This is an archive of past discussions about Monty Hall problem. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page.
"As Rick confirms we have agreed almost everything and are happy for you to use your discretion and judgement to decide the rest." (Martin Hogbin)
I have kept the following RfC as close as possible to the agreed-upon version. The differences between the above-mentioned RfC and the version I am posting can be seen here.
In the next day or so, I will be publicizing this RfC so as to attract more opinions.
Probability exists on the information level, not on the physical level
According to the Wikipedia Probability page lede paragraph, "Probability is a measure of the expectation that an event will occur or a statement is true" and in that, the word "expecation" is key. It is simply not possible to form expecations without information (whether real or imagined). Accurate expecations are based on accurate information; inaccurate expecations are based on inaccurate information (such as false information, imprecise information, no information, or a guess). Suffice it to say then, expecations based on something other than accurate information, are nothing more than a WAG or as best a Scientific Wild-Ass Guess (SWAG). In the end, all the consternation about this "paradox" is foolish. The Monty Hall problem is no "paradox" at all. Rather, there is typically a communication defecit whereby the player is under the impression that he must guess. And if he does, he leave out of his equation the information derived from the certain fact that the removed door does not have the car. It's a cheap carnival trick - not a true paradox. If you don't believe me, read the Paradox article. 98.118.62.140 (talk) 20:03, 9 September 2012 (UTC)
Take a look at that Paradox article you linked, and search the page for "veridical". This problem is one of those. Even if it weren't, despite what the Wiki article seems to suggest, dictionary definitions of paradox confirm that it is not just about impossible facts, but rather about the perception of something as impossible. OED & Merriam Webster use words like "seems", "seemingly", and "apparently". Monte fits the textbook dictionary definition of "paradox".--Tom Hulse (talk) 07:15, 10 September 2012 (UTC)
Comment from a reader
The crux of the problem is why the extra information given by the host's choice of door does not alter the probabilities or equivalently why the probabilities that the car is behind door i and the player chooses door j are statistically independent. However the article does not emphasise that this is real difficulty and why this is not intuitively obvious. Moreover, once this is understood, it becomes clear why the Bayesian argument is more accurate - it actually explicitly includes information about the Host's choices. The article could easily be improved if the probability space (which only has 24 elements if you distinguish between the goats; it has 12 if you do not) was listed in a table, with the associated probabilities of each event, as this would make it clear what is going on.
Put another way: there are two related problems with the Monty Hall problem being the more complicated version. In the first problem, the Host does not reveal a door but simply says to the contestant "would you like to take whatever is behind the two doors that you did not choose, or would you like to stick with your initial choice of door?" For this problem, vos Savant's table accurately describes the probability space if the contestant chose door 1. However the Monty Hall problem includes information regarding the Host's choice of door. And this affects the probability space. The reason the Monty Hall problem is difficult is because the probability measure on the probability space for the Monty Hall problem is not the naive probability, but is a weighted probability. It is weighted so that the probability that the car is behind door 1 and the contestant chooses door 1 and the host chooses door 2 is equal to 1/6 (and likewise the probability that the car is behind door 1 and the contestant chooses door 1 and the host chooses door 3 is equal to 1/6), but the probability that the car is behind door 2 and the contestant chooses door 1 and the host chooses door 3 is 1/3, etc. So the crux is to understand why the correct probability is the weighted probability. The reason this article suggests (but does not emphasise) is that it is because in the Monty Hall probability space the probability that the car is behind door j and the probability that the contestant chooses door i are statistically independent. However the article offers little detailed explanation of why this should be the case. (Apologies if I have put this comment in the wrong place on the talk page). 121.218.54.187 (talk) 22:38, 10 September 2012 (UTC)
Rules for Monty
Thanks to Tom Hulse for clarifying that I had misunderstood the RfC. That does appear to be true, so I'll open a new section for this point, which is apparently orthogonal to the RfC.
I think it's a serious flaw that the question of the rules constraining the host is not mentioned at the very top. The quoted passage does not give any indication that the host had to do what he did. But if he was not required to — especially, if he was not required to open a door at all — then the analysis changes completely. This point is not mentioned at all until section 4, where it is called a "variant", in spite of being a perfectly reasonable (not "tortured", Tom!) reading of the text as given.
Moreover, this point is of special interest precisely because, at first glance, it does not appear to matter, when in fact it is crucial. In this way it is parallel to the fact that the host knows what is behind the doors, which vos Savant is at pains to tell us, but somehow she neglects this equally central point (maybe, because she didn't notice that it mattered? but of course I'm speculating). --Trovatore (talk) 10:09, 9 September 2012 (UTC)
The "standard interpretation" is given in the first section, "Extended description of the standard version", and is unambiguous. At least a couple of years ago (when I gave up on the "discussion" here going anywhere) almost everyone here agreed that the standard version is the main topic. Other interpretations generally served only to sidetrack discussions which were already doing perfectly well at going nowhere on their own. Guessing from the RFC discussion, this hasn't changed. The reason for the other interpretations being treated as also-rans relegated to the variants section is simple enough: the standard interpretation is almost universally the one that is meant — it is the MHP.
One of many patterns for "hilarity" to ensue is this: The "unconditionalists" sometimes insist that the unconditional solutions are correct because they produce the correct answer. The "conditionalists" argue back that it's only due to a (un?)happy coincidence, because it isn't so generally when the host chooses with different probabilities. The unconditionalists then argue that that's a variant problem and so irrelevant. The conditionalists argue that that wasn't the point, the point was that the method of "proof" is not valid because it sometimes produces wrong results. Then the unconditionalists argue that it is correct because in this particular case it always produces the correct result, and [back-and-forth ad nauseam], while in the sidelines there are still people (basically, anyone new to the discussion) who don't realize that nobody is actually interested in the variants, at least until the issues with the main topic have been resolved. Just warning you about a possible can of worms you might be opening.. -- Coffee2theorems (talk) 12:20, 9 September 2012 (UTC)
Not quite so. I am one of the unconditionalists, and argue that it is correct because this particular case is symmetric, not "because in this particular case it always produces the correct result". Details are buried deeply in the archive to this talk page, but I retain a copy, if you like... Boris Tsirelson (talk) 14:05, 9 September 2012 (UTC)
+1 Therefore the article is to present the MHP scenario first, with its clear answer, showing also Bayes theorem in the easy to understand "posterior odds = prior odds times likelihood" version. And, showing why probability to win by switching is NOT 1/2, immediately followed by the forgetful host who, by showing the car in 1/3, makes the chance of both still closed doors 1:1 indeed, as per the very common intuitive appraisal. A clear structure of the article is most essential. Gerhardvalentin (talk) 14:33, 9 September 2012 (UTC)
Trovatore, because of the argument that has bugged this page for years the article has fallen a little into disarray. It is almost universally agreed that the host must open a door to reveal a goat and I do not think there will be much opposition to stating this early in the article once the longstanding dispute has been settled. Martin Hogbin (talk) 17:01, 9 September 2012 (UTC)
As Coffee2theorems mentions, the constraint that the host must open a door is already mentioned in the first section following the lead (the section presenting the problem in detail) so I'm somewhat perplexed what the issue is here.
Boris - is an "unconditional" solution that makes no symmetry argument responsive to the question? My reading of Proposal 1 is that the article will begin with solutions like vos Savant's which show only the odds of a strategy of switching (all players switch) is 2/3 - without attempting to connect this in any way to the odds for a player who has picked door 1 and has seen the host open door 3. Yes, these are the same if the problem is symmetrical, but is it sufficient to leave both the symmetry and the consequence of the symmetry completely unmentioned? -- Rick Block (talk) 17:20, 9 September 2012 (UTC)
I don't think "the first section following the lead" is good enough. It needs to be immediately after the quote from Parade. Otherwise the rest of the lead may be misunderstood. I think most of the participants have been thinking about the "standard" interpretation for so long that they maybe don't notice that there's nothing particularly obvious about it from a basic understanding-of-English-language point of view. The condition is critical, and it is simply not there in the text. --20:40, 9 September 2012 (UTC)
Only "strategy"? – Not correct. Symmetry already is implied from the outset, by the unambiguous scenario where the paradox appears (as per Krauss and Wang, Norbert Henze and others), by the honest scenario where the host of course will maintain secrecy regarding the car-hiding door. So door numbers are completely irrelevant from the outset, in this unambiguous scenario where the famous paradox "appears". No deviance between the actual game and a "strategy" in case of recurrence. For a better understanding we should show Bayes rule in odds form for the reader, just at the beginning.
To distinguish "before" and "after" by showing conditional probability theory is not necessary for the paradox itself. You can show this where it makes sense, in later sections of quite aberrant scenarios of some whistle-blowing host who doesn't keep secrecy about the car-hiding door. But the article must stop to be degraded to a lesson in probability theory. This article is on the famous paradox, not on probability theory. The words of "before" and "after" and of "strategy" regarding the unambiguous scenario just do mirror the mingle-mangle mixing of different incompatible scenarios, that the unclear article for years has been suffering from, and that the conflict still is about. The article has to show what the sources say. The article is on the paradox that appears in symmetry, as per the introduction of the article. Quite aberrant scenarios of a whistle-blowing host etc. have to be shown in later sections. Gerhardvalentin (talk) 20:11, 9 September 2012 (UTC)
In no way! I never wished the symmetry to be unmentioned! I wish it to be used, to the benefit of the reader, explaining him why an asymmetric strategy cannot be preferable in the symmetric game. Boris Tsirelson (talk) 18:05, 9 September 2012 (UTC)
Then you may want to carefully read proposal 1 again. Perhaps Martin could clarify his intent as well. In proposal 1, will the initial "Solution" and "Aids to understanding" sections explicitly mention symmetry and because of symmetry the overall odds must be the same as the odds for a player who has (say) picked door 1 and has seen the host open door 3? Or, will symmetry and this connection (between the overall and conditional probabilities) be relegated to a later section "giving a full and scholarly exposition of the 'conditional' solutions"? -- Rick Block (talk) 19:42, 9 September 2012 (UTC)
I did read it again. Well, I wish the symmetry to be used somewhere in the article. Where exactly, this is another question (and I do not have a strong opinion here). But still, I believe that the intuitive understanding of symmetry and its implications is available to most readers, and therefore some top segment of the article may rely on it. And some later segment may explain it more formally (for those readers who still read). Boris Tsirelson (talk) 21:22, 9 September 2012 (UTC)
Exactly! That way the general reader is satisfied, the pedant is satisfied, the mathematician is satisfied, and no animals are harmed. Martin Hogbin (talk) 22:20, 9 September 2012 (UTC)
Are you saying yes or no to the question above? This sounds like your intent is the initial sections (including the section called "Solution") will include solutions only addressing the overall odds with no mention that because of symmetry the overall odds must be the same as the odds for a player who has picked door 1 and has seen the host open door 3 (all of this will be deferred to a later section). Is this correct? -- Rick Block (talk) 22:37, 9 September 2012 (UTC)
Quite the contrary. Because of symmetry there cannot be any difference regarding special door numbers, and because of symmetry there cannot be any difference to the overall odds. The explicit mentioning of the host who maintains secrecy on the car-hiding door in the unambiguous scenario where the paradox does "appear", assures symmetry. And, not to befog the readers, this should clearly be mentioned just from the start. My demand: stop nebulizing, stop obfuscating the readers. Gerhardvalentin (talk) 23:21, 9 September 2012 (UTC)
Rick, just as Boris says, the symmetry is so obvious, natural, and intuitive that it can initially be assumed without comment. Later on, for the benefit of those with a perverse streak, the assumed symmetry can be discussed along with the many other possible complicating and obfuscating issues that it is possible for anyone who puts their mind to it to devise. Martin Hogbin (talk) 23:30, 9 September 2012 (UTC)
Indeed and a comment preferably that not simply name drops the term symmetry but explains its (precise) meaning in the given context. Which of the various assumption are "natural" or "obvious" is up for the reader to decide for himself. We should primarily stick simply report and explain their consequences. In addition we can mention which assumptions some authoritative source might as "natural", if it says so explicitly.--Kmhkmh (talk) 07:03, 10 September 2012 (UTC)
I suppose you could structure this in a way similar to the 0.999... featured article, by starting with an unconditional solution section that ends with a discussion subsection saying that it actually solves a slightly different problem and can be converted to a solution of the original problem by a symmetry argument. Then there would be another section on complete proofs right after (no "aids" section in between), such as the decision tree proof and the symmetry proof.
Note that the 0.999... article is on a more elementary topic than the MHP and is just fine with putting unnecessary academic pedantry such as limits and infinite series in the second section. Conditional probabilities are simpler than those, so there should be no need to dumb down this article either. -- Coffee2theorems (talk) 10:08, 10 September 2012 (UTC)
Answer to Rick Block's edit today, discussed above: Sorry Rick, I do object to the edit. :) Your edit, and the use of "always" makes the false assumption that vos Savant's solution is about multiple cases and also that it is necessary for the basic solution to be about multiple cases. Neither is true. Since her problem (the most common and "standard" one) is clear that it is about a single occurrence, and the host does open a door showing a goat, and does make the offer to switch; your worry about whether that means he "always" makes the offer is irrelevant false reasoning, since the case and the question is specifically about only one occurrence. She asks 'should you switch', not 'should you always switch'. --Tom Hulse (talk) 03:11, 10 September 2012 (UTC)
How about "must" rather than "always" - i.e. "This version omits various details of the host's behavior considered to be part of the standard problem, such as that the host must open a door showing a goat and must make the offer to switch." Or are you arguing that these host constraints are not considered part of the "standard" problem? -- Rick Block (talk) 03:17, 10 September 2012 (UTC)
No, you would still be assuming Marilyn's problem is about multiple cases, and it is not. I am saying that those constraints are already built into the standard problem (Marilyn's), since the standard is about one case, not multiple cases. The standard problem says that the host does those things one time, and that is all that is necessary. Are you postulating that there might be additional trys at the problem by a contenstant, and then it must be specified for those extra trys that the host must also do what he did on the first try?--Tom Hulse (talk) 03:28, 10 September 2012 (UTC)
Tom, you're wrong on this, but I agree that it doesn't need to be mentioned in the lead. The problem as vos Savant analyzed it makes no sense without the additional assumptions, but those are already described in the first section after the lead. Without the assumptions, it might be the case that Monty is trying to get you to switch because he knows you've chosen the correct door, and that he would not have given you the choice if he had known you were losing already; that is not the interpretation intended, as it's not a problem in probability if that's what might be happening. Dicklyon (talk) 04:01, 10 September 2012 (UTC)
I think it does need to be mentioned in the lead. Otherwise the problem as stated is just not the one being treated.
Anyone who doesn't think it's necessary, which, if any, of the following do you disagree with?
The stipulation that the host must open a door is not part of the plain meaning of the text. I think this is undeniable — it just isn't there.
It is also not a natural common-sense assumption — unless you think people do only what they are required to do.
For those familiar with the actual show, it is not an obvious extrapolation of Monty's actual behavior — someone correct me if I'm wrong; it was a long time ago, but if memory serves, Monty did this sort of thing, to all appearances, ad lib, though of course it may in fact have been scripted.
The only context in which it might be considered a natural assumption is in a sort of stylized puzzle-solvers' paradigm, on the grounds that you need some extra assumption to make the problem well-specified, and among those extra assumptions that are sufficient, this is the simplest one to state. For example, the alternative stipulation that Monty is free to open a door, or not, ad lib, would be sufficient if you assume that Monty is a rational player whose goal is to deny you the car, but that's not necessarily very natural either (the simplest way to deny you the car is just not to offer it in the first place). --Trovatore (talk) 04:34, 10 September 2012 (UTC)
Both is correct. The famous question is put in a one occurrence game show. No log-list about a special host's "bias" available. But the world famous crystal clear "paradox" (not 1:1 but 1:2) definitely "appears" in its well-defined proper scenario. Gerhardvalentin (talk) 04:30, 10 September 2012 (UTC)
Dicklyon, here's why that fails: assume your proposed case is true and Monty only offered you the switch because he knew you picked the car in door 1. Imagine it's really you, in real life. What would you really, actually do? Is there any possible way for you to be aware that he only only offered you the switch because he knew you nailed it? No! There is no possible way for you to know that. If you did, it would give you the car. It would invalidate the whole game. Monty would never allow you to know that. It must always be unknowable to the contestant. So in real life, without any knowledge of his motivation, you, Dicklyon, would have to switch. You can't read his mind, so you can only use the stated facts of the case. Cases like yours wrongly approach the problem from Monty's perspective and what he is thinking, when the problem is defined instead as only being from the contestant's perspective, and knowable facts that the contestant could know.
Completely separate proof is that anyone who actually watched the show (I did often), knows Monte certainly did not only offer the switch only when the initial guess was right, as you postulate. If he did, it would take no time for guests to figure that out make a mockery of his system. He is therefore required not to let himself get in a rut of only offering when the first pick is right, to avoid ruining his show. --Tom Hulse (talk) 05:13, 10 September 2012 (UTC)
Tom, am I correct in reading that it never can be of relevance "which one" of his two doors the host just has opened? Or do I need to apply a mathematical cure-all approach regarding each and every imaginable circumstance? Gerhardvalentin (talk) 05:52, 10 September 2012 (UTC)
It's relevant to Monte, but he isn't making the crucial switch-or-stay decision. It's relvant to God if he was wanting to compute probabilities with all secret thoughts included in the formula. But the problem is by definition from the contestestant's perspective only, and there is no way to know what Monte's methods or intentions are. So all the info that is available to the contestant is that the two doors look the same and there is no way to know What Monte is thinking. --Tom Hulse (talk) 06:40, 10 September 2012 (UTC)
But Tom, that's what I was saying to Elen, at the time when (as you correctly pointed out) I was misinterpreting the RfC. If you don't make any assumptions about Monty, then there just is no answer. It becomes an ill-specified problem.
