the MHP, amongst them Morgan et al's. Nijdam (talk) 13:52, 11 December 2009 (UTC)

Please pardon the intrusion. Nijdam, you wrote this above: "I consider this as insufficient." It seems to me that your opinion contrasts with many reliably published sources. I think Wikipedia's policy is pretty clear on this, and I imagine Rick Block would agree. Are you suggesting, insisting actually, that the article be edited to your opinion, rather than the reliably published sources? And although you did not ask me, I think your argument in the above paragraph is a rather weak insistence on the necessity of the so-called 'conditional' problem statement. How can a 'conditional' problem statement be a Monty Hall problem statement, when it doesn't even follow the 'words' of the actual problem statement from the published sources? Glkanter (talk) 11:52, 9 December 2009 (UTC)

Boris, Nijdam - As Boris has suggested, the concept of a condition is to some degree arbitrary. What is the chance I will get a head if I toss fair coin given that I clap my hands before tossing it. Strictly speaking we could call that a conditional problem but nearly everyone would intuitively say that clapping your hands will not make any difference to the outcome so it can be ignored, and who can disagree? There has to come a time where it is permissible to say that a condition is unimportant and can be ignored. It is intuitive to most people that if a random legal goat door is opened it will not matter which one, this is also a true fact. It seems right to me, therefore, that we can ignore this detail in our initial presentation of the problem and its solution. Martin Hogbin (talk) 12:18, 11 December 2009 (UTC)

(1) I agree with Martin (as usual).

(2) About finding me on this or that side... Asked (by Nijdam) what is my point I have formulated it (in two concise items, you know). If each of us will do so, then it will become clear what are the possible coalitions (and the old "yes-no" list will become obsolete). I would be especially glad to reach a common position of Rick, Nijdam, Martin and myself. Boris Tsirelson (talk) 13:55, 11 December 2009 (UTC)

I am not in favor of just 4 editors hammering this out. It seems inconsistent with my understanding of Wikipedia's policies. Or, is this a commonly used approach to consensus on Wikipedia? Glkanter (talk) 15:57, 11 December 2009 (UTC)

Sorry Martin and Boris (I discoverd you reacted in the mean time). We have discussed these ideas over and over a long time ago. I'd rather see you coming to some insight. I write the conditioning completely out; chosen is door 1, door chosen by the host to be opened in boldface:

door* 1	door 2	door 3	probability
car	goat	goat	1/6
car	goat	goat	1/6
goat	car	goat	1/3
goat	goat	car	1/3

The host opens door 3 showing a goat:

door* 1	door 2	door 3	conditonal probability
car	goat	goat	1/3
goat	car	goat	2/3

I hope you admit there is factually conditioning and not some handclapping or whatever. Nijdam (talk) 14:25, 11 December 2009 (UTC)

There is some conditioning in so far as a door has been opened, I think this is obvious to everyone. But we should allow the case that either door 2 or door 3 is opened to reveal a goat (even though in Whitaker's question he gives door 3 as an example), thus it is not important which door has been opened. Martin Hogbin (talk) 17:40, 11 December 2009 (UTC)

Okay, door 2 might be opened too. Here it comes:

The host opens door 2 showing a goat:

door* 1	door 2	door 3	conditional probability
car	goat	goat	1/3
goat	goat	car	2/3

Again factual conditioning. You permanently "forget" that the player is on stage and sees which door is open. In fact the player might have chosen door 2 or 3 as her initial choice. Any combination of choice and opened door leads to the same analysis. That's why, the given combination serves as an example for the complete problem. Nijdam (talk) 09:20, 12 December 2009 (UTC)

Boris, please comment. Nijdam (talk) 10:16, 13 December 2009 (UTC)

Comment Boris

(I added a section break for easy editing)

At the moment when X is already known (but H and C are not) we can say the following. If C=X then H is distributed uniformly on {1,2,3}-{X}.

>>>Okay, let us start from X given; then P(H=h|C=X)=1/2 for h!=X Nijdam (talk) 22:44, 14 December 2009 (UTC)

>>>>Yes; this is exactly what I wrote. To be uniform on a two-point set means just probabilities 1/2 at each of these two points. What is the problem here? Boris Tsirelson (talk) 05:59, 15 December 2009 (UTC) >>>>>No problem hereNijdam (talk) 14:22, 15 December 2009 (UTC) Otherwise, if C differs from X, then still, H is distributed uniformly on {1,2,3}-{X}. ("Given C != X", not "given C"!

>>>P(H=h|C!=X)=1/2 for h!=X ???? In my opinion: P(H=h|C!=X)=0 for h=C,XNijdam (talk) 22:44, 14 December 2009 (UTC)

>>>>No! Again: "Given C != X", not "given C"! It cannot be 0 for h=C simply because C is not given! This conditional probability does not depend on C. Boris Tsirelson (talk) 05:59, 15 December 2009 (UTC)

>>>>>Okay,I see what you mean.Nijdam (talk) 14:22, 15 December 2009 (UTC)

>>>>Use the force, do formulas not words! Words are sometimes ambiguous; formulas are not. The formula P(H=h|C!=X) means P(H=h|A) where A is the event C!=X. This event includes both values of C. In words: the condition "C != X" does not disclosure the value of C; it gives only a partial knowledge about C.

>>>>To be honest, these are not the true formulas, because one premise remains words: "At the moment when X is already known (but H and C are not)". But it is easy to get rid of the words completely, inserting the condition on X everywhere. That is, we consider P ( H = h | X, A ) where A is the event C != X. Or, if you prefer longer style formulas, P ( H = h | X=x, A ). In any form, it is important that the given condition leaves to C two possible values. Boris Tsirelson (talk) 06:54, 15 December 2009 (UTC)

The (3!) symmetry is already broken, but a (2!) symmetry persists.) That is, the (conditional) distribution of H does not depend on the event A = {C=X}. In other words, A and H are independent.

>>>>Before it was A={C!=X}?Nijdam (talk) 14:22, 15 December 2009 (UTC)

>>>I don't understand Nijdam (talk) 22:44, 14 December 2009 (UTC)

>>>>See above. And think again. Boris Tsirelson (talk) 06:38, 15 December 2009 (UTC)

>>>>It's clear now what you mean. Nijdam (talk) 14:22, 15 December 2009 (UTC)

>>>>You may also think about the following (rather standard) exercise. A fair coin is tossed 10 times. What is the (conditional) probability of "head" in the first trial given that there are exactly 7 "heads" in the total? Boris Tsirelson (talk) 07:10, 15 December 2009 (UTC)

Thus, the distinction between P(A) and P(A|H) is rather academical (like the distinction between (100-1)+25 and 100+(-1+25) in my example elsewhere); I think so. On the other hand, Rick says that somehow this distinction helps to many people to avoid errors. Maybe; this is beyond my expertise. Boris Tsirelson (talk) 12:40, 13 December 2009 (UTC)

>>>I did show in the above tables the factual conditioning. Where am I wrong? Nijdam (talk) 22:44, 14 December 2009 (UTC)

>>>>I never told there is no conditioning. I always agree that P(.|B) is not the same probability measure as P(.). I only say that the conditioning is ineffective on the considered event, due to independence. And the independence is not a numerical coincidence, but a consequence of the symmetry. Boris Tsirelson (talk) 06:43, 15 December 2009 (UTC)

>>>>>Okay, but I doubt whether these are the arguments the (I wrote adversaries, but meant) advocates of the simple solution have in mind or whether this makes it better understandable for the interested readers. I also cannot imagine you use this reasoning in your introductory course. Nijdam (talk) 14:22, 15 December 2009 (UTC)

Well, we now agree on the mathematics (modulo a miserable problem of denoting by A different things on different days of discussion), and I feel my mission finished. About my introductory course: I had no reason to use this argument, but I could, and it would not make more troubles than other topics; however, this is hardly relevant to WP. About adversaries I do not know; I only feel that arguments of symmetry are quite easily guessed by many non-mathematicians (they do not vast effort to the trouble of formalization...). Boris Tsirelson (talk) 15:23, 15 December 2009 (UTC)

[outdent]I do not forget that fact. You seem to forget that we are addressing the formulation in which the host chooses randomly between legal goat doors. There is therefore a strict conditioning, just as there is in my urn problem, but it is quite obvious that this conditioning can be ignored. If the host chooses at random, it cannot matter which door he opens. It is just like my hand clap, is does physically occur but it is obviously irrelevant.

It is relevant. Look at the probability the car is behind the door to be opened by the host. Before opening this equals 1/3, after opening 0. They are different probabilities. The same with de door left closed: before 1/3, after 2/3. How do we calculate these? By looking at the chosen door: before 1/3, after (also) 1/3, but although the same value, different probabilities! Before we have the (unconditional) distribution: 1/3, 1/3, 1/3; after (conditional): 1/3, 2/3, 0. Different distributions. If you're not convinced, please give a proper analysis, in formulas (with words to explain), not just in words. Nijdam (talk) 10:33, 12 December 2009 (UTC)

It is not relevant to calculating the probability of interest, which is that the car is behind door 1. Martin Hogbin (talk) 10:51, 12 December 2009 (UTC)

Please, show me the calculation, in proper terminology. Nijdam (talk) 10:16, 13 December 2009 (UTC)

Terminology and choice of sample set

The above heading is essentially for ease of editing - this section follows on from the thread above.