This is counterintuitive, of course. It seems like a well-specified problem, but it isn't. That's one of the subtlest and most interesting aspects of the entire affair, and it deserves to be taken note of in the lead. --Trovatore (talk) 07:30, 10 September 2012 (UTC)
To put my comments above, then, in the context of your post, I would say that the "assumption" about Monte that we can make, or know, is that yes he must always offer a switch according to Marilyn's simple problem, since it pictures a one-time problem where his offer is past-tense. There is no "always", just this one time. He did offer. It's already done according to the problem. This occurrence stands alone and there is no Monte pattern to analyze or fabricate. --Tom Hulse (talk) 07:53, 10 September 2012 (UTC)
I agree that there is no "always", that it's just the one time. That doesn't change anything. If you make no assumptions about Monty beyond the facts reported, the problem is ill-specified and has no answer. --Trovatore (talk) 07:56, 10 September 2012 (UTC)
What other assumptions or facts about Monte do you need to know to get an answer, other than that we already know he was forced to reveal the door by the conditions of the problem?--Tom Hulse (talk) 08:04, 10 September 2012 (UTC)
Oh, no, that one is sufficient. But it's not in the facts reported. At least, not in the text quoted from Parade in the lead. That's exactly why I think it needs to be dealt with, immediately after the quote from Parade. --Trovatore (talk) 08:08, 10 September 2012 (UTC)
Ok then, next step, Marilyn's quoted text in the article says "the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat", so it says he actually does open it for the problem to be valid. So please stipulate why you think actually opening a door has different odds from being forced to open the same door on a one-time problem. --Tom Hulse (talk) 08:30, 10 September 2012 (UTC)
All it says is that he did open the door. It does not say — it does not even remotely hint — that he was forced to.
Of course you don't really need to know whether he was forced to. All you really need to know is whether his decision to open it or not was influenced by whether or not your original choice was correct. Saying that he was forced to is nothing but the simplest way of saying that it was not influenced. However, the problem as stated does not tell you anything at all about whether the decision was thus influenced. Maybe it was, maybe not. Therefore the problem is ill-specified and has no answer. --Trovatore (talk) 09:36, 10 September 2012 (UTC)
When you claim that it is necessary to know Monty's influence to solve the problem, you're not understanding that you're looking at the problem from a completely unrelated angle, call it "God's perspective", instead of the regular contestant's perspective with facts that a normal contestant could know. Your God's perspective is one where esoteric facts are viewed as vital to knowing the "true" odds. But you're loosing focus on the real problem. The real problem is only the simple "should you switch" posed to the contestant who is only provided with regular contestant information. "True" probability with all factors in the universe accounted for are not only a different problem, but impossible to ever reach on almost all probability questions. Probability is almost always viewed "based on these simple facts that we know".
The additional proof that it is incorrect (I wanted to say pedantic, but that implies a tiny bit of correctness) is in the thousands of simulations run all over the world by people of all types. I know you'll say they (all!) set up the simulation wrong, so I'm not referring to their results. Rather, I'm pointing to the fact that 2nd graders, 3rd graders, 4th graders, etc... all the way up to groups of grad students, math professors, Los Alamos National Laboratory, dumb people, smart people... everyone, had no trouble interpreting the simulations in the same way, with the same conditions and the same result. It's obvious that they assumed there was no mysterious hidden influence in Monte's mind. So if there is any ambiguity in the instructions, it's in your head, lol! ;)
Ok, third proof: Assume Monty DID have bias and was influenced to try to trick you however he could within the rules specified by Marilyn. You be Monty and I'm the contestant. What could this influence lead you to do (specifically!) that would change the odds? I pick door one without your influence, now what? What could you possibly do under this influence that could change the odds? The problem specifies that you do open a door with a goat, so you have no choice. Even if you are evilly influenced you still must open a goat door to satisfy the conditions of the basic problem. Monty's bias is irrelevant since he is locked into a course of action by the conditions of the problem.
Fourth proof I gave above, for us to assume this biased Monty is a possibility, you propose that he might decide to always only offer a switch if the contestant was correct with their first pick. However this introduces easily spotted predictability that would massively favor the contestants (opposite of the desired effect). It would just never happen and is not a reasonable assumption. --Tom Hulse (talk) 10:39, 10 September 2012 (UTC)
The problem is that what you state in the "third proof" is wrong. The problem says that I do open a door. It does not say I have to, just that I do. The problem begins from there, from where I did. You don't know why I did, and it does matter. If you want to answer the question, then you have to start bringing in your analysis of why I might have done certain things, and how probable those are.
On your "fourth proof", you're starting to engage the issue a little bit. But the option you consider and reject is not the only alternative. Try this: I'm Monty (a standin for the interests of the TV producers and TV station). My interests are as follows: On the one hand, I want to increase viewership, and thereby, advertising revenue. On the other hand, though this involves less money, I do (or the station does) have to pay for the cars awarded. So I have some complicated payoff function that takes into account the audience reaction to what happens, minus the cost of the cars. My strategy will be based on maximizing that function.
Now, you're the contestant. You're not God, but you can nevertheless figure out that I have such a payoff function. How do you proceed? I don't know — it depends on your estimate of my payoff function, or the probabilistic distribution of that function. But in any case it's not the "simple" analysis. --Trovatore (talk) 10:58, 10 September 2012 (UTC)
OK, concrete proposal: It has filtered through to me from things I've read here and from the article that vos Savant specifically addressed this issue in the second Parade column. I propose that we report whatever it is she said about it, immediately after the quote from the first Parade column. Something like
In a followup column in Parade, vos Savant clarified that.... (whatever it is she exactly 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's wrong of course; the original column does not specify that condition. But that's OK; this isn't bash-Marilyn day; maybe that is indeed what she intended to say. It's enough for us that the text exists. So let's use it, immediately after the Parade passage:
In a followup column in Parade, vos Savant clarified that "the host always opens a losing door on purpose"
Is that clarification acceptable? Ideally, I would also like to interpolate a remark on why it matters:
In a followup column in Parade, vos Savant clarified that "the host always opens a losing door on purpose". (To understand why this is important, see the variants section below).
The exclusive focus on vos Savant's version (in this talk thread) is inappropriate. Her version is not "the one and only true MHP". It originated with a pair of letters from Steve Selvin to the American Statistician (in 1975 - 15 years before vos Savant published it in Parade). It has been widely analyzed in numerous forms in hundreds of publications (academic and not) since 1975. The "standard" version (whether it's exactly vos Savant's version or not), includes the constraints that the host must (or always, or however you want to phrase this) open a door revealing a goat, must make the offer to switch, and must choose between the remaining two doors randomly (uniformly) if the player happens to initially pick the door hiding the car. How much of this to put in the lead section is a matter of editorial discretion, see Wikipedia:Lead.
We could say The standard version of the problem includes additional constraints on the host's behavior, such as the host must open a door showing a goat and must make the offer to switch. This conveys the notion that there is a "standard" version (beyond vos Savant's) which the article will explain in detail (currently does, in the very first section following the lead). -- Rick Block (talk) 12:54, 10 September 2012 (UTC)
Here is the problem description of Steve Selvin. The host there is completely free in his actions. And he was never committed to open a door revealing a goat, and never he made the offer to switch. And it was Monty Hall himself who demonstrated in 1991 that the 2/3-solution for the problem formulation of Marilyn vos Savant was wrong. But it was this formulation - curiously often changed in the "wrong" direction - which was going around the world. And by an encyclopedia it is roughly misleading to divide the topic into a "serious" thread starting with Selvin and ending with the "standard version", and a subordinate thread around Marilyn vos Savant; and without emphasizing which version caused the "protest storms". This strategy hides the fact that it is a widespread error to claim the 2/3-solution only based on the blank fact that the host opens an unchosen door with a goat; followed by explanations which can only be nebulous because the crucial rule is missing. Apropos "standard version": This pedantic formulation hides this crucial rule within several obvious rules. Granberg lists seven rules; the last one reading: G. The host is truthful. But before I end up here with cabaret I stop.--Albtal (talk) 10:59, 12 September 2012 (UTC)
I just saw a convenient solution based on the second column and thought we could cut the Gordian knot. But your version is also fine with me. The stuff about randomness strikes me as a second-order tweak to the main point of whether the host must open a door at all. --Trovatore (talk) 19:23, 10 September 2012 (UTC)
Trovatore, you said " The problem says that I do open a door... You don't know why I did, and it does matter. If you want to answer the question, then you have to start bringing in your analysis of why I might have done certain things, and how probable those are." No, no, no, lol! This is absolutely incorrect. Rick, you should look at this also, as it is relevant to your desire to describe Marilyn's problem as anything but perfect. There is an assumption here that Monty's intent is relevant, that it could change the odds, but this assumption is false. Trovatore, you are still Monty, you didn't answer the question: what exactly can you do with your evil intent to change my odds if the problem stipulates a goat door IS actually opened? Nothing! Play it out, he has very limited options. Evil intent Monty, Neutral Monty, and Helpful Monty all end up with exactly the same actions here. No difference between them. No odds can change with intent of the host. No matter what you do within Marilyn's standard problem, I am still going to take your car 2 out of 3 times and there is nothing you can do to stop me. Host intent is irrelvant.
Rick, no one said anything about "the one and only true MHP", I know I certainly don't believe that. Take it easy on the straw man arguments my friend. :) Our article uses it for the lead because it is by far the most widely used & recognized version (more than Steve Selvin's by a factor of more than 4:1). That's why it is the "standard" version. Your opinion as to which is the standard is just as irrelvant as mine. Use the sources. It's overwhelming. --Tom Hulse (talk) 03:48, 11 September 2012 (UTC)
It's true that all three Montys have the same (observed) actions; nevertheless, the answer for the contestant is different. This is subtle and counterintuitive, but it's true. I have limited patience for explaining why to you personally; other participants understand why, and so do various sources. I'm not asking for any major disquisition on the question, just for an unambiguous statement of the problem up front (Kraus and Wang's seems satisfactory, if used to replace vos Savant's), and ideally for a pointer to the variants section where the reader can see why it matters. --Trovatore (talk) 04:27, 11 September 2012 (UTC)
Just generic "sources" on this particular problem has a unique meaning, as evidence by the 1000 PhD's who wrote in to insist the problem can't possibly be true, most of whom later recanted. You can find anyone to say anything on this problem because so many really smart people misunderstand it. Don't use the old 'limited patience to explain' thing to worm out of answering the very, very basic question you keep avoiding: what exactly can you do with your evil intent to change my odds if the problem stipulates a goat door IS actually opened? You listened to someone else on the internet who said intent was relvant and you didn't run it through yourself to see if that was actually true. Here is an important statement of fact that you can not challenge and proves you are wrong: there is nothing you can do as host, within Marilyn's problem, to prevent me from winning the car 2 out of 3. Try it! It's proof. --Tom Hulse (talk) 04:45, 11 September 2012 (UTC)
No, I did not "listen to someone else on the Internet". I am perfectly capable of figuring this out myself.
You are just simply wrong on your last claim. In fact, if you adopt a strategy of unconditionally switching, I can choose a strategy whereby your chance of getting the car is zero. My having adopted that strategy is perfectly consistent with the statement in Parade. --Trovatore (talk) 04:54, 11 September 2012 (UTC)
Well then quit telling me it's possible and actually tell me the strategy. Is this the 5th time I've asked for it now? Why are you stalling? Marylin says Monty does open a goat door, so you get to pick which one, and you know where the car is hidden. What is your strategy to produce zero wins? --Tom Hulse (talk) 05:10, 11 September 2012 (UTC)
I didn't tell you because I assumed it was obvious. My strategy is, if you pick the car, I offer to switch, and your strategy causes you to pick a goat. If you don't pick the car, I don't make the offer, and you wind up with a goat.
Nothing in the statement in Parade rules out the possibility that Monty was using this strategy. As you pointed out, it's not very plausible that the real Monty would use this strategy, but it is sufficient to refute your claim that you can always win 2 out of 3, independently of Monty. --Trovatore (talk) 05:20, 11 September 2012 (UTC)
C'mon man. That is, honestly, ridiculous. Did you not read in there where it simply said "the host, who knows what's behind the doors, opens another door". Tovatore, he opens another door. He opens it. How can you not plainly understand that? Why are you trying to torture out exotic interpretations that just aren't there? The exotic interpretation comes in first pretending as if there is more than one of these occurences. Then saying "I know I was required to open the door by Marilyn's problem, but if we repeated it, it's not clear that I should follow the same rules and open the door again", or "Even though she says the host opens the door, how can I possibly know if she intends for that to be repeated if I want to run the problem again, even though she says nothing about repeating it?" That is outrageous logic. No one really thinks that way.
Marilyn says that Monty "opens another door". Even a young child or someone of very low intelligence can understand that if the problem is repeated then you must repeat it with same rules.
Look at even the basic premise of the Parade problem: boil it down to its simplest most basic part. What is that? It's question at the end, 'should you switch'. That is what the problem IS. You can't pretend like you are allowed the option to not offer the switch, since that is the very crux of the problem. You are obviously required to offer the switch if you want to ask 'should you switch'. Can't you see that? --Tom Hulse (talk) 05:52, 11 September 2012 (UTC)
I can't understand how you cannot understand this, as it's perfectly clear. There is nothing obvious about whether the host is required to offer the switch. All you, the contestant, know, is that he did. Anything else is up for grabs, including whether he's playing the strategy I described. And if he is, you definitely should not switch, which as you say is the most basic question. --Trovatore (talk) 06:49, 11 September 2012 (UTC)
@Trovatore: You are completely right. And it was Marilyn vos Savant herself who wrote: When I read the original question as it was sent by my reader, I felt it didn't emphasize enough that the host always opens a door with a goat behind it, so I added that to the answer to make sure everyone understood. (MvS, The Power of Logical Thinking, p. 15). And Martin Gardner, Persi Diaconis and Monty Hall himself ... (see John Tierney, New York Times, Juli 21, 1991 (see References)). The task set presented had no 2/3 solution, and the uniqueness of the MHP as a "paradox" is due to the fact that the rules of the game are disclosed after the game ... Now "researchers" tell us that people assume the necessary rules for the 2/3-solution automatically; that is they would think: The host has now to open a not chosen door with a goat and to offer a switch. We simply can falsify this by showing a friend the original problem and explaining the solution saying: The host now has to open another door with a goat. And the friend asks: Why must he do so? - And the 2/3-solution collapses like a house of cards ... The German article One Car and Two Goats - The Wonderous History of a Mathematical Problem in detail goes to this aspects. For example we can read there: In my opinion many people within the "2/3 fraction" up to today don't see the difference between the blank fact that the host opens a not chosen door with a goat after the first choice, and the enforcement by the rule of the game which leads to this action. And this enforcement is crucial for the 2/3-solution, especially too for reenactments and "computer proofs".--Albtal (talk) 10:15, 11 September 2012 (UTC)
Addition: Often the famous mathematician Paul Erdös is given as an example for a person who didn't believe the 2/3-solution. But strangely no one tells which task set was presented to him. The "nearest" source I could find is this.--Albtal (talk) 12:21, 11 September 2012 (UTC)
This discussion is far beside the point of the RfC. Widely accepted as THE MHP is the standard version, say the Kraus & Wong formulation. And for good reasons. This formulation is easy understandable, and shows the intended paradox. The original formulation presumably was not intended to be a mathematical rigorous one. However (almost) anyone will accept for instance the assumption that the car is hidden at random, as that is what normal people will understand when they hear that a car is hidden behind one of three doors. However nothing in the wording formally says it is. By questioning all the assumptions, you've got nothing, also no MHP, so why bother, nothing useful can be said. Should the contestant switch? Well, if the car is noit behind his chosen door: YES! Interesting!Nijdam (talk) 11:37, 11 September 2012 (UTC)
The crucial rule for the 2/3-solution - that is, for the whole problem - is equivalent to this: The contestant has to determine two doors, of which the host has to open one with a goat. Any other problem does not have a 2/3-solution. And this rule is simple, short and clear.--Albtal (talk) 12:02, 11 September 2012 (UTC)
The article should start with a concise definition of the MHP using the Krauss and Wang version, not the vos Savant version. The lead should define the problem and ask "is it better to switch" without any answer, solution, or explanation. Save that for later.
Next should be a history section, still not saying what the answer is but instead documenting what letter writers to the American Statistician wrote in 1975, Marilyn vos Savant's answer, the answer of the 1,000 PhDs who disagreed with her, etc. The key here is to avoid saying what the answer is, but rather reporting (with citations) what various notable people say it is.
Next should be a section on simulation using plastic cups (better than playing cards; goats are indistinguishable) and then computers. This should be placed above all solutions because simulations don't provide solutions or explanations, just raw data that you can use to draw your own conclusions.
This structure separates what the answer is from why. When you are someone who is convinced that there is no advantage to switching, both simple solutions and conditional probability are pretty much useless; they are explaining why the right answer is the right answer before the reader accepts that it is indeed the right answer. As before, we shouldn't give the answer but rather should report (with citations) on the result of people running simulations.
Finally, after the Lead, History and Simulation sections, the other approaches should have a section or sections. Here, for the first time, we should reveal what Wikipedia says the answer is, along with answers to different but related questions. As to whether this section or sections should emphasize simple or conditional solutions, I will leave that for the RfC above to decide.
Could you please quote the Krauss and Wang version, or give a convenient pointer? I might support this if their version is indeed unambiguous. --Trovatore (talk) 01:31, 11 September 2012 (UTC)
"A formulation of the Monty Hall problem providing all of this missing information and avoiding possible ambiguities of the expression "say, number 3" would look like this (mathematically explicit version):
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. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one randomly. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door. Imagine that you chose Door 1 and the host opens Door 3, which has a goat. He then asks you “Do you want to switch to Door Number 2?” Is it to your advantage to change your choice?"
Source: Krauss, Stefan and Wang, X. T. (2003). "The Psychology of the Monty Hall Problem: Discovering Psychological Mechanisms for Solving a Tenacious Brain Teaser,'" Journal of Experimental Psychology: General 132(1). http://www.usd.edu/~xtwang/Papers/MontyHallPaper.pdf
Yes, this appears to work. It's unfortunate that making the statement unambiguous also makes it harder to read, but I guess that's life. Unless perhaps one of the pre-vos Savant sources has something pithier? --Trovatore (talk) 04:46, 11 September 2012 (UTC)
(Oh, as to the rest of your proposal — honestly I haven't fully digested it. But I do think it's a little odd to leave the answer out of the lead — after all, the lead is supposed to summarize the article, and the answer is one of the most salient points to summarize. How about a friendly amendment — give the answer in the lead, but not the rationale. Then the rationale can be discussed in a leisurely fashion that can hopefully accommodate everyone.) --Trovatore (talk) 04:50, 11 September 2012 (UTC)
Wikipedia pages about puzzles/problems that do not give the answer in the lead:
WP:LEAD says that the lead should summarize the article, should be able to stand alone as a concise overview, and summarize the most important points. A stand-alone overview of the MHP that does not say that the answer is 2/3 clearly does not fit the bill. If you had to write an article about the MHP for some Concise Encyclopedia of Very Short Articles, would you omit the answer? I think not. If someone asked you to read the MHP article and give them an executive summary of it in under 500 words, would you omit the answer? I think not. Many people read only the lead, so Wikipedia does double as that kind of concise encyclopedia and the lead as an executive summary. Inadequacies of other articles does not change this. -- Coffee2theorems (talk) 08:20, 11 September 2012 (UTC)
Please do not claim that WP:LEAD requires something without addressing the point that the editors of Missing dollar riddle, The Hardest Logic Puzzle Ever, Bridge and torch problem, Zebra Puzzle, Prisoners and hats puzzle, Wine/water mixing problem, Fox, goose and bag of beans puzzle and Missionaries and cannibals problem -- who presumably have also have read WP:LEAD -- did not come to the same interpretation of what is required. WP:LEAD does not require what you think it requires. Defining it as a puzzle meets the requirement. Giving the answer in the lead is not required. the " 'summarize the most important points' requires cramming everything into the lead" argument could just as easily be used to argue that the history section must be in the lead -- is not the history an important point? WP:LEAD also says "the lead should normally be no longer than four paragraphs" and "The lead often becomes cluttered with parenthetical details (often to the point of absurdity)." Some things cannot be summarized in a single paragraph. The history of the problem is one of them. The answer is another. --Guy Macon (talk) 14:03, 11 September 2012 (UTC)
As someone who was tricked by this, then refused to accept the answer, then relied on simulation to see why (and is still baffled by the fact it's still being argued although it appears to have shifted from "switching is not beneficial" to working out why switching -is- beneficial) I support this proposed layout change. The theory and probabilistic-heavy sections are left to the end, where people who really want to know (and can understand them) can find them but most people can stop reading after simulation or "simple" answers. 83.70.170.48 (talk) 08:49, 11 September 2012 (UTC)
The lead should definitely not start with the Krauss & Wang definition, because it is precisely the ambiguity of the original question (as in vos Savants column) which led to much of the fuzz of the problem and plenty of sources/publications (including Krauss & Wang deal with that issue). So imho there is no question that the original problem and a pointer to the ambiguity belongs into the lead. Now Krauss & Wang can (and maybe should be) used in the first section dealing with solutions/explanations and could in a longer lead in addition, but not as a replacement for the original question. Skipping the original question in th e lead would imho be gross misrepresentation of the MHP complex and depriving the reader of essential information (by it is a violation of the suggestions in WP:LEAD as well).--Kmhkmh (talk) 08:54, 11 September 2012 (UTC)
I'm open to that argument, but I would ask that the ambiguity be clarified immediately, within the next paragraph after the Parade quote. The article will be seriously misleading if the lead simply gives the Parade quote and the answer, without fixing the ambiguity. To be honest, I would take issue with the word ambiguity — it's not so much that the Parade quote is ambiguous as it is that it unambiguously does not give enough information to result in a unique solution. The assumption that vos Savant seems to think is obvious is simply not in the text at all. That's not a defect in the problem; actually it makes it a better problem, and I would love to have the article explore that aspect of it more, but it does not appear that this is the topic most editors think is most important. --Trovatore (talk) 09:17, 11 September 2012 (UTC)
You are not alone in your opinion that "the ambiguity of the original question" is what causes so many people to get it wrong, but Krauss and Wang argue that this about less ambiguous version: "Even though the Bayesian solution is now wholly justified, fleshing out the problem in this manner would fail to foster insight into its mathematical structure. The problem is that people still do not have access to an intuitive solution. We argue that most of the criticisms of the standard version regarding its unstated assumptions are mathematically relevant, but not psychologically relevant, because participants still assume the intended rules, even if those rules are not stated explicitly. Evidence supporting this claim comes from the observation that the vast majority of people wrongly regard the stay and switch choices as equally likely to result in winning." You need a better reason that that to use a description that everyone agrees is inferior. --Guy Macon (talk) 14:03, 11 September 2012 (UTC)
That's fine with me basically, I just don't want to see the "original" problem dropped from the lead for the reasons stated above. I agree, that the lead should not simply provide a unique ("right") answer without resolving the ambiguity first. To the very least needs to mention that the resolution of various ambiguities/additional implicit assumptions are need to arrive at the "standard" solution. I also agree that ambiguity from some perspective makes it an even more interesting problem. In fact one of the first (and sometimes most important steps of (math) problem solving) is to take the original (vague) problem and to render it more precise or even redevise it into a new version that allows a(n unique) answer.--Kmhkmh (talk) 09:53, 11 September 2012 (UTC)
Yes — if you don't already know him, Google "be less helpful" to find the works of a brilliant in-the-trenches math-ed theorist, Dan Meyer. I think it's a great approach to education. Unfortunately it kind of obviously doesn't fit the paradigm of an encyclopedia. --Trovatore (talk) 10:07, 11 September 2012 (UTC)
No. The lead does not need to "mention that the resolution of various ambiguities/additional implicit assumptions are need to arrive at the 'standard' solution." You have four paragraphs, no more (see WP:LEAD). We already have two paragraphs too many. We need to reduce the lead section, not expand it. The insistence on cramming everything into the lead part of the problem here. --Guy Macon (talk) 14:03, 11 September 2012 (UTC)
In the following article of 2004, Stefan Krauss and Silke Atmaca don't use the problem formulation which is here called the "standard problem" but the Parade version with a remarkable supplement:
Wie man Schülern Einsicht in schwierige stochastische Probleme vermitteln kann; Unterrichtswissenschaft 32(1)/2004, S. 38-57
The supplement reads Ich zeige Ihnen mal was (Now I'll show you something) and is placed just before the host opens another door with a goat. The whole problem set is that of Gero von Randow in the German newspaper DIE ZEIT, July 19, 1991, which made the problem known in Germany. The supplement is not compatible with the "standard problem", although the authors claim in a footnote that these intended rules are implicitly assumed by almost all experimental subjects. This is a blank claim, and - an important point - answers that it doesn't matter if switching or not can in no way be compared with thousands of letters written to authors who presented the problem without the crucial rule together with the 2/3 solution and attemts to prove it. I don't know "storms of protest" following an article with the correctly formulated problem and corresponding explanations; and with that we even reached the topic going on here: If the contestant has chosen door 1, he will win the car by switching in two of three cases: If the host opens door 2, he will open door 3, and if the host opens door 3, he will open door 2. So he will win the car, if it is behind door 2 or door 3. This is completely true, and at the beginning of the game the probability for the "switcher" to win is 2/3. (There may be fine additions which extend this true statement to a proof of the fact that the 2/3-probability holds just before the second choice. But I think that they all are a little too abstract for a "simple solution".) Seamlessly we may write that a mathematical proof for the 2/3-solution just before the second choice (which is asked for) reads as follows ("Bayes light"): The host opens door d with probability 1 if the player has chosen a goat, and with probability 1/2 if the player has chosen the car. Therefore the final chance for the switching player is (1/3 * 1)/(1/3 * 1 + 1/3 * 1/2) = 2/3 . We also may add some lines with "cases" to improve understanding of this short calculation.--Albtal (talk) 17:18, 11 September 2012 (UTC)
lede
it says For example, if Monty only offered the contestant a chance to switch when the contestant had initially chosen the car, then the contestant should never switch.. Umm... isn't that sort of obvious? Does it really need to be stated? VolunteerMarek00:23, 7 September 2012 (UTC)
Actually, it does. The Monty Hall problem has an interesting attribute; many engineers, scientists and mathematicians get it wrong at first and are adamant in claiming that that they are right. Then, when it becomes obvious that they are wrong, there is a strong tendency to claim that the problem description was misleading or ambiguous. Because of this, a lot of "obviopus" details need to be explicitly stated. --Guy Macon (talk) 01:27, 7 September 2012 (UTC)
It did from the beginning: as a proof that the problem in its original Parade version had no 2/3-solution. It is exactly the wording of Martin Gardner's argument in John Tierney's article in the New York Times 1991 (see references). The problem is not that this is obvious, but the possibility of this behavior of the host in the original formulation of the problem: The blank fact that the host opens a not chosen door with a goat does not lead to a 2/3-solution. It is only true if the host is committed by the rules of the game to do so. The example is a counter-example for the 2/3-solution for the problem wich was going around the world.--Albtal (talk) 20:22, 14 September 2012 (UTC)
Proposal
As the simple solution is plain wrong, but the lot will never be able to understand this, I suggest we formulate a simple solution as follows:
Due to the random placement the car is with probability 1/3 behind the chosen door 1. After the host has opened door 3 the original probability of 1/3 for door 3 has now clearly changed into a new probability 0. The new probability that door 1 hides the car however has the same value 1/3 as the original probability. Hence the remaining door 2, having originally a probability of 1/3, must have a new probability of 2/3 to hide the car.
Such a formulation is fully correct and, in my opinion perfectly understandable for everyone. It is not OR, as it is just a rephrasing of the correct solution.Nijdam (talk) 09:59, 10 September 2012 (UTC)
This has the advantage that it leaves a clear and compact piece to be proven later, namely that the probability of door 1 hiding the car does not change. The conditional probability section currently doesn't make it clear to the average reader why it is needed, but saying that this one thing was left unproved would provide a simply explainable raison d'etre, and people who think it is obvious can skip it with reassurance that they aren't missing anything they care about. You could even explain what "probability changing" means in clear terms using a simulation (rejection sampling), and the easily understood change of one probability from 1/3 to 0 helps in grasping the idea. I don't know if this is the best approach, but it would allow for a nice flow for the article. -- Coffee2theorems (talk) 10:44, 10 September 2012 (UTC)
I like this idea. It is a clear but simple phrasing, and you can expand on the components - why is it still 1/3 for the door the contestant is holding, why has the other probability changed - in the next couple of paragraphs, still simply (maybe using some modeling), then move on down into Bayes (since folks seem to like it), then move on into the issues caused because the original formulations of the problem contained hidden assumptions that could impact the outcome, and therefore need to be pulled out for a full explanation. --Elen of the Roads (talk) 12:36, 11 September 2012 (UTC)
I've just been told by a mathematician that probabilities never change. Hence this explanation cannot be correct. :) --Lambiam04:12, 12 September 2012 (UTC)
That's just linguistic jiggery spookery Nijdam. Outside of mathematical rechercheness (which isnt a word but I like it anyway), a woman who goes upstairs wearing a pink dress and comes down wearing a blue dress has changed her dress, except for mathematicians who would say that she is now sporting a new dress and the old dress hasn't changed at all. If a recalc of the probabilities shows that the probability of the car being behind the other door is now 2/3, in common English it has changed. Articles have to be written in common English. If they needs to use words with their specific meaning in maths (or Evangelicalism, or french cookery) then those words need to be explained to the common reader. Elen of the Roads (talk) 09:51, 12 September 2012 (UTC)
Interesting this ' linguistic jiggery spookery'. I don't mind if you think about it this way, as long as you understand that not only the probability for the opened door has "changed" from 1/3 to 0, but that also the probability for the chosen door has "changed" from 1/3 to 1/3. As long as you do not understand this, you will not know what the struggle here is about. Nijdam (talk) 10:35, 12 September 2012 (UTC)
Open door 0 (we know it has a goat). Chosen door 1/3 (remains in base state - if there had been 4 doors it would still be 1/4, 10 doors would be 1/10 etc). Other door now 2/3 (because the probability of the car being somewhere must be 100%). Elen of the Roads (talk) 12:47, 12 September 2012 (UTC)
What do you mean by: remains in base state? And then you progress by saying other door NOW 2/3. I know where your problem lies, but do you? Therefor I also invite you here to make the exercise. Nijdam (talk) 21:10, 12 September 2012 (UTC)
Nijdam, I'm not (as in 'I refuse to', rather than 'you may have an erroneous belief that I am' - a good example of ambiguous language) playing word games with you. I am always saying the same thing - but since I am using natural language, I have a number of ways of expressing it. See below, where I said that one of your ways of expressing it was elegant. I have used in the course of the whole discussion two terms used by quantum physicists - who also deal with probabilities. One is base state, the other is collapsed. The contestant's chosen door remains in base state - the probability is still what it was at the start of the game, when the probability of their being a car in the system is 1 and the base state of all doors is a 1/3 probability that the car is behind that door. However, the probability of Monty's goat door has collapsed - the probability that the car is behind it has upon observation collapsed to zero. If it were Schroediger's cat, it survived this time. However, as the probability of there being a car in the system remains at 1, the probability of the car being behind the untouched door must now be 2/3. I'm just using different language options to explain (not prove) the outcome.
If you want to play a different game, imagine one played by Doctor Who on the TARDIS. Instead of opening a door to reveal a goat, the Doctor jettisons one of the doors (chosen at random), along with whatever is in the room behind it (a useful feature of a TARDIS). The probability that there is a car in the system is now less than 1. Is it still advantageous to switch? Elen of the Roads (talk) 22:15, 12 September 2012 (UTC)
Nijdam, you write: As the simple solution is plain wrong, but the lot will never be able to understand this, ...: but may I point out that this is a very silly thing to say, as of course the simple solution is just fine. I think you discredit yourself by saying things like this. I understand that you may want to change what the article says, but you can't win an argument by saying something wrong, and then pressing the point. linas (talk) 22:45, 12 September 2012 (UTC)
Other proposal
A lot of sources give (in variant wording) the "solution" (explanation S00): A player whose strategy is to switch loses if and only if the player initially picks the car, which happens with probability 1/3, so switching must win with probability 2/3.
This is not sufficient, but may be completed by adding something like:
Although this in general true, it doesn't take into account the knowledge of the chosen door and the door opened by the host. With an appeal to the symmetry of the situation, i.e. the problem is essentially the same for every combination of door numbers, the overall 2/3 probability of winning when switching also holds for any particular case, for instance when door 1 has been chosen and door 3 is opened.
The current evolving discussion is unfortunately a demonstration, why it is so hard (and probably immpossible) to come to satisfying conclusion. First of the the proposals in the RFC seem seem to be almost exactly what Martin and Rick where proposing in 2009 already. So after 3 years, various conflict resolution tools (moderators, arbcom) and discussion volume of several books we're still left with the same 2 proposals unable to make a decision. Even more so before the RFC even completed or some conclusion were attempted the discussion has moved on to other issues/changes (completely redesigning the lead) and opening new battlegrounds there. So it looks like we are on our way to trigger a new round of endless debates having settled/achieved nothing at all.--Kmhkmh (talk) 09:14, 11 September 2012 (UTC)
I have learned something from this fiasco, never propose a compromise, it basicaly gives away ground with nothing in return. A lesson learned. Martin Hogbin (talk) 21:35, 11 September 2012 (UTC)
The solution is clear: all frequent editors between say 2009 and 2011 inclusive don't touch the page for the next two years. These people did great work! It even had spin-offs in new research publications. Surprising new results in the Annals of Monty Hall Studies. Let a new generation have a go. At least they won't be hindered by having put a powerful emotional stake in a particular way the article should be written, and by allergic reactions to some of their old opponents.. The old guard should keep off the MHP article talk page too, except possibly to mention links to material on their personal talk pages. That way we won't completely forget the past achievements. Richard Gill (talk) 11:34, 15 September 2012 (UTC)
I'd agree and most would serve WP better if they spend elsewhere on articles where urgent completely undisputed things need to be fixed. However I'm afraid that's never going to happen and even if it were my guess would be that the new ones will engage in similar (endless) battles, so we'll end up with new actors but the same old show.--Kmhkmh (talk) 11:43, 15 September 2012 (UTC)
Morgan et al., million doors, and assorted thoughts
Arbitrary section heading inserted here, because the thread above has gotten completely off the topic of the opening post. ~ Ningauble (talk) 20:26, 15 September 2012 (UTC)
I don't know whether it's known here that Morgan et al. (1991, cited in the article) cite Marilyn's problem description wrong just at the place where door numbers come into play:
Morgan: Suppose you're on a game show and given a choice of three doors. Behind one is a car, behind the others are goats. You pick door No. 1, and the host who knows what's behind them, opens No. 3, which has a goat. He then asks if you want to pick No. 2. Should you switch?
Vos Savant: 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, say number 1, and the host, who knows what's behind the doors, opens another door, say number 3, which has a goat. He says to you, "Do you want to pick door number 2?" Is it to your advantage to switch your choice of doors?
Falsifying biased they have removed "say number" for door 1 and "another door, say number" for door 3.
Also Marilyn's problem description in the article here has a supplement which is not in the original text: [but the door is not opened]. It seems that somebody didn't realise the meaning of the original text here.