I do see your point, Nijdam, it is not easy to produce neat diagram (or formula) that proves my point, but I will have a go. To some degree the choice of initial sample set is arbitrary and this choice defines how the calculation proceeds. It is the choice of initial sample set that has to be defended with words. It may be that with your choice, my point is not easy to make. The problem is that other choices (say based on goat doors and car doors) may be difficult to justify. Let me think about it. Martin Hogbin (talk) 11:23, 13 December 2009 (UTC)

But I took the challenge to answer to Nijdam in his own terms, see above. Boris Tsirelson (talk) 13:00, 13 December 2009 (UTC)

What about this? Not the terminology that you choose but it implicitly includes the symmetry of the situation in that the action of the host makes no difference to the answer.

1/3		1/3		1/3
Goat		Goat		Car
The host opens door 3 to reveal a goat
Stick	Swap	Stick	Swap	Stick	Swap
Goat	Car	Goat	Car	Car	Goat

Martin Hogbin (talk) 12:14, 13 December 2009 (UTC)

Which is door 3? Nijdam (talk) 22:46, 14 December 2009 (UTC)

Hey, Nijdam, why don't you quit ignoring my edits and questions? It's pretty rude. If Huckleberry had used your method instead of Devlin's method, what would have been different in the game play that went on infinitely? Would he win a 'different' 2/3 of the time, or the 'same' 2/3 of the time? Wouldn't you agree it made no difference whatsoever to Huckleberry which door Monty opened? Isn't this known as 'indifference' in Mathematics? And, there is no Wikipedia policy stating story problem paradoxes must be solved using formal notation. Wouldn't be such a popular paradox in the real world with that requirement, would it? Glkanter (talk) 13:08, 12 December 2009 (UTC)

There is also a words, not formulas explanation - which I think would be accessible to a layperson - at Talk:Monty Hall problem/FAQ. And, to address a point Glkanter attempts to interject above - this argument is firmly based on wp:reliable sources, not wp:or. Nijdam is reiterating what Morgan et al. and Gillman, and Grinstead and Snell, and most elementary probability textbooks (as claimed by Kmhkmh) say. Glkanter definitely, and Martin to a lesser extent, misinterpret the main point of the Morgan et al. paper. I believe Glkanter genuinely does not understand that there is a difference between asking about the probability of winning by switching (in general, for all players) and asking about the probability of winning given knowledge of which door the host opens, i.e. is having trouble with the basic concepts of conditional probability (see, for example, this edit). I think Martin understands the difference but insists that a complete explanation of the "notable" MHP can be made without mentioning this difference and it should therefore remain unmentioned until (essentially) a variant section. JeffJor understands this difference and insists that the MHP is explicitly asking about the former rather than the latter and therefore any source that says the MHP is asking about the latter is WRONG and should be excluded from THIS article.

You told me not to speculate on your likely actions regarding this consensus. How then is it appropriate for you to speculate on my personal knowledge of the subject matter, given that we have had no communications whatsoever other than on the various Wikipedia pages? Glkanter (talk) 16:44, 11 December 2009 (UTC)

I agree with both of Boris's "position points" above, but would add for the 2nd point that co-existence CANNOT mean excluding one in favor of the other (per FUNDAMENTAL Wikipedia policy, i.e. WP:NPOV). I've been referencing this version of the Solution on other threads as well, which I think is far closer to adhering to both of these points than what Glkanter, Martin, and Jeffjor are suggesting. -- Rick Block (talk) 16:24, 11 December 2009 (UTC)

I guess I do agree with you, Rick, that I do not understand the main point of the Morgan paper, but that is because it is so badly written that it is impossible do divine what the main point is. I can assure you, however, that I do understand the subject and the issues involved. As you will see from my discussions with Nijdam, I accept that, strictly speaking and in some formulations, the problem is conditional, but in the symmetric case (host opens a legal goat door randomly) the condition (which door the host opens) is clearly irrelevant to calculating the probability of interest, thus it can be ignored. I also agree with Jeff, that the MHP can be reasonably interpreted to ask an unconditional question.