Another passage of Morgan et al.:She then went on to give a dubious analogy to explain the choice. But they don't write what analogy is meant. - It is the variant with 1000000 doors. Maybe someone here can explain them the meaning of this analogy.--Albtal (talk) 22:48, 14 September 2012 (UTC)
Ah, the million door analogy. It appears to be an attempt to make readers realise (pace Nijdam above) why the probability of the car being behind the door you first picked remains at 1/3, and doesn't move to 1/2. If one could imagine picking from a million doors with only one car, the chance of you having the car is vanishingly small (although not as vanishingly small as winning the UK lottery with a single ticket). As Monty patiently opens more and more doors to reveal more and more goats, the hope is that it becomes clear to the reader that the door that he leaves closed after opening 999,998 doors revealing 999,998 goats, has a hugely higher chance of having the car behind it, than the contestant's door picked from the original pool does. Elen of the Roads (talk) 00:57, 15 September 2012 (UTC)
Yes, the many door analogy is for many people the eye-opener. Nothing dubious about it. And one of the main tools whereby a creative mathematician solves problems (indeed, tools for problem-solving in general): vary the conditions! Investigate extreme cases! See George Polya's wonderful book "How to solve it". I only ever met one person whose first answer to my question "would you switch" was the right answer. (My own initial answer when I first learnt of the problem was also the wrong answer "it doesn't make a difference, so I'd stay"). That person first of all astutely realized the question must be a trick question, and secondly intuitively generalized the problem to 100 doors, 99 goats, one car. The person was actually my mother who has no mathematical education though she was one of Alan Turing's "computers" at one of the Bletchley Park outstations (a computer was a young lady who operated the bombe, a calculating machine built for a unique task, namely to crack the enigma code. They needed an awful a lot of computers...). Richard Gill (talk) 07:43, 15 September 2012 (UTC)
'Eye opener' an excellent description of what is needed here. We know from the letters that vos Savant got that most people do not believe the correct solution when presented to them. If we do not convince people at the start of the article they will think the writers are idiots and read no further, thus we fail in our encyclopedic purpose.
I think that it is initially worth taking a few mathematical liberties so that readers will stay with us and, if they wish, then read the details. I have a suggestion which is in line with my original proposal but which might be acceptable to others. Martin Hogbin (talk) 09:11, 15 September 2012 (UTC)
Yes; but I think that maybe we don't need mathematical liberties and not necessary additional details, and not the K&W rules; and that we don't have to solve the problems in the table above. Our "simple" 2/3-solution holds for all possible questions and at every time before we know the position of the car. For we have to answer the questions based on the knowledge we have. And we know exactly already before the game what will happen. And the 2/3-solution which we can "compute" is exactly based on what can happen "in the future". And it would be a real "paradox" if we change our chance to win by switching if, what we exactly knew will happen (the host opens another door with a goat) now really happens. And I think we all agree that it would be a bad "consultant" of the player who said "No" to the 2/3-solution. And my thesis is that there is mathematics which says this "average" or "overall" probability of 2/3 is exactly that what matters here. For if not mathematicians would be worse "consultants" than laypersons. And the little "gap" in the "Richard Gill thread" here could be closed by a (maybe very "general") mathematical source (game theory, decision theory ... maybe Richard has already done it). And this in my opinion is the only place where we need a "source" for the solution of the MHP - but please in a "detail" section. We surely need sources for the "story" around MHP; and for the versions who had been presented; for example to Paul Erdös ...
Important hint: Don't use the 1/3 stays argument for the solution for it is exactly what is to bring to proof; and it is often used by people who didn't really understand the 2/3-solution; and it may be a reason why the wrong task set was going around the world - because this "argument" seems to hold also if the critical rule is missing - and a reason for the "shock" for people (see sources) who once realised that their "argument" doesn't hold if the host opens another door without knowing where the car is, and accidentally shows a goat. Take the easy to understand "operative" argument: If the contestant chooses door 1, he will win the game by switching in two of three cases: If the host opens door 2, he picks door 3, and if the host opens door 3, he picks door 2.--Albtal (talk) 11:08, 15 September 2012 (UTC)
I don't see why it is necessary to take any mathematical liberties at all. We just have to be careful that what we write is both sourceable and true. That's not difficult if we are careful with words. Every correct mathematical argument uses some assumptions and comes to conclusions which are conditional on those assumptions. If we focus on presenting notable but moreover correct arguments in notable sources (who did not actually make big mistakes), and do that in a careful and responsible way (ie wikipedarian way) the reader will automatically see what assumptions are actually used, and what the conclusion of the argument actually is.Richard Gill (talk) 11:17, 15 September 2012 (UTC)
I was invited to comment
In preface, I started a new section not from arrogance but separation. This has gone on too long IMO to reach a consensus for resolution using traditional means. I suggest an out of the box approach. I was going to suggest a few but after more thought I believe I'll suggest just one. If a simple majority agrees then I suggest binding arbitration for this article (this is not the formal arbitration committee). If this is permissible then I have further suggestions on how to proceed after arbitration of the current problem. Since I'm not certain this is permissible or if it will be ratified, I'll reserve further comment until then. Jobberone (talk) 23:11, 9 September 2012 (UTC)
I think that it is a good basic policy to give whatever dispute resolution method is in play a fair chance instead of deciding beforehand that it won't work The RfC will close after 30 days, at which point we can look at alternatives if the RfC doesn't do the job. BTW, I have already made a rather radical proposal that, if implemented, is 100% sure to resolve the problem: Talk:Monty Hall problem/Archive 29#Ten Years And A Million Words --Guy Macon (talk) 01:34, 10 September 2012 (UTC)
That's fine. I've never seen that approach anyway but you don't run into ten year disputes that often either. Let me know if I can help. Jobberone (talk) 02:36, 10 September 2012 (UTC)
From a discussion with Boris Tsirelson I noticed it is not clear what we are talking about. I want to present here some definitions as a help in the discussion:
MHP
The MHP is the problem as specified by Kraus & Wong, with the extra assumption that the original choice of the player is independent of the placement of the car.
Simple00
The simple solution 00 reads: An intuitive explanation is to reason that a player whose strategy is to switch loses if and only if the player initially picks the car, which happens with probability 1/3, so switching must win with probability 2/3 (Carlton 2005).
Simple01
The simple solution 01 reads: Due to the random placement the car is with probability 1/3 behind the chosen door 1. As the opened door 3 shows a goat, this door has clearly probability 0 to hide the car. Hence the remaining door 2 must have probability 2/3 on the car.
Simple02
The simple solution 02, also called combined doors solution, reads: Another way to understand the solution is to consider the two original unchosen doors together. Instead of one door being opened and shown to be a losing door, an equivalent action is to combine the two unchosen doors into one since the player cannot choose the opened door (Adams 1990; Devlin 2003; Williams 2004; Stibel et al., 2008).
Or extensively: As Keith Devlin says (Devlin 2003), "By opening his door, Monty is saying to the contestant 'There are two doors you did not choose, and the probability that the prize is behind one of them is 2/3. I'll help you by using my knowledge of where the prize is to open one of those two doors to show you that it does not hide the prize. You can now take advantage of this additional information. Your choice of door A has a chance of 1 in 3 of being the winner. I have not changed that. But by eliminating door C, I have shown you that the probability that door B hides the prize is 2 in 3.'"
Sometimes this solution is simplified as: The chosen door 1 has probability 1/3 on the car. The doors 2 and 3 together have probability 2/3 on the car. As door 3 is opened, door 2 must have chance 2/3.
Yes, it is an elegant approach. In "real life" this is my favorite solution for non-technical people, as I mentioned at the end of this thread last year. In my experience using MHP to entertain and educate several groups, partitioning the doors between "your door" and "Monty's doors" is the most persuasive explanation, and just suggesting the partition is often enough of a hint. (Simulations are also persuasive, but they don't explain or offer insight.) Unfortunately, I am not aware of any citable source on this method that completes a formal proof that the probabilities for the two partitions do not change when a goat is revealed, only the probabilities within Monty's partition. (The proof I use is Bayesian, but most people around here, apparently having a passing familiarity with undergraduate probability, seem to consider this anathema, so it just doesn't work in "wiki life".) ~ Ningauble (talk) 21:28, 12 September 2012 (UTC)
Yes, it has a common sense obviousness - you have a 1/3 chance of having the car, and Monty has a 2/3 chance, and that doesn't change when he opens the door on the goat. But I'm not aware of anyone ever having cited a source for it. Elen of the Roads (talk) 21:57, 12 September 2012 (UTC)
You can find a proof in the collected works of yours truly ... under, of course, the extra assumption which is needed to make it work: symmetry. Richard Gill (talk) 18:59, 16 September 2012 (UTC)
Richard, you know as well as I, that the basic idea behind the combined doors solution is that the not chosen doors have together a chance of 2/3 to hide the car, and that by showing that one of them hides a goat, the other door must have probability 2/3. Looks very obvious, but is nonsense, together the not chosen doors have 2/3 chance to hide the car, but the reason is that each separate has 1/3 chance. And, yes. most of the erroneous solutions can be fixed, one way or the other. It doesn't make them less erroneous. Nijdam (talk) 19:19, 16 September 2012 (UTC)
Solutions?
For your information: all the "solutions" S00, S01 and S02 are erroneous. — Preceding unsigned comment added by Nijdam (talk • contribs)
No, they are only 'erroneous' according to some very tight definitions of 'erroneous' and 'correct'. They are not mathematical proofs. They are explanations for a lay reader, to help them grasp the apparently paradoxical outcome. In this context, "Well, it's all down to Sooty's magic wand" is an erroneous explanation. --Elen of the Roads (talk) 22:21, 12 September 2012 (UTC)
Sorry Elen, maybe you'll change your opinion after you have read my comment on why you still wear the same dress. I don't reject them because they're not mathematical proofs, no, I reject them as they are logical wrong, and no way of arguing can be accepted as an explanation if it's logically flawed. Nijdam (talk) 08:41, 13 September 2012 (UTC)
See my comments above. They are not 'erroneous' at all - the statement that the probability remains 1/3 (or that it is also 1/3 after Monty shows his goat) is absolutely correct mathematically. The puzzle causes such consternation because even those of considerable intelligence and deductive reasoning, come to the conclusion that the odds of the car being behind either door is 1/2 - a 50/50 chance. There is no mathematical reason why this would be true - I think personally that the two doors just fool the brain. Explaining that the odds on the first door haven't changed doesn't require maths, because it isn't a maths problem. Rather it requires some method that unfools the brain - which is why so many styles of explanation have arisen. If the explanation (would it help if we stop calling them 'solutions' - then you might stop confusing them with mathematical proofs) doesn't explain this well, then it would be fair to call it a 'poor' explanation, but it is certainly not an 'erroneous' solution. --Elen of the Roads (talk) 18:55, 13 September 2012 (UTC)
No matter what else gets decided, I think replacing "solutions" with "explanations" is a brilliant idea that should go into the article.
I think the "method that unfools the brain" observation is spot on. I also think it explains a lot of the arguing here - different people get unfooled in different ways, and naturally think that whatever unfooled them is the best thing since sliced bread and that anything that failed to unfool them (or worse, that they estimate/guess would not have unfooled them) is worse than useless. Then the next editor comes to the opposite conclusion and the fight is on.
Another issue is that we know that many very intelligent and highly educated people get fooled, and when they get unfooled some of them start looking for "not my fault" reasons why they were fooled. Thus the arguments about how the problem is described and the claims that some explanations are wrong - and thus nobody can blame them for being fooled - it's all somebody else's fault.
And that is why we have over had over 1,300,000 words written on this talk page (all seven Harry Potter books together have 1,084,170 words) plus multiple trips to various dispute resolution venues, all without any agreement. Meanwhile, the page just gets worse and worse because so many cooks are following different recipes.
If, as I expect, this RfC doesn't solve the problem, I propose the following solution. (Yes, this is a serious proposal, and yes, I do realize that it cannot be decided here.)
[A]: Apply a one year topic ban on everyone who has made over 100 edits, (which includes me) with it made clear that the topic ban is because of an extraordinary problem that requires an extraordinary solution and absolutely does not imply any wrongdoing.
[B]: Reduce the article to a stub.
[C]: Let a new set of editors expand it as they see fit.
[D]: If anyone tries to invoke a policy that would not allow this, invoke WP:IAR.
Nothing else has worked. Although I hope to be proven wrong, I don't believe that anything else is going to work. Our best mediators, arbcom members etc. have utterly failed to solve this problem. My proposal will solve it. --Guy Macon (talk) 20:01, 13 September 2012 (UTC)
A counsel of despair. Perhaps just revert the article to the FA version. I am convinced that a lot of the issues here are semantic, and bugger all to do with maths. Someone else above mentioned that we should use 'approaches' instead of 'solutions' for the same reason. I think a lot of the objection seems to be coming from parties who are treating the explanations as if they are mathematical proofs. Elen of the Roads (talk) 21:53, 13 September 2012 (UTC)
Correspondence between the point of the player's decision, probabillity of interest, solution, and simulation
Would the esteemed professors of statistics and probability in our midst kindly comment on the following. Isn't there a rather close correspondence between when in the sequence of events the player is asked to decide whether to switch, the probability of interest, the form of the solution, and an appropriate simulation? Assuming the doors are distinguishable (assuming doors are not distinguishable seems like kind of a stretch - part of the charm of this problem is that it so easily maps to real world constructs which certainly does not include indistinguishable doors), it seems to me these are all related, such that
If the player must decide whether to switch
then the probability of interest is
which is determined by this sort of solution
which is shown by this sort of simulation
after the car is hidden but before picking an initial door
P(win by switching)
If the car is initially equally likely to be behind each door and you always switch you'll get the car 2/3 of the time
rules per Krauss and Wang, count every sample
after picking a door, say door 1, but before the host opens a door
P(win by switching|player picks a door), e.g. P(win by switching|player picks door 1)
If you pick door 1 and stay with your initial choice you win 1/3 of the time (only when the car is behind door 1), but if you switch you win 2/3 of the time (if the car is behind either door 2 or door 3)
rules per K&W, count all samples where the player picks door 1
after picking a door, say door 1, and seeing the host open a door, say door 3
P(win by switching|player picks a door and host opens a door showing a goat), e.g. P(win by switching|player picks doors 1 and host opens door 3)
If you pick door 1 and the host opens door 3 you win if the host opened door 3 because the car is behind door 2 (1/3 chance) but lose if the car is behind door 1 and the host randomly chose to open door 3 (1/3 * 1/2 = 1/6 chance). Winning with a 1/3 chance and losing with a 1/6 chance means you double your chances of winning by switching, i.e. the conditional probability you win by switching is 2/3.
rules per K&W, count only samples where the player picks door 1 and the host opens door 3
Certainly if you assume symmetry all of these decision points, formal probabilities, solutions, and simulations are equivalent - but if you assume distinguishable doors (distinguishable to player and host) don't these all line up exactly in the rows I've put them? I believe this is the point Nijdam is insistently making, and is the point that Morgan et al. and the other sources I've cited are making. Please climb down out of your ivory towers and look at this from a real world perspective (where doors are inherently distinguishable). If I pick door 1, and the host opens door 3, and now I must decide whether to stick with door 1 or switch to door 2 what probability do I care about, what solution is most meaningful to my situation, and what simulation is most relevant?
Rick, I can answer, but I do not want to be repetitive. As far as I remember, this was discussed, and answered, in 2009 (if not before that). Maybe you know the answer but reject it for some reason? Then please add: "I remember the answer that ... but I reject it since ..." in order to avoid loops. Otherwise say: "no, I do not remember any answer; what do you mean, Boris?" Boris Tsirelson (talk) 06:38, 14 September 2012 (UTC)
Rick, you ask about "the probability of interest". But it is only an auxiliary quantity. We have to decide whether or not to switch. Our decision has to be made at a certain point, right, given certain information. It's a theorem that deciding according to conditional probability will give the best overall success rate and vice versa: the best overall success rate means that our decisions are the same as according to conditional probability. If all doors are initially equally likely to hide the car then always switching gives the best overall success rate, 2/3. Hence there is simply no need to look at conditional probabilities at all.
Secondly, where did you get the idea that doors are indistinguishable? The point of symmetry in connection with a subjective Bayesian approach where we know nothing about the way the car is hidden and the way the host chooses a door to open is that by symmetry, the actual numbers written on the doors are irrelevant. Richard Gill (talk) 10:05, 14 September 2012 (UTC)
About indistinguishability: I know of only two contexts where this notion is meaningful. The first is, combinatorics; if you want to calculate the number of ways to choose something, indistinguishability can insert some factorial(s) to the denominator. However, in probability it does not, since here these factorials influence the numerator and denominator so that ultimately nothing changes. The other context is, quantum probability. Photons, and other quantum objects, can be really and provably indistinguishable, which really has striking consequences... but this never happens to cars and goats. Boris Tsirelson (talk) 11:09, 14 September 2012 (UTC)
Re. Distinguishability: Most statements of the problem do label the doors, which makes them distinguishable; but bear in mind that things which are distinguished only by unique labeling are, by definition, symmetric with respect to that distinction. The symmetries in the K&W problem statement are not assumed, as you say. The statement that a choice is equidistributed between alternatives that are distinguished only by unique labeling fairly shouts that the situation is symmetric with respect to that choice.
Re. "point of the player's decision" and "probability of interest": The sources that have been cited as making an issue of the distinctions in columns 1 and 2 have not generally been addressing the rules per Krauss and Wang used in column 4. They have been discussing much less constrained host behavior that breaks the symmetries inherent in the K&W rules.
I went looking for the discussion Boris refers to above, and suspect it is this (and subsequent threads that immediately follow in the archive). In this discussion Boris goes so far as to prove (given the problem is symmetrical) that all the conditional probabilities must be the same as the unconditional probability - which makes everything in the table above equivalent. And, yes, I already knew this (I even said it above). Whether all of these are the same given symmetry is not the question I'm asking.
The question I'm asking is whether the point of the player's decision (before picking a door, after picking a door but before the host opens a door, after the host opens a door) logically determines the probability we're interested in, and whether the solutions and simulations in the table accurately reflect the specific probability mentioned such that within each row any one of these pretty much implies all the others. For example, if you're given an answer that says (in its entirety, no mention of symmetry) "If you pick door 1 and stay with your initial choice you win 1/3 of the time (only when the car is behind door 1), but if you switch you win 2/3 of the time (if the car is behind either door 2 or door 3)", isn't this answer addressing P(win by switching|player picks door 1) which is logically the probability of interest in a problem where the player's choice occurs after picking a door but before the host opens a door which we'd simulate by running repeated samples according to some rules and determine the proportion of winning by switching in all samples where the player initially picks door 1?
Yes, this particular solution produces the correct numeric probability if the problem is symmetrical (in which case each of the probabilities has the same numeric value as any of the others, any of the decision points is the same as any of the others, etc.) - but if we don't establish symmetry, with this solution the probability we're saying is 2/3 is P(win by switching|player picks door 1) - right? Another interpretation is that someone offering this solution is not attempting to say anything about P(win by switching|player picks door 1 and host opens door 3), but is interpreting the question to be about the unconditional probability. If such a source itself says nothing about symmetry and nothing about the specific probability they're attempting to show - should we believe they're assuming symmetry without mentioning it and therefore conditional and unconditional answers can be (but aren't) shown to be equivalent?
Boris says above, in response to my question about whether it is sufficient to leave both the symmetry and its consequences completely unmentioned, "In no way!". But yet, he says he favors Proposal 1. Many sources say nothing about what specific probability they're addressing, say nothing about symmetry, and proceed with an unconditional solution. I believe the intent of Proposal 1 is that the article do this as well in the initial "Solution" section. However, many other sources clarify they're talking about the conditional probability given what door the player picks and what door the host opens (even if the host constraints ensure symmetry), and show this probability is 2/3. Establishing symmetry and showing symmetry means the unconditional and conditional answers are the same is one way to do this. Another is to directly show a 1/3 chance of winning by switching and a 1/6 chance of winning by staying (e.g. the "Decision tree" section). which by definition means the conditional probability of winning by switching is 2/3. Another is to use Bayes Theorem. What "Proposal 2" says is not that the unconditional solutions must be criticized, but that conditional solutions are so prevalent in the literature that they should be included in the initial "Solution" section as well. Doing this preserves editorial neutrality with regard to whether "conditional" or "unconditional" (i.e. "simple") solutions are better. The fact that "simple" solutions are criticized is one reason to carefully preserve a neutral stance, but the criticism itself would not be presented until a much later section of the article (this is common to both proposals). -- Rick Block (talk) 21:34, 15 September 2012 (UTC)
"are criticized", "must be criticized", and – where to present "the criticism itself" ... All of that is a less important side-show here. Because it's just only a narrow view of "how to define the actual probability". Where probability itself, you know, is just an unneeded side-show only, as other methods – without needing probability theory at all – already give the clear answer "switch here and now", "switch in each and every case, as staying will diminish your chance to win." Gerhardvalentin (talk) 22:20, 15 September 2012 (UTC)
Rick, I have only one correction to the facts and logic you present: about "But yet, he says he favors Proposal 1" − this is obsolete; my comment was completely rewritten at 15:48, 13 September 2012. :-) We agree on facts and logic. But we answer somewhat different questions.
What is "correct"? It depends on context. In the context of a proof assistant, it means formally correct in a very strict sense. In the context of a student on math exam, it is less strict. In the context of a math research article it means: such that the reader hopefully can easily fill in every gap separately (but not all gaps simultaneously since this would take too much time). In our case, I believe, it should mean: basically correct, but not pedantic; some details may be addressed later, if needed.
Example. A coin is tossed 10 times. The first two outcomes are "head, tail"; further outcomes are yet unknown. What is the (conditional, of course) probability that the first and last outcomes are both "head"? For a proof assistant, it definitely is But a human can easily take likewise, a human asked to substitute into writes rather than What is bad for a first-year math student on exam, can be pretty good for a non-mathematician. It is a matter of pedantry level. For me, this is an example of the appropriate pedantry level. I quote Nijdam: "Although this in general true, it doesn't take into account the knowledge of the chosen door and the door opened by the host. With an appeal to the symmetry of the situation, i.e. the problem is essentially the same for every combination of door numbers, the overall 2/3 probability of winning when switching also holds for any particular case, for instance when door 1 has been chosen and door 3 is opened."
Also, I feel, you bother to "give grades" ("correct", "incorrect") to various sources, while I only bother to explain a solution to a reader.
Tree showing the probability of every possible outcome if the player initially picks door 1
The conditional probability of winning by switching given which door the host opens can be determined referring to the figure below, or to an equivalent decision tree as shown to the right (Chun 1991; Grinstead and Snell 2006:137-138). In each case, it is assumed that the contestant initially chooses door 1. The probability is also formally derived as in the mathematical formulation below.
Assuming the player picks door 1, the car is behind door 2 and the host opens door 3 with probability 1/3. The car is behind door 1 and the host opens door 3 with probability 1/6. These are the only possibilities given the player picks door 1 and the host opens door 3. Therefore, the conditional probability of winning by switching is (1/3)/(1/3 + 1/6), which is 2/3. (Morgan et al. 1991)
Car hidden behind door 3
Car hidden behind door 1
Car hidden behind door 2
Player initially picks door 1
Host must open door 2
Host randomly opens door 2
Host randomly opens door 3
Host must open door 3
Probability 1/3
Probability 1/6
Probability 1/6
Probability 1/3
Switching wins
Switching loses
Switching loses
Switching wins
If the host has opened door 3, these cases have not happened
If the host has opened door 3, switching wins twice as often as staying
S: the number of the door Selected by the player, and
H: the number of the door opened by the Host.
Then, if the player initially selects door 1, and the host opens door 3, the probability of winning by switching is
as due to the random placement of the car and the independence of the initial choice of the player and the placement of the car:
and due to the host's behaviour:
Odds
The above calculation with the use of Bayes' formula is equivalent to calculating the odds for switching versus staying. These are 2:1, irrespective of particular actual door numbers:
As
the calculation comes down to calculating one of the mentioned probabilities.
And the same formal solution for the standard Monty Hall problem in written words, given that the player has selected door 1:
The car is behind door 2 and the host opens door 3 with probability (1/3)x1=1/3, while
the car is behind door 1 and the host opens door 3 with probability (1/3)x(1/2)=1/6.
These are the only possibilities in which door 3 can be opened. Altogether, the host opens door 3 with probability 1/3+1/6=1/2.
Given that the host opens door 3, the car is behind door 2 (1/3)/(1/2) = 2/3 of the time.
For the standard version of the Monty Hall problem, that means in turn:
Given that the host opens door 2, the car is behind door 3 (1/3)/(1/2) = 2/3 of the time.
So in the standard Monty Hall problem the odds for switching versus staying are 2:1, irrespective of any particular actual door numbers.
When I see a condition probability I expect to see something of the form P(A|B) but it does not appear until the Formal solution section. This not immediately clear from either the tree diagram, the second paragraph or the table with the goats that we are dealing with condition probability at all.
The formal solution section has much more complicated notation than is needed. With P(C=2 | H=3, S=1) = horrendous formula. What does that mean? Why not write this out in full P(Car is behind door 2 GIVEN THAN the the player chose door 2 AND the host opened door 3)? Isn't this all backwards though? The cars position is set at the start, why are we calculation the probability for the car being behind a door. I am being deliberately dense here to illustrate the difficulty readers will have with this section. I like to see something like P(win the car | some set of actions).--Salix (talk): 16:53, 15 September 2012 (UTC)
Proof via Bayes theorem (under K&W conditions) can be so, so simple. Let's fix the door chosen by the contestant as door 1. Bayes theorem says posterior odds equals prior odds times Bayes factor (aka likelihood ratio). Let's compute the posterior odds on car being behind door 1 versus car being behind door 2 "posterior to" getting the information that the host opens door 3 to reveal a goat.
Prior odds (= odds on door 1 to door 2, before anything has happened) = Prob(car behind door 1):Prob(car behind door 2) = 1/3 : 1/3 = 1 : 1.
Bayes factor = Prob(host opens door 3 given car behind door 1) : Prob(host opens door 3 given car behind door 2) = 1/2 : 1 = 1 : 2
Therefore posterior odds = 1 : 2 , in other words, after seeing the host open door 3 the car is two times more likely behind door 2 as behind door 1. And the reason is simple: because he was two times more likely to open door 3 when the car is behind door 2 as when the car is behind door 1. (And for Elen of the Roads: this proof is given by many reliable sources).
Note how the prior odds depends on (and depends only on) the assumption that the car is initially equally likely behind every door. The Bayes factor depends on (and depends only on) the assumption that the host's choice of door to open is forced when the contestant chose door 1 and the car was behind door 2, and that it was completely random - 50/50 - when the contestant chose door 1 and the car was behind door 1. Richard Gill (talk) 18:17, 16 September 2012 (UTC)
If the guest by chance did select the car, then the host got two goats to show. And as MvS surely and without any doubt did require a host who *maintains strict confidentiality* regarding the car-hiding door, you have to assume that his choice of door in that case is completely random - 50/50. Moreover: Marilyn's simple problem pictures a one-time problem, so no other possibility admissible than to maintaining secrecy on the car hiding door. K&W and Henze confirm that fact. Gillman and MCDD obviously were wrong to impermissibly "condition on the door opened by the host", as this only makes sense in learning and exercising probability maths in classrooms, but completely irrelevant to the probability to win by switching, pointless for the MHP. Gerhardvalentin (talk) 19:54, 16 September 2012 (UTC)
Simpler (if longer) description of the conditional solution by simulation
I don't think the article does its best to explain the conditional solution and its relevance to MHP, or its difference with the unconditional solution. Simulations are easily understood, and I was wondering if something vaguely like the following description (maybe somehow condensed; the simulation results could also be put in figures the same way as diagrams) could be used explain things so that everyone understands, even those who don't know anything about probability theory.
Simulate multiple runs of the game by the following procedure:
Place the car behind a random door using a fair die (1 or 2 = door 1; 3 or 4 = door 2; 5 or 6 = door 3).
The player picks door 1.
The host picks a door as follows:
If the player picked the car door, the host randomly picks one of the two goat doors by flipping a fair coin.
If the player picked a goat door, the host picks the other goat door.
The player switches to the other door and either wins or loses depending on what's behind it.
The first twenty runs of one such simulation went as follows (ignore the bolding for now):
Player chose door 1, host opened door 2, player won
Player chose door 1, host opened door 3, player won
Player chose door 1, host opened door 3, player won
Player chose door 1, host opened door 3, player lost
Player chose door 1, host opened door 2, player lost
Player chose door 1, host opened door 2, player won
Player chose door 1, host opened door 3, player lost
Player chose door 1, host opened door 2, player won
Player chose door 1, host opened door 2, player lost
Player chose door 1, host opened door 2, player won
Player chose door 1, host opened door 3, player won
Player chose door 1, host opened door 3, player won
Player chose door 1, host opened door 3, player won
Player chose door 1, host opened door 3, player won
Player chose door 1, host opened door 2, player won
Player chose door 1, host opened door 3, player won
Player chose door 1, host opened door 2, player won
Player chose door 1, host opened door 2, player won
Player chose door 1, host opened door 3, player lost
Player chose door 1, host opened door 3, player won
The player won 15 of these 20 runs, giving a winning rate of 75%. It is shown in section [reference to unconditional solution here] that as the simulation length grows, this winning rate almost surely approaches 2/3, or about 66.7%. This means that 2/3 of the people who go on the show (some who see door 2 opened, some who see door 3 opened, as above) win using the switching strategy.
The problem description, however, states that the player is one of those who see door 3 opened. What happens to players who see door 2 opened is irrelevant to his situation. The lines in bold match the situation in the problem description: the player has chosen door 1 and the host has opened door 3. The winning rate in this situation is obtained by collecting only the runs marked in bold and discarding the rest:
Player chose door 1, host opened door 3, player won
Player chose door 1, host opened door 3, player won
Player chose door 1, host opened door 3, player lost
Player chose door 1, host opened door 3, player lost
Player chose door 1, host opened door 3, player won
Player chose door 1, host opened door 3, player won
Player chose door 1, host opened door 3, player won
Player chose door 1, host opened door 3, player won
Player chose door 1, host opened door 3, player won
Player chose door 1, host opened door 3, player lost
Player chose door 1, host opened door 3, player won
The player won 8 of these 11 runs relevant to his actual situation, giving winning rate of about 73% in that situation. It is shown in section [reference to conditional solution here] that as the simulation length grows, this winning rate almost surely approaches 2/3, or about 66.7%. This means that 2/3 of the people who go on the show and see door 3 opened win using the switching strategy.
One can similarly show that 2/3 of the people who go on the show and see door 2 opened win, by collecting the cases corresponding to that situation instead. It therefore turns out that the probability of winning is the same for the player whether the host opens door 3 or door 2, and the specific door number is therefore entirely irrelevant. This irrelevance depends crucially on the host using a fair coin to choose. If he chooses using another method, then the door number is no longer irrelevant and the probabilities are different; such non-standard variants of the problem are covered in section [reference to variants section here].
Without something dead simple like this, I don't think we're going to be able to explain everything to most people in a way that they understand. -- Coffee2theorems (talk) 17:36, 11 September 2012 (UTC)
Strong agreement. I would add that the simulation should be of the problem as described by Krauss and Wang. The vos Savant description is ambiguous and allows for several different simulations, some rather silly. I would also really like to see any objections to the above be objections to a simulation of the problem as described by Krauss and Wang, not objections that require invoking some other set of rules. --Guy Macon (talk) 21:04, 11 September 2012 (UTC)
I have no objection to this simulation, it is pretty much what vos Savant proposed, in fact the simulation also neatly shows that the door numbers chosen by the host make no difference. Martin Hogbin (talk) 21:29, 11 September 2012 (UTC)
The table in en.citizendium.org, section "Explicit Computations" shows a good simulation. Scenario: the paradox (staying:switching "not 1:1" but 1:2) clearly "appears" in its proper fringe scenario. The premature "biased host" consideration should be relativized in the article, for "assuming" some unknown host's bias ["q"<>1/2] is incomplete unless allowing for the complementary bias as well, reducing "q" to 1/2 again). So a proper clear scenario should be shown, including the fact that any door preference of the host cannot be supposed (Norbert Henze e.g.: The host will keep absolute secrecy regarding the car-hiding door).
And imo the section "Criticism of the simple solutions" is extremely biased and unclear for presenting the clean "paradox". It should also be shown *there* that, in order to "solving" the problem, neither "condditional probbility" nor "probability theory at all" is necessary, as shown by various other methods (game theory, strategic dominance etc.) showing that the chance to win the car can never be improved by "staying". Gerhardvalentin (talk) 22:59, 11 September 2012 (UTC)
OK, so people like the simulation approach. Would this be an acceptable way to structure the article?
First section: the extended description, just as it is now.
Second section: explain the simulation somewhat like above, noting that it is a very direct implementation of the Krauss and Wang description. It would reference later sections that essentially prove things about the simulation, tying it all together. It would limit itself to giving the reader just the facts and the means to verify those facts for themselves, no proofs or arguments.
Third section: The unconditional solutions showing that 2/3 of all switchers who go on the (simulated) show win in the long run. People who are convinced by section 2 should be able to skip this if they want.
Fourth section: The conditional solutions showing that 2/3 of all switchers who go on the (simulated) show and see door 3 opened win in the long run. This would also include the symmetry solutions. People who are convinced by section 2 should be able to skip this if they want.
Optional fifth section if really really necessary: aids to understanding. (I think that if we need this section, it's an indicator that the previous sections weren't clear enough)
Sixth section: Sources of confusion. Note that understanding some of the cognitive psychology material (e.g. that referenced to Falk) requires understanding the unconditional/conditional issue. The second section alone should be clear enough to enable the reader to fully understand this section. We could explain things in terms of the simulation here, too, if necessary for clarity. This cognitive psychology part should probably be quickly referenced to in the second section as well (at the "crucial dependence on a fair coin" point).
Seventh section: History.
Eighth section: Variants. Also, if some of the more mathematical content (e.g. possibly the "formal solution", "dominance solution" and "alternative derivations") didn't fit in sections three and four, it could be placed here, or in an extra section before this one.
Placing the variants section before the sixth section would make sense as well, because part of the sixth section mentions some variants. However, I think that a short mention of the issue in the second section ("crucial dependence on fair coin" above) along with a reference to the variants section suffices, and that both the common confusion and the history are more important than the variants, and should therefore come first. Putting sections 2-5 as subsections of one "Solution" section would also be a possibility, but the difference is largely cosmetic. -- Coffee2theorems (talk) 15:30, 12 September 2012 (UTC)
The problem with this approach is WP:NOR a simulation made by us become OR. If someone else has published a simulation we can report that, but we can't do our own.--Salix (talk): 22:00, 12 September 2012 (UTC)
In my opinion, history should be moved up to be between sections 1 and 2, and the simulation section should be structured around multiple simulations that we have sources for, not something we put together - this is, after all, an encyclopedia. We should definitely include the elementary schools running manual simulations if we can find sources. --Guy Macon (talk) 20:09, 13 September 2012 (UTC)
I don't think this is OR unless you read the policies way too strictly. There are zillions of sourceable simulations essentially the same as this one for the unconditional problem, and how to do the conditional one is mentioned in the Morgan et. al. paper: "The correct simulation for the conditional problem is of course to examine only those trials where door 3 is opened by the host." This way of estimating conditional probabilities in general is also probably easy to source to zillions of probability 101 texts if needed (I think the Morgan reference suffices).