Because of the both the facts presented above, I believe that the article should start with a complete section in which the problem is treated unconditionally (as in many reliable sources). After that (or even at the end of that section) I am happy to point out that even if the host chooses randomly the problem might be treated conditionally, with a reference to Morgan. This could then lead on to a variations section which would include the Morgan scenario (we know the door opening policy of the host). Martin Hogbin (talk) 11:09, 12 December 2009 (UTC)

(unindent) back from weekend Here is a proof that 99+25=124:

99+25=100-1+25=124.

Is it a correct, complete proof? Or should I show that I understand the conceptual difference between (100-1)+25 and 100+(-1+25) and use the associative law in order to overcome the difficulty? Boris Tsirelson (talk) 15:33, 12 December 2009 (UTC)

People never write complete proofs. Yes, absolutely never. If in doubt, look at Mizar system; there you can find source texts and programs that allow to generate some complete proofs (but probably you'll never have enough paper in order to print such a proof).

The more so, we should not insist on complete proofs in the encyclopedic context. Symmetry arguments are usually treated by non-mathematicians as too evident for being proved. Boris Tsirelson (talk) 15:40, 12 December 2009 (UTC)

Thus, we'll never be able to convince the majority of our readers that the conditioning is really relevant in the symmetric case. Likewise, we could not convince children that 99+25=124 cannot be accepted if the associative law is not involved.

And still, I think, it is not bad if conditioning will be mentioned in the article, as far as we'll not be insistent about its relevance in the symmetric case. Boris Tsirelson (talk) 15:48, 12 December 2009 (UTC)

Isn't the "host forgets' case also symmetric? If we're using "symmetry" to say the host's action does not affect the player's initial chance of selecting the car, why is it that this same argument does not apply in this case? I know the reason and you know the reason but I'm willing to bet many people who "understand" the popular solutions to the MHP do not understand why these solutions DO NOT apply in this case. See, for example, this exchange on the arguments subpage.

Rick, we've argued this many, many times. The problems are greatly different. Just one example, there are times the forgetful host doesn't offer the switch. Because he revealed the car. Two stated premises of the MHP are that he always reveals a goat, and always offers the switch. I'd hate to see any editor invest time by responding to this oldy, moldy filibuster. Glkanter (talk) 18:36, 12 December 2009 (UTC)

(I've basically said this before - in this section, above) In a fairly recent column, vos Savant addresses the "host forgets" variant. In her analysis of this version she laments [1] [unfortunately now a dead link, and not available on the Internet Archive ] "Back in 1990, everyone was convinced that it didn’t help to switch, whether the host opened a losing door on purpose or not. ... Now everyone is convinced that it always helps to switch, regardless of what the host knows. But this is just as incorrect!" This is absolutely true. And, IMO, it's precisely because the popular sources do NOT address the "classic" MHP using conditional probability. By avoiding addressing the problem in this way, the popular sources have simply replaced one incorrect notion (two unopened doors always means each has equal probability) with another (whatever the host does he can never change the 1/3 probability of the player's initially selected door). As a featured article on Wikipedia, IMO this article must not make the same mistake. -- Rick Block (talk) 18:17, 12 December 2009 (UTC)

Really? Well, if the conditioning helps to not make errors then of course it is useful. Then we only have to explain to the reader why it is important (namely, demonstrate him typical errors) before bothering him by conditioning; then hopefully he will not be disturbed. Boris Tsirelson (talk) 18:58, 12 December 2009 (UTC)

Especially if we deal with the problem in more detail later on in the article for those who are interested. Martin Hogbin (talk) 13:25, 13 December 2009 (UTC)

Informal mediation requested

I have filed a request with the mediation cabal, see Wikipedia:Mediation Cabal/Cases/2009-12-06/Monty Hall problem‎. -- Rick Block (talk) 18:20, 6 December 2009 (UTC)

What is the procedure now? Should we prepare a statement of what we definitely all do agree on? Martin Hogbin (talk) 18:56, 6 December 2009 (UTC)

This mediation is completely informal and there's no guarantee anyone will accept this case (trying this first is generally a requirement before proceeding to formal mediation handled by the Mediation Committee). For now, I think we continue as best we can. -- Rick Block (talk) 19:17, 6 December 2009 (UTC)

No responses so far, unfortunately. How long do we wait, and what comes next? 'Cause you all are just OR-ing and interpreting to beat the band. It's pointless, really. Glkanter (talk) 16:57, 8 December 2009 (UTC)

The informal mediation request has been closed, with a recommendation for formal mediation.

Martin or Glkanter - would one of you like to file the request for formal mediation? See Wikipedia:Requests for mediation/Guide to filing a case. -- Rick Block (talk) 19:36, 17 December 2009 (UTC)