The only way this could possibly be OR is through WP:SYN, but even that one says "Do not combine material from multiple sources to reach or imply a conclusion not explicitly stated by any of the sources.", and the latter part manifestly does not fit. The conclusions and the reasons for those conclusions are in the sources. What is not in the sources is the explicit log of such a simulation, because those sources have space restrictions and are not written to lay audiences. I'd say the simulation here is also an obvious calculation in the sense of the policy: the way to do it is described in the sources and anyone who knows how to roll dice (even school children!) can do it. We routinely do much more complicated calculations for illustrations in maths articles — most people wouldn't be able to draw the figures in e.g. the Lambert W article given only a pen and graph paper — and yet such things are not considered OR. That's probably because those, and this simulation as well, do not cover any new ground. They are interpolations, not extrapolations. They illustrate to laypersons what something like
means in concrete terms. Strict interpretation of OR in such cases leads to us being unable to write things at a more widely accessible level than is present in the sources. In this case, that would mean that over half of the content of the article would be inaccessible to people who do not know what conditional probability means. -- Coffee2theorems (talk) 16:00, 17 September 2012 (UTC)
Combining doors - a suggestion
A solution that is convincing to many but which has been criticised here is Devlin's combining doors solution. Its weakness is that it implicitly assumes that the (numerical value of the) probability the car is behind door 1 after the host has revealed a goat is the same as it was initially. My suggestion is that we initially state this assumption without proof or explanation. After all it does, in fact, turn out to be correct. This at least makes clear the assumption on which the solution is dependent but leaves it 'clean'.
We can then move on to the other surprising, and often questioned fact, that it matters that the host knows where the car is. At first sight it might seem to the reader that the 'combining doors' should apply to the case where the host chooses at random, however, we can explain that the information given by the host revealing a goat in this case changes the probability that the car is behind door 1. We can then compare this with the standard MHP and cover our initial mathematical embarrassment. Martin Hogbin (talk) 10:21, 15 September 2012 (UTC)
There is no weakness in the argument that the probability that the car is behind Door 1 (contestant's initial choice) is unchanged by the host opening *a* door revealing a goat. That's going to happen, with certainty. It *can* happen *with certainty* because the host knows where the car is. That's why this matters. Then, to continue: moreover, *under symmetry*, there is no change to this probability on being informed *which* door was opened. By symmetry it can't make a difference!
Devlin got it nearly right but got confused when someone pointed out that his argument (in which a step was missing) wasn't watertight. So under pressure of the Bayes police he backed off and said you should do it by Bayes because then even though it was tedious and computational instead of quick and intuitive, at least you would be guaranteed to get it right? (if you are careful and have time to spare and know the definitions). The poor man is not a probabilist and it's clear to me that he was quite out of his depth. Fortunately it's known how to fix his argument (and you've seen how easy that is). Once fixed, it becomes quick, intuitive and watertight. Richard Gill (talk) 10:54, 15 September 2012 (UTC)
See my "parallel" comment above - surprising ... And indeed my "Important hint" there in my view is very important, because it concerns the central point of the MHP furor: The people who are surprised that their argumentation doesn't hold if the host can open a not selected door with a goat by other reasons ("motivation") than by the commitment by the rule of the game did not understand the problem. They carelessly agreed to the 2/3-solution although their understanding of the problem should lead to a fifty-fifty-solution - creating what is called the "Monty Hall Paradox". And the pretended watertight "proofs by simulation" definitely concealed their error. For the not formulated rule slips implicitly into the simulation. I think the situation which is still up-to-date is well described in this German article: In my opinion many people within the "2/3 fraction" up to today don't see the difference between the blank fact that the host opens a not chosen door with a goat after the first choice, and the enforcement by the rule of the game which leads to this action. And this enforcement is crucial for the 2/3-solution, especially too for reenactments and "computer proofs".
It is not because this is a surprisingly good moment for me to start a longer pause here, but because most discussions here are unnecessary tough and time-consuming. You know my arguments, and I don't have to repeat them. I wish the "Richard Gill Thread" success in filling the small "gap". I also would wish the "Trovatore Thread" success, but this for a longer time seems to be hopeless; so the vast majority here (the best argument I have ever heard in mathematics) again grows to a relaxing and creative 100%-majority.--Albtal (talk) 11:10, 15 September 2012 (UTC)
I think people accepted it because of familiarity with game shows in general, where "the reveal" always takes place at the end of the show (to answer a comment made by User:JohnSRoberts99, the game show appears never to have worked in the manner described in the problem, so there are no statistics from the show), so they just assumed that Monty 'must' open a door with a goat, because if he didn't, the show would end at that point, and the show always ends when the contestant opens their door/box/curtain. And of course if Monty picked his door at random and showed a goat, the odds on either remaining door containing the car do collapse to 50/50, because now you have two doors and one definitely has the car behind it. Nijdam did not respond in the Doctor Who example above - it would be impossible to answer that question because no information is given as to how he picked his door, and one simply cannot make any kind of guess about Dr Who's motivation. Elen of the Roads (talk) 13:20, 15 September 2012 (UTC)
"1:1, because now you have two doors and one definitely has the car behind it" — attention (two goats, one car): There's always "two doors and one definitely has the car behind it", even in the 1:2 situation, where the host opened a goat-hiding door intentionally, on purpose. Gerhardvalentin (talk) 21:17, 15 September 2012 (UTC)
Would it help if it went something like the version below. The introductory approach parallels Zebra Puzzle, which also has errors in the original version of the puzzle, meaning that the original proof makes some unstated assumptions (which is basically what the problem with Vos Savant's version is - it is her statement of the puzzle that is 'erroneous', not her solution, which is accurate for the version in her head.) In Zebra Puzzle, rather than saying that the solution is wrong, the article highlights the errors in the statement of the puzzle, and the assumptions that are therefore necessary for the published solution to be correct. --Elen of the Roads (talk) 14:06, 15 September 2012 (UTC)
Lede
"The Monty Hall problem is ... much older Bertrand's box paradox.
The problem was originally ..."Ask Marilyn" column in Parade magazine in 1990 (vos Savant 1990):
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, say No. 1 [but the door is not opened], and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
Vos Savant's response was that the contestant should always switch to the other door. After the Monty Hall problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine claiming that vos Savant was wrong (Tierney 1991).
While Vos Savant's answer is only correct if one assumes that the game show has the usual format, whereby the "reveal" (ie the final outcome) is only apparent when the contestant opens the door of their choice, many people still do not accept that switching is the best strategy in those circumstances, even when given explanations, simulations, and formal mathematical proofs.
The Monty Hall problem has attracted academic interest because the result is surprising, the problem is interesting to formulate, and it is possible to show proof of the outcome using more than one method. Furthermore, variations of the Monty Hall problem are made by changing the implied assumptions, and the variations can have drastically different consequences.
The first iteration
The puzzle as set by Vos Savant lacks a key piece of information - what governed Monty's choice of door. Vos Savant's solution assumes that the show's host will not show the car, based on his knowledge of where the car is. (write her solution out in full)
However, if Monty can pick a door at random and might have shown the car even though he does know where it is, the result is quite different, and there is no advantage in switching. (write that out in full)
The second iteration
The confusion caused by the omission in the original published version and solution has led to an expanded version of the problem, which sets out to clarify which set of conditions apply. (give full version)
Despite this, many people still refuse initially to believe that switching is beneficial. According to the French magazine Le Figaro Magazine, Paul Erdős, the most prolifically published mathematician of modern times, found it very hard to believe the result of his own calculations, and admitted he was not fully convinced until a simple computer simulation confirmed the predicted result.
As a result, a number of different ways of explaining the solution have been put forward.
Explanations, thought experiments and simulations
(already in the article)
Mathematical proofs
1. Bayesian
2. Etc
(already in the article, I gather the Bayesian one could be better written)
Variations
(already in the article)
There is something to this proposal, but it leaves out an iteration.
Iteration 2 added the requirement that Monty must open an unchosen door revealing a goat, i.e. he must open a door, he may not open the contestant's chosen door, and he must not show the car. Vos Savant apparently assumed readers would understand this from the "game show" context. Without this understanding the whole "paradox" is moot.
Iteration 3 added the explicit conditions that the car is initially placed behind one of the three doors at random, and that Monty's choice of goat to reveal behind an unchosen door is random. Two observations about this amendment are significant:
From a strictly frequentist perspective, Iteration 2 does not permit a closed-form solution for the probability of winning.
From a Bayesian (subjectivist) perspective, Iteration 1 and Iteration 2 are equivalent.
Note that your proposal to have the lede say vos Savant's answer "is only correct if one assumes..." depends on whether one assumes Iteration 2 or Iteration 3 and on whether one uses "probability" in a frequentist or Bayesian sense. Therein lies the crux of a large part of over a million words of disputation.
I agree that it is a good idea to explain at the outset that people have been talking about different problems. I have also long thought that having a section entitled "Criticism of the simple solutions" at the end is a completely non-neutral way to present what should be a section on the historical evolution of the problem. Unfortunately, attempts by editors to contextualize solutions with respect to problem versions are usually met with loud objections, in no small part due to the fact that the legitimacy or completeness of a solution for a particular problem version depends on what you mean by "probability" (particularly since sources are rarely explicit about this), how you interpret any ambiguity in the problem statement, and on what level of detail you expect a solution to provide about the underpinnings of the logic involved. ~ Ningauble (talk) 22:55, 15 September 2012 (UTC)
Thanks for the comment. Yes, you are right that there are three iterations. The second is necessary to explaining the outcome, the third is necessary to attempt a mathematical proof, and so should be included immediately above the 'proof' section. I am in favour of separating the 'explanation' or 'demonstration' phase out from the 'mathematical description' or 'proof' because the volume of readers who do not have maths post 16 will vastly outweigh the volume who did any kind of maths after that. It is trivial to demonstrate that switching is preferable (the original question) provided one assumes that Monty won't show the car. This must come at the top of the article so people can get their head round it. Those who want to move on to either 'exactly how much of an advantage is is' or 'prove that mathematically' or 'but what happens if instead...' can then move on to discover how precisely a question has to be set in order to answer either of these questions. Elen of the Roads (talk) 14:11, 16 September 2012 (UTC)
Adding, I also think that editors need to stop inserting their own explanations, demonstrations, objections, Bayesian expressions etc. The article must contain only what is published if it wishes to meet Wikipedia's standards for verifiability. Elen of the Roads (talk) 14:13, 16 September 2012 (UTC)
It is not a good idea to say or imply that the third iteration is necessary to attempt a mathematical proof. The proofs by symmetry do not employ or depend on the stipulated a priori frequency distributions.
My idea of a good "formal derivations" section (not "mathematical proofs" please) would contain one derivation (only one) using Bayes' theorem (noting that it uses the frequency distributions of Iteration 3), and one derivation using symmetry (noting that it solves Iteration 2 without a priori frequency distributions). A game theoretic derivation (noting that it compares fixed strategies) would be ok, but I am less enthusiastic.
I agree that "explanations" should come first, with "formal derivations" later. People who understand the formal derivations can connect the dots back to the various explanations; but people who to not understand them are not well served by interjecting technicalities amongst the explanations. Personally, I would be happy to call the explanations "solutions", but I accept that this makes some people unhappy. ~ Ningauble (talk) 17:38, 17 September 2012 (UTC)
I do not understand how this resolves the long-standing dispute over the validity of the 'simple' solutions. There is little argument about the standard version of the question or what vos Savant said. Martin Hogbin (talk) 15:23, 16 September 2012 (UTC)
If I understand Nijdam correctly (and I may be wrong here, as we've seen, I do have some trouble sometimes) apart from Vos Savant's original answer being wrong because she hadn't stated even enough key conditions to make a scratch explanation, every other simple 'solution' is wrong because it doesn't address part of the problem, or doesn't address it correctly. But equally, you have to include Vos Savant's answer, and the other sourced demonstration type answers. Rebadging them as 'explanations' or 'demonstrations', intended for those unfamiliar with the higher maths might take make it clearer that these are not mathematical proofs, while demonstrating to people how the phrasing of the question can alter the outcome. Having settled on a sufficiently precise question, sourced 'solutions' can be offered, and the controversy over how to construct a solution discussed, separately from how to demonstrate to disbelieving folks that if you know Monty wont show you the car, it's always better to switch. Elen of the Roads (talk) 12:45, 17 September 2012 (UTC)
Former discussion with Richard Gill
To make clear where Richard Gill stands, I copy here a discussion I had with him:
And Richard, I really am fed up by having to repeat this over and over, without getting an answer. So once and for all, answer to the following, and restrain yourself to what I say. One form of the simple solution reads: Due to the random placement the car is with probability 1/3 behind the chosen door 1. As the opened door 3 shows a goat, this door has clearly probability 0 to hide the car. Hence the remaining door 2 must have probability 2/3 on the car. it is this formulation I'm fighting. Especially because it looks quite plausible and is copied by lots of school pupils and students who gets assignments about the popular MHP. Now just comment on what I wrote here. Nijdam (talk) 22:06, 7 January 2011 (UTC)
@Nijdam, you are absolutely right. "Hence" is wrong. What is said is a nonsequitur. The fact of specifically Door 3 being opened could change the likelihood that the car was originally hidden behind Door 1. So far we only used "all doors initially equally likely" and that isn't enough to get a conditional result, while this argument is about a conditional result, since it talks about how probabilities change (or don't change) on getting the "Door 3" information. There is a missing step where "no host bias" has to be explicitly used. Eg, appeal to independence: by symmetry, the number of the door opened doesn't change the probability the car is behind door 1. Or if you prefer, use Bayes' rule and show by explicit computation that the odds on Door 1 hiding the car isn't changed by the information *which* specific door is opened. Next task, find a reliable source which explicitly makes this point, because I'm afraid that some editors could consider this observation of yours "OR". (Or write it yourself: why not submit a small note to Statistica Neerlandica?) Richard Gill (talk) 07:24, 8 January 2011 (UTC)
This is about the good old Devlin solution. Unfortunately as correspondents pointed out to Devlin, he got it wrong (skipped one step) and he retracted his argument. Unfortunately he could have fixed it, so easily. The fix has since been published in several reliable sources. Devlin solution (fixed): The car is initially behind door 1 with probability 1/3. Opening *a* door to reveal a goat doesn't change that. Being informed *which* door was opened, door 3, doesn't change it either (symmetry). Hence the probability the car is behind door 2 jumps to 2/3.Richard Gill (talk) 18:41, 16 September 2012 (UTC)
Thank you! Another contributor has repeatedly referred to Devlin's "retraction" without responding to repeated requests for a citation. This is very helpful. To be clear, Devlin is saying that the combined-doors solution does not work for "a slightly modified version" (his words) in which the goat is revealed by happenstance, without knowledge of whether there is a goat or a car behind the door. This is in distinct contrast to the unmodified problem, of which he says: "In the original Monty problem, Monty knows from the start where the prize is, and he uses that knowledge in order to always open a door that does not hide a prize. Moreover, you, the contestant, know that Monty plays this way. This is crucial to your reasoning."
In my opinion it is highly inappropriate to claim this means he is saying the combined-doors solution is categorically wrong: It is a matter of problem definition, not a matter of false reasoning. This is exactly the sort of selective misreading I was referring to at #Comments from Ningauble in the RfC above. ~ Ningauble (talk) 20:57, 18 September 2012 (UTC)
If you accept that the probability of the player's initial choice being correct does not change by Monty's action (which indeed it doesn't), then the result ("Hence") follows. --Lambiam21:29, 9 September 2012 (UTC)
Anyway, for your information: A probability never changes, the probability for the car to be behind door 1 is 1/3, and stays 1/3, even when door 1 is opened and shows a goat! I guess you're not a mathematician, otherwise you would have known this. When the host opens door 3 what happens is that new probabilities have to be considered. Instead of looking at the probability for the car to be behind door 1, we have to look at the CONDITIONAL probability for the car to be behind door 1. If you wish you may say, the former probability has changed from unconditional to conditional. Such things happen. What's puzzling many of the editors is that both probabilities have the value 1/3. The point however is that the simple solution does not consider the conditional probability and hence it is plain wrong, as a way of reasoning! And so are all the people defending the simple solution, simply because they do not understand the flaw. Nijdam (talk) 09:15, 10 September 2012 (UTC)
Nidj, the probability for a given set of circumstances never changes. It's disengenious to make the sweeping statement that probabilities never change, because the probability of an event clearly changes with changes in the environment. "What's the probability of my house being struck by a meteorite?" "Oh, well over a million to one" "Oh, look at the shooting st...." WHOOSH! BANG! "Ok, what's the probability of my house being struck by a meteorite now?" Elen of the Roads (talk) 10:05, 12 September 2012 (UTC)
Gradually, Elen, you seem to see some light(ning). Do you notice the use of the word "now" in your example? A probabilist would use the term "conditional". In the article we may of course use plain English, like the word "now"or "new" (as I did in my proposal). Nijdam (talk) 10:42, 12 September 2012 (UTC)
Yes, I do understand that you are using specialist language, which is fine, but in sentences that look like they contain ordinary language. You just need to keep bearing in mind the audience, which is likely to include general readers at least at the start of the article. In this case, in general English something 'changes' if it is different the next time you look at it. "That house has changed" is a perfectly applicable comment to make in English if they have pulled down the prewar prefab and put up a condo. You don't normally say "the house hasn't changed, a new house has been put in place of the old one." I have often thought that a good deal of the problem with Monty is semantic, not mathematical, where it is not clear whether precise or imprecise language is being used. Elen of the Roads (talk) 12:43, 12 September 2012 (UTC)
Your remarks on the language I use, bear no significance. The language here is just for the discussion here, and therefor necessarily a little technical. As you may notice: The proposal I made for a simple solution is formulated in plain English with no technical terms. So, what do you mean? The point is that if you want to say the probability for the opened door 3 has changed from 1/3 to 0, that's fine with me, but!!!! you should then also make clear that the probability for the chosen door 1 has changed from 1/3 to 1/3. Can you find nice simple wording to explain this? If so, please let me know. And, Elen, to see whether you understand this, make the exercise I formulated a little further down. Nijdam (talk) 21:06, 12 September 2012 (UTC)
No, I seriously think a lot of this is down to not knowing when language is intended with a precise meaning and when with a looser, natural language meaning. You switch from one to the other a lot, then leap in like Perry Mason when someone misinterprets what you meant. It's rather tiresome sometimes, although I don't for a minute think you're doing it deliberately. As to why the probability that the car is behind the door I picked has "changed" from 1/3 to 1/3 - I'm still wearing the dress I went out to work in this morning, I haven't changed. What you want to know is why I am still wearing the same dress, and didn't change for dinner when the rest of the family did (into sweats, pjs etc, not penguin suits I hasten to add). ie - what the reader needs to have explained to them is why the probability that the car is behind the door they chose has not changed, and is still 1/3, when most people (not familiar with the problem) fall into an erroneous assumption that it has changed to 1/2 - a fifty-fifty chance. Elen of the Roads (talk) 23:02, 12 September 2012 (UTC)
@Elen: Right, now you're hitting the nail on the head. You need to explain why you're still wearing the same dress. And there it is where the simple solutions S01 and S02 fail. No, worse, this solutions simply do not "know" of this need. The people defending this solutions argue that the probability for door 1 just IS 1/3, most, or all of them, do not have the slightest clue this needs explanation. Once you know this, you cannot accept this simple solutions as correct solutions, also because readers of such solutions learn to accept this arguing as correct (fact!). Nijdam (talk) 08:34, 13 September 2012 (UTC)
See, there you go again Perry. You're being unintentionally insulting again. This doesn't need explanation mathematically, it needs explaining conceptually to the frood who's reading this for the first time. So the simple solutions are not 'erroneous' or 'failed' - they are quite correct in saying that the probability remains 1/3. But they may omit sufficient contextualisation to sufficiently explain to the satisfaction of the reader, why the probability for the first door is still at 1/3, and hasn't changed to the 1/2 that most casual readers erroneously (there's the error) assume it has done. Elen of the Roads (talk) 16:25, 13 September 2012 (UTC)
Well, you disappoint me, I thought you came to understand. Would you say you do not need to explain why you still wear the same dress? Please, just answer this question. Nijdam (talk) 21:15, 13 September 2012 (UTC)
Please stop being condescending. We don't have to express things in exactly the words you want us to use in order to be right. An explanation (not a solution) which does not leave the reader with an understanding of why the odds on the door he picked at the start having the car behind it are still only 1/3 is a poor explanation, but it is not erroneous. Do you have a good way of explaining this to the common reader? Elen of the Roads (talk) 21:28, 13 September 2012 (UTC)
Nijdam, some of your behavior is becoming problematical. Consider the following quotes:
"Don't be naive"[1],
"So once and for all, answer to the following, and restrain yourself to what I say."[2],
"Do you think you know better than Richard and me, and many others??"[3],
"The lot will never be able to understand this"[4],
"I hope you also restrict yourself to this. Even then a lot of missionary work has to be done. "[5],
"I know where your problem lies, but do you?"[6],
"You don't need sources, of course, just simple logic will do."[7],
"Well, you disappoint me, I thought you came to understand... Please, just answer this question."[8]
And that's just from the last couple of days! You are belittling anyone who dares to disagree with you. You are barking orders at us, telling us how we must answer and refusing to engage anyone who does not follow your script. Your insistence that you are right and that everyone who disagrees with you must not understand is arrogant and condescending. Free clue: several people who are intelligent and educated have carefully examined your arguments, understand them just fine, and for some unfathomable (to you) reason do not find your arguments to be compelling. Deal with it. --Guy Macon (talk) 22:05, 13 September 2012 (UTC)
Dilligence
I have been away for a short time and have to catch up for about a week.
@Richard: The British are famous for their humour, but twisting someone's words and taking them out of context aren't the best instances. I had a smile though, and looked up what you wrote me some two years ago (quotes); On wikipedia you will never be able to convince most of the editors that only the full conditional solution is fully correct. And also: It seems almost impossible to explain to ordinary people why the unconditional solution should be seen as incomplete. Remember? Nijdam (talk) 21:22, 15 September 2012 (UTC)
Nijdam, in the first quote, I do not say that only the full conditional solution is fully correct. I say that you will never convince most editors that only the full conditional solution is fully correct. In the second quote I do not say that the unconditional solution should be seen as incomplete. I say that you won't be able to convince ordinary people that it should be seen as incomplete. Subtelty of British use of English language? If you want to know what I think about MHP, read my paper in Statistica Neerlandica. My preferred solution is "pick your initial door at random and thereafter switch". That's the very best advice I can give to any would-be contestant. You'll get the car with probability 2/3. The conditional probability you would get it if your pick happened to be door 1 and if the host happened to open door 3 is not well defined, and hardly interesting. Richard Gill (talk) 07:04, 16 September 2012 (UTC)
Looks to me like some dubious lawyers try to reason their way out. (Of course it just is about the probabilistic approach and the difference between unconditional and conditional.) Why not written it would be impossible to convince someone of an incomplete solution? Matter of English language? And what about the second quote? And what about your comment I presented above in "Former discussion with Richard Gill"? Concerning your preferred solution, it's no solution to the problem. It is indeed, as you write, an advice to someone who's going to be player in the MHP. Does it help the player who, not choosing randomly, but for some reason did choose door 1? You mentioned it already, they are interested in the conditional probability. And you're right, without some additional assumptions, not numerically determined. But hardly interesting? I think it is better understandable to make some quite reasonable assumptions, than to look away and solve something else, or present an incorrect solution, or even not solve the problem at all but formulate some advice, and give readers the impression it is a solution to the problem. Nijdam (talk) 08:04, 16 September 2012 (UTC)
Nijdam, the only thing that's important for Wikipedia are Wikipedia policies and Wikipedia-sense reliable sources. What I might have said in discussions on talk pages several years ago is irrelevant. If only because my own opinion might have evolved in the mean time. You have to abandon your personal mission to save the world, and do some collaborative editing instead. First of all, the wikipedia page has to help people understand why MHP is an interesting problem at all. Richard Gill (talk) 09:23, 16 September 2012 (UTC)
@Martin: What in the world has happened with you? Amazing! Apparently you do understand now the flaw in the simple and combined doors solutions.Congratulations. Let's work together to show the ordinary reader a simple, but correct, explanation. Along the lines I set out with my proposal or almost equivalent by your suggestions. BTW, better extend a little the simple solution, than the (equivalent) combined doors solution. Nijdam (talk) 21:29, 15 September 2012 (UTC)
Nijdam, if you do that, I will ask an uninvolved admin to drop a discretionary sanction on you for inserting Original Research. Everybody has got to stop putting their own explanations, formulations, distillations and extrapolations into this article. If it isn't published and you want to include it - go write a paper and get it published in an academic journal. Elen of the Roads (talk) 14:18, 16 September 2012 (UTC)(sorry Nijdam. see below Elen of the Roads (talk) 12:21, 17 September 2012 (UTC))
I'm taking Yoda Nijdam at his word here. He keeps saying that all the simple solutions are erroneous, and now he's saying that him and you can develop one that isn't erroneous. Maybe he's right, and maybe you can, but you can't put it into the article, because it would be WP:OR. All the versions of the problem, and all the explanations and solutions, must be sourced to an outside source. If anyone (not just aimed at you or Nijdam) starts developing their own solutions on talkpages and tries to insert it into the artice, I will ask for a WP:AE enforcement under the discretionary sanctions that this article is subject to (see the top of this talkpage). Elen of the Roads (talk) 22:20, 16 September 2012 (UTC)
I do agree it was a bit Queen of Hearts, and I apologise as it seems that what Nijdam meant was not the way it came over. I have struck my initial edit to him as a result. I have left the general statement about sources. I'm seeing a huge pile of OR on this talkpage and while I do understand that mathematicians like to draw diagrams for each other, the danger that the OR gets into the article is always there. Elen of the Roads (talk) 12:21, 17 September 2012 (UTC)
Thank you. It is all very frustrating. One of the 'problems' here is that people have generally been civil and committed to WP policies, which is to our common credit. Both sides of the original dispute claim to have logic and WP policy on their side. Martin Hogbin (talk) 08:52, 18 September 2012 (UTC)
Explanations, not solutions
Someone pointed out that if only people wouldn't use the word "solution" all the time, we wouldn't need to spend so much time arguing here.
It's so simple: just don't call a simple solution a *solution*. Call it an observation, insight, argument, guide to understanding, or explanation why one's initial feeling (that there's no need to switch) is incorrect. If you can, add the extra line (perhaps in parentheses) which completes the argument, when an argument can be easily strengthened. Moreover, let's give a simple Bayes solution, sorry, explanation, among the collection of simple arguments.
The whole mess starts because people decided to create a contrast between what they see are simple solutions which are mathematically incomplete, and the one and only true mathematically correct solution via a lengthy computation of conditional probabilities from first principles (thus not even making use of Bayes theorem; instead reproving it in this special case!). Morgan et al were responsible for creating this false image.
Three examples:
Simple argument 1. (Strategic thinking). If you always switch you'll win the car 2/3 of the time (assuming the car is equally likely behind any of the three doors) is a perfectly true statement, and extremely helpful to people first meeting MHP. Add to it the observation than 2/3 can't be beaten and there is no longer any reason to consider staying in any situation at all.
Nijdam, if you like one could add, perhaps parenthetically (since only for the cognoscenti), "therefore, under symmetry, the chance of winning by switching given contestant's choice was door 1 and host opened door 3 equals the overall chance of winning by switching, 2/3". But what I said before under (1) did not assume symmetry. I assumed less and got just as good a conclusion, ie, a compelling argument for "switch". Richard Gill (talk) 06:49, 17 September 2012 (UTC)
Simple argument 2 (Devlin fixed). The car is initially behind your door with probability 1/3 and this is not changed by the host opening a door to reveal a goat because we know he's certain to do so. Being informed which door is opened can't change it either, by symmetry.
This *is* the combined doors solution (host offers you Door 1 vs Doors 2+3), but now watertight as regarding probability statements (host's specific choice is uninformative). Devlin missed one step in his argument. Richard Gill (talk) 06:49, 17 September 2012 (UTC)
Before others are explaining what I think: combining doors is misleading, because by combining the not chosen doors and stating the probability is 2/3 they hide the car, is no more than saying each door separate has probability 1/3 on the car. I do not see in what way this information is helpful in understanding. Opening one of these doors, say number 3, does not change this probability, but implies that the new probability is now 0 for this door. But there is in principle no guarantee the sum of the original probabilities has the same value as the sum of the new probabilities. And that's the strong suggestion imposed by this explanation. So it just comes down to either show that the new probability for the chosen door is 1/3, or to show the probability for the remaining door is 2/3. The combining bears no significance whatsoever for this last step, meaning it bears no significance at all. (Or it must be to tell the reader that 1/3 + 1/3 = 2/3.)Nijdam (talk) 08:30, 18 September 2012 (UTC)
For the purpose of finding coherence in this thread, it may be noted that the foregoing post "Before others are explaining what I think..." was posted after several responses below. ~ Ningauble (talk) 21:13, 18 September 2012 (UTC)
The contestant says: Now together with the host I shall open door 2 and door 3, and I shall win the car if it is behind one of them. So I have exactly the same chance of winning as if I would throw a die winning with "1,2,3 or 4". And I have thrown a die to decide which doors to open. And if there is somebody who disbelieves that I have a 2/3-chance I shall play the game with closed eyes only saying "door 1", and if the host asks if I switch: "Yes". - Or I say to the host: "I choose door 1 and shall switch. The rest you may do yourself." - So the "only one" who "decides" completely whether I shall win by switching or not is the die at the very beginning who "tells" me which door I have to choose: With two outcomes I shall win, with one lose: p = 2/3.--Albtal (talk) 10:00, 17 September 2012 (UTC)
Yes. Overall, the always-switcher wins 2/3 of the time, the always-stayer 1/3 of the time; these are exactly complementary events. Misleading? No help in understanding? Nijdam's opinion, following Morgan et al, is that it is misleading and no help in understanding. He maintains that you should be focussed on the success rate within those special cases when the contestant chose door 1 and the host opened door 3. I think that this is a second-order concern. It's easy to see that you can't beat 2/3 overall, so you can't do better by altering your behaviour in special circumstances. Therefore it's a waste of time to compute conditional probabilities - you'll find out that they too always favour switching. But a heap of reliable sources think this distinction is absolutely crucial, so we are still wasting time fighting about that here, and to be sure, the wikipedia article will have to pay some attention to this issue. Richard Gill (talk) 13:04, 17 September 2012 (UTC)
Yes; and this strange "second-order concern" was initialised by a purposeful misquotation of Marilyn's statement of the problem in the paper of Morgan et al.: They removed "say number" for door 1 and "another door, say number" for door 3 (see above).--Albtal (talk) 14:25, 17 September 2012 (UTC)
Complementary hints (see above: The contestant says...): 1. Does it matter who opens his door first? Yes, the host (with a goat). But why does the "combining doors argument" doesn't hold if the host opens his door without knowing where the car is? Because then it is possible that the host opens the car which changes the chance too for the case where he shows a goat. It's exactly my knowledge that there can be no additional informations during the game which makes the "combining doors" argument true. 2. Can the contestant who closes his eyes have other chances than those who doesn't? No, because the one with open eyes has the same information as the one with closed (both knew before what will happen). 3. If the host "does it alone": Should he first open a door with a goat? Yes, so that nowbody can say quickly (see also 1.): "This is not the correct game ..." 4. We see: We have to formulate carefully ("symmetry" or "better"), so that the simple solution is watertight even if considered hairsplittingly. But I think that the reader has no problem with the fact of "symmetry" (but maybe with the formulation about it like "What I knew before that it will happen cannot change my chance if it really happens" ...). I think that there is enough know-how here in the "Richard Gill Thread" (and no problem to find "sources") to "optimize" simplicity combined with "watertightness". Maybe my remarks can help. And checking my arguments for what I regard as the main problem with MHP may also help. Good bye.--Albtal (talk) 20:40, 17 September 2012 (UTC)
Simple argument 3 (Bayes, odds form). The prior odds on the car being behind doors 1 and 2 are 1 to 1. Now, if the car is behind door 1 (chosen by the competitor) there is a 50% chance the host will open door 3, but if the car is behind door 2 there's a 100% chance the host will open door 3. So the host is twice as likely to open door 3 if the car is behind door 2 than if it is behind door 1. So the odds of the car behing behind door 2 to being behind door 1 change to 2 to 1 after the host has opened door 3. Richard Gill (talk) 19:36, 16 September 2012 (UTC)
The Brits (and Ozzies and... ) love gambling. The Dutch think it's a sin so they don't do it. Instead they invented the Dutch book in order to cheat foolish English sailors dropping by Dutch ports. Richard Gill (talk) 06:49, 17 September 2012 (UTC)
Simple argument 1: If you always switch you'll win the car 2/3 of the time (assuming the car is equally likely behind any of the three doors, {I'd add} and because the solution cannot depend on the door numbers, also in the specific case of the player he'll win the car with probability 2/3. {I wouldn't even mind to call this a solution}.
Simple argument 2: If you mean the combined doors idea, I have objections, as the basic idea behind combining door is nonsense. Combining doors with each 1/3 chance gives a combined chance of 2/3, but is not very helpful in explaining why a door with chance 1/3 also should have chance 2/3. The combining does not explain anything. However, your addition works as well for the simple solution, calculating the chance for the remaining door (after the host opened door 3) as the complement of the chance for door 1.
Simple argument 3: Seems rather complicated for the common reader. (Or are British accustomed to think in odd terms, I mean terms of odds.)Nijdam (talk) 20:04, 16 September 2012 (UTC)
The reader understands that two doors have double chance compared to one door. Proven by maths, proven even by simulation. But if you address "additional info on the car hiding door", and if you "assume" that the host does never keep *secrecy* regarding the car-hiding door, and that he definitely has opened his *strictly avoided* door because the car is behind his *preferred door*, indicating that his second still closed door is most likely to hide the car, then you are in an outlandish scenario, quite "outside" the MHP. Did you forget that the host keeps *secrecy* regarding the car hiding door, as per the sources? Gerhardvalentin (talk) 20:45, 16 September 2012 (UTC)
Dr Gill, I do believe as I've said that distinguishing explanations from proofs is extraordinarily helpful. The simple explanation of why Marilyn Vos Savant's answer was wrong (or was only right if one included something she didn't say) should immediately cause most people to see why switching is an advantage if Monty is trying to ensure that the game goes out on a bang. It should also immediately alert people as to why the exact specification of the problem is so important.
Nijdam, yes I do think most British people think in terms of odds rather than probabilities. Possibly based on our long history of gambling on anything worth a bet. I also agree with you that the combined doors explanation (while it gives the same answer) does not ultimately take the problem too far forward. Elen of the Roads (talk) 22:34, 16 September 2012 (UTC)
The core of the conflict
What is the "core" of the conflict? It's on diametrically opposed sources. From my humble horizon view, imho it's the disputed validity of a claim made by (for example) Morgan et al. (MCDD), that – leaving the world famous *paradox* outside – the answer to the question of Marilyn vos Savant, who surely excluded a host that does not observe *secrecy* regarding the car hiding door, and who pictured a one-time problem, that the answer to her question "should you switch?" has to be based on *nothing other* (!) than on the exact knowledge of the exact actual rate of "probability to win by switching" (pws) in each and any "actually single game" (in each and any "repeated show"), and that a correct solution / mathematical proof of "pws" not only has to pay regard to the fact that just "one" door actually having been opened by the host to show a goat, but also and especially *must* pay regard to "which one" of the host's two doors he actually has opened ("say," the leftmost or the rightmost). The reasoning of MCDD was that additional information on the actual location of the car *could* have been revealed by "which one of his two doors" the host actually did open. And therefore a valid mathematical evaluation has to condition on "which one" of his two doors the host has opened. The sole rationale was the peradventure of the rate of "pws" for each and any game-show:
Some "different" pws could be *supposed* (host's bias could yield pws=1 if he exceptionally did open his strictly avoided door "because" his favoured door actually hides the prize), e.g.
As such "bias" will forever be unknown, not able to give any answer to any *actual* situation other than "pws=2/3", but claiming this rate "could differ" if you *assume* sth, giving a rate corresponding to your respective *assumption* (not being able to address the actual situation though), so quite unable to give any valid pws for the "actual" game — other than 2/3.
But, irrespective of "different pws", other authorities (especially textbooks in teaching and learning conditional probability theory) confirmed that you can get a conditional solution, and if you want to get a full "valid conditional solution" you *must to pay regard* to "which one" of his two doors the host just has opened. Although unable to show sensuousness other than to comply with being high principled. Very suitable for educating purpose in conditional probability theory. And forgetting about other arguments to "always swith", in face of fully valid quite other approaches. That never was intended by those textbooks on conditional probability theory.
I challenge all the participants to solve the following simple exercise:
Imagine the MHP scene (as it seems not to be clear, when I mention the MHP scene, it is the standard MHP, say according to Kraus & Wong. No need for another version) . The car is with probability p=1/3 behind the chosen door 1. The host has opened door 3 showing a goat, so we may state that now the probability for door 3 to hide the car is q=0. Calculate the probability for door 2 to hide the car. Simple, isn't it. Who dares? Nijdam (talk) 10:57, 12 September 2012 (UTC)
calculate? If you "assume" that the malevolent biased telltale host has opened door 3 showing a goat only BECAUSE the car is behind the chosen door 1, then the probability for door 1 to hide the car was never 1/3 (from the host's perspective), and the probability for door 2 to hide the car "for you" is zero now.
But if you "assume" that the benevolent biased telltale host has opened door 3 showing a goat only BECAUSE the car is behind his strictly favoured door 2, then the probability for door 2 to hide the car is 1.
So you can assume a whistle-blowing malevolent biased host, who by his action does not keep secrecy regarding the car hiding door, and you can assume a whistle-blowing benevolent extremely biased host, who by his action also does not keep secrecy regarding the car hiding door. Once more: secrecy regarding the car hiding door.
You need not calculate, the probability is zero or 1 in those cases, it's that easy.
Please don't forget to note: this is true but only within the field of your (maybe false) private assumptions, and nowhere else.
But if the host keeps absolute secrecy regarding the car hiding door (Norbert Henze e.g.) then you need no "q", nor conditional probability anymore. You CAN use it, but it is not necessary to do so. It's completely futile, for any "closer probability" you unavailingly expect to get, will be fatuous. Will be illusive and phantasmal. For you always will just get what you WANT to get, completely offside reality, completely offside the famous paradox. A mere subject of training conditional probability maths. So: All you will ever "know" is the average probability. Any "closer actual predictand" forever is illusory, a completely fruitless struggle. All senseless and in vain. You never will "know" better.
So my proposal: Such meaningless offside stuff that never will be a tangent to the clean paradox, has to be shown in very "later" sections, while just in the beginning we should show the Bayes theorem in the "posterior odds = prior odds times likelihood" version, easy o grasp by the reader, followed by the "forgetful" host who, by showing the car in 1/3 eliminates the chance to win by switching in that 1/3, while in the remaining 2/3 the car with equal probability is behind one of the two still closed doors.
And btw: Imo Marilyn for sure was definitely excluding a host who does not keep secrecy regarding the car hiding door, and so her answer was completely right, but she did not mention that "naturally" (without saying - certainly - for sure - of course) the host keeps / kept secrecy regarding the car hiding door. MvS was not wrong. Wrong is only the "unnatural" creation of quite another perverse scenario, where the host is NOT keeping absolute SECRECY regarding the car hiding door, as per MCDD. Gerhardvalentin (talk) 12:59, 12 September 2012 (UTC)
I challenge Nijdam and all others who make similar arguments to accept the following quite reasonable ground rule: Use the far superior Krauss and Wang problem description, not the known-to-be-flawed Marilyn vos Savant description. If your arguments are sound, they shouldn't depend on purposely picking a flawed, ambiguous description just to make things turn out the way you want them to.
Here, once again, is the Krauss and Wang problem description:
"A formulation of the Monty Hall problem providing all of this missing information and avoiding possible ambiguities of the expression "say, number 3" would look like this (mathematically explicit version):
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. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one randomly. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door. Imagine that you chose Door 1 and the host opens Door 3, which has a goat. He then asks you “Do you want to switch to Door Number 2?” Is it to your advantage to change your choice?"
Source: Krauss, Stefan and Wang, X. T. (2003). "The Psychology of the Monty Hall Problem: Discovering Psychological Mechanisms for Solving a Tenacious Brain Teaser,'" Journal of Experimental Psychology: General 132(1). http://www.usd.edu/~xtwang/Papers/MontyHallPaper.pdf
Given the above quite-reasonable ground rule, I calculate the "probability for door 2 to hide the car" in Nijdam's statement above as being 2/3. --Guy Macon (talk) 17:33, 12 September 2012 (UTC)
As I added above, Of course I mean the standard (K&W) version. I'm very much interested in your calculation, not particularly in the numerical answer. Nijdam (talk) 21:16, 12 September 2012 (UTC)
+1 Yes. And Norbert Henze adds: "Bei allen diesen Betrachtungen ist natürlich entscheidend, dass der Moderator die Autotür geheimhalten muss, aber auch verpflichtet ist, eine Ziegentür zu öffnen." ("For all of these considerations it is naturally essential that the host must keep secrecy regarding the car-hiding door, and is also obliged to open a goat-door.") Imo this explicit reference to "secrecy", once and for all excludes any imaginable difference of "before and after".
And this very secrecy excludes the malevolent host also, who offers a switch "only if" the guest should have selected the door hiding the car. Imo "secrecy" is a very important reference to picture the scenario where the clean paradox is appearing. Gerhardvalentin (talk) 17:56, 12 September 2012 (UTC)
Nijdam, one "classic" way of computing this is an appeal to symmetry. Its classic, because it predates Blaise Pascal. Symmetry arguments predate the concept of frequentism, and certainly predate Bayes. Symmetry worked just great for coin tosses, and for dice roles. Frequentism was developed because symmetry arguments are not enough to solve many problems in probability. The reason that Bayesian methods became popular is that frequentism itself is incomplete and inadequate. But for problems possessing symmetry, you do not need to appeal to these modern developments; the classical symmetry arguments work just fine. (There is a book: "The Science of Conjecture: Evidence and Probability before Pascal", Franklin, charming read. Does not talk about MHP, but does examine ancient notions of probability). linas (talk) 23:06, 12 September 2012 (UTC)
OK, let's assume full symmetry. I will in particular assume that the contestant makes his initial choice uniformly at random. By full symmetry, the actual numbers on the doors are irrelevant when having to choose between switch or stay: the choice is between staying with initial choice and switching to the other door left closed by the host. Now, the car is behind the door he chose first with probability 1/3, by symmetry. Since we may ignore door numbers we have no new information by which that chance might have to be revised (another door wiill be opened revealing a goat, anyway). Therefore the other closed door hides the car with probability 2/3, he should switch. He'll win with probability 2/3. (And of course the actual numbers on the three doors concerned don't change the chance of winning by switching: but we knew this right from the start, by symmetry). Richard Gill (talk) 12:03, 15 September 2012 (UTC)
Nijdam, yes it is. If you are unable to perform these calculations on your own, just say so, instead of posing "challenges" and muddying the debate. But really, if you want to learn how to compute using probability theory, this is the wrong place to ask for help. If you don't have the patience to teach yourself from books and online sources, try enrolling in classes at the local community college. I don't know that they'll be very good, but it might get you enough of the basics, and allow you to understand some of what is going on here. linas (talk) 03:59, 19 September 2012 (UTC)
Please not Kraus and Wang
The Kraus and Wang formulation is too heavy for the general reader and unnecessary for simple and at the same time mathematically rigorous and complete solutions.
Marilyn asks for a decision, not a probability. We know the right decision is "switch". The first thing to get across to the new reader is why. There are many good reasons why. Not one.
Conditional probability is one way to give an answer, and it requires many assumptions which not everyone will be happy with, and it requires mathematical formalism, which not everyone is happy with. There are easy reasons why which do not require conditional probability at all and which make less assumptions. They even make conditional probability superfluous.
For instance: if the car is initially equally likely behind each door and you'll always switch you'll get the car 2/3 of the time. There is no way to do better than 2/3.
What more needs to be said?
If you actually want mathematical proof of the statement "no way to do better" you might like to consider Gnedin's dominance observation: however you and the quizmaster play there's always a door such that you'd miss a car behind that door. (I don't think it would ever occur to 99.9% of our readers that this is a fact needing proof, but a mathematician does realize that also the "obvious" needs to be proved).
And for all those lovers of conditional probability out there, it follows from what I said so far that conditional probability always favours switching. No need to compute anything. (Of course in the symmetric situation, no need either).
MHP is supposed to be fun, it's supposed to be entertaining. It is not supposed to be an instrument of torture for reluctant students of Probability 101. Richard Gill (talk) 19:45, 12 September 2012 (UTC)
Yes! Regretfully, Nijdam opposes (rather unconvincingly, I'd say): "Firstly, let us not divert from the probabilistic approach. That definitely is the approach Wikipedia will show its common readers. I'm not interested in a game theoretical analysis, from which I BTW am not convinced it is appropriate." Boris Tsirelson (talk) 19:59, 12 September 2012 (UTC)
The MHP is quite general considered to be a probability puzzle, a teaser. The vast majority of sources do so. And the controversy has not been about presenting it different. And as such it is not solved by this argumentation (as Richard writes some lines above): if the car is initially equally likely behind each door and you'll always switch you'll get the car 2/3 of the time. There is no way to do better than 2/3. If you know your sources, Richard, you know this way of reasoning is in itself correct, but does not address the problem. You don't need sources, of course, just simple logic will do. Nijdam (talk) 21:31, 12 September 2012 (UTC)
(EC) I find the treatment given by Gill in his eprint [9] very nice, its cleared a lot of questions up for me. Various different attempts have been made to define the problem; vos Savant, Kraus and Wang and I'm sure others. These should be given appropriate coverage according to WP:WEIGHT. The lead is not the place to give a rigorous problem statement a simple one is appropriate here together with a note to say that there are different formulations discussed later in the article. If a particular formulation has problems then that should be mentioned.--Salix (talk): 21:46, 12 September 2012 (UTC)
Two separate issues here. You are discussing whether Kraus and Wang should be in the lead (which I happen to favor, but acknowledge that there are good arguments against doing that). The other issue is whether Kraus and Wang should be used as the definition when someone wants to engage in an involved mathematical argument on this talk page. That one is a no-brainer. We are seeing argument after argument based upon assuming all sorts of weird Monty behaviors, and when anyone points out how silly this is the cry "but it is allowed by the (flawed) description!!" is raised. If you need to invoke a description that you know to be ambiguous in order to argue your position, and if your arguments disappear in a puff of smoke when you start with the more rigorous Kraus and Wang description, then you are just engaging in sophistry rather than making a logical and evidence-based argument. --Guy Macon (talk) 01:57, 13 September 2012 (UTC)
Guy, I don't like citing my own publications on wikipedia, that's why I did not give a reference to one place where I know that this elementary and correct argument is written out. But see also the recent papers by A.V. Gnedin. My description is not *ambiguous*. it's minimalistic. I state exactly what needs to be assumed in order to draw correctly the conclusion I want to draw. My argument has logic and evidence behind it, 100%. Why make superfluous and discutable assumptions? What is going on here is the fight between those who want to situate MHP *primarily* in a first year university probability course, or other academic environments where probability and statistics are primary, and those who want to see it *primaily* on Wikipedia as a popular brainteaser. Secondly the academic probability focus blinds everyone to the decision theoretic component of the problem. Thirdly, as a mathematician, I am bound to say that it simply is not true that the only way to mathematically respectably completely solve MHP is through Bayes theorem (much as I love Bayes!). Some other editors here seem to think so. But there is no reliable source behind that. It is merely an opinion. I know that on Wikipedia the truth is irrelevant, but I have to say also that it is an opinion which is false. Richard Gill (talk) 07:30, 13 September 2012 (UTC)
I was wondering whether there was a description that doesn't have the ambiguity and nonsimulatability of the vos Savant description without the rather turgid prose of Kraus and Wang. Would someone uninvolved be so kind as to post the description Dr. Gill mentions above, and any other candidate descriptions that you know of? --Guy Macon (talk) 08:08, 13 September 2012 (UTC)
Description: Perhaps it would suffice to mention that the famous "not 1:1 but 1:2" paradox clearly exists in a simple scenario where it exists, but that there are other scenarios where the paradox never exists, those variants being usually used in textbooks on conditional probability theory. Gerhardvalentin (talk) 16:33, 13 September 2012 (UTC)
If you want to simulate MHP you are going to have to simulate the quizmaster's behaviour and the player's behaviour. There are two components on the quizmaster side: how is the car hidden, and how does the host choose a door to open if he has a choice? His second choice could in principle depend on where the car is and which door was chosen by the player. There are two choices on the player's side: how he picks his initial door, and what his behaviour should be (switch/stay/ toss a coin to decide...) when offered the option to switch. His second choice could in principle depend on which door he chose first and which door was opened by the host.
Moreover, if you simulate MHP, you have to decide what it is you want to measure. Are you interested in how often a player who always switches wins, overall? Or are you interested in how often a player wins who always switches, split up according to the six different situations he might be in when he has to decide whether to switch or stay?
In mathematics we can often completely solve problems which are not completely specified. But if you want to simulate some problem you will have to specify everything. You may think you've solved a problem by simulation (actually you just get insight into the problem), but really you have only got insight into the special case which you simulated.
Here are three solutions and accompanying assumptions. The more you want to conclude, the more you need to assume.
(1) In order to be able to claim that always switching wins the car 2/3 of the time, it suffices to assume that your initial choice is correct 1/3 of the time.
(2) In order to be able to claim that always switching wins the car 2/3 of the time and that that cannot be improved, it suffices to assume that any of your three possible initial choices would be correct 1/3 of the time.
(3) In order to be able to claim that always switching wins the car 2/3 of the time that a particular door was initially chosen and a particular door was opened, whatever two doors these are, it suffices to assume that any three of your possible initial choices would be correct 1/3 of the time and that when the host has a choice, he equally often makes either choice.
In other words, the more you put in, the more you get out. The three statements that I made cover all the usual assumption sets written down by different writers (different writers use different sets of assumptions). All three statements can be proved with simple logic which your grandmother can understand; no need to appeal to Bayes theorem. But sure, teachers of elementary probability courses like to introduce the fullest set of assumptions and get the strongest conclusions, and they like to do this using Bayes' theorem. Nothing wrong with that. It's just not the only way. It's not the only way to formalize the problem, and it's not even the only way to solve the problem so formalized.
Steve Selvin, who wrote the first two published papers on the Monty Hall problem, in effect derived solution (1) to problem (1) and solution (3) to problem (3) in his two papers without even remarking on the difference. A discussant (I think his name was Bell) of the infamous paper Morgan et al. pointed out that (3) followed from (1) by symmetry and hence that there was not a big deal. The present fight we are having here goes right back to the origins of the problem and though it is a big deal for some academics it is not a big deal for most amateurs, and not a big deal for many academics either. Several probability textbooks even introduce the problem with the specific door numbers left out altogether, since by symmetry the actual numbers painted on the doors are irrelevant: the only thing that is relevant is how each door is used by the host and his team, and by the player (of course, that's an assumption: made by many, not by all). Richard Gill (talk) 13:26, 13 September 2012 (UTC)
It's easier said that the door numbers are irrelevant, than understanding what that should mean. The doors are distinguishable, that's for sure. Do we have to number them? Well, no, we could paint them in different colours, label them B, C an D, or just speak of the left, the middle and the right door. All equivalent to numbering the doors. Do the numbers play a role? Yes, the numbers let us distinguish between the six possible situations. And as the player is offered to switch after the host has opened a door with a goat, the player is in one of the six situations. Do the numbers play a role in the solution? No, in the sense that the solution is independent of the door numbers, i.e. for all combination of chosen and opened door the solution is the same, provided some assumptions are made like Kraus and Wang. Nijdam (talk) 14:44, 13 September 2012 (UTC)
No assumptions necessary. We do not need to distinguish. Only teachers in probability theory have to do that. Marilyn's simple problem pictures a one-time problem. The standard is about one case, where the host obviously did keep secrecy regarding the car-hiding door. Not multiple cases. But textbooks on conditional probability theory and maths-teachers do not talk about MvS, but about some outlandish contradictory nonstandard variants where the host did NOT keep secrecy regarding the car hiding door. But this is no more the simple problem as per MvS. Gerhardvalentin (talk) 16:05, 13 September 2012 (UTC)
Response to Nijdam: What does it mean? It depends on what you mean by probability. If you are a Bayesian and have no prior information distinguishing one door from another, and also no information concerning how the quizmaster chooses a door when he has a choice, then for you, the problem is symmetric in the door numbers. This makes the actual numbers irrelevant, all that is important is their roles.
A clever Bayesian with no prior knowledge can reduce a problem by symmetry before introducing probability distributions for everything. No need to use symmetry to figure out what all his priors are, if symmetry means that some aspects of the problem are independent of his task at hand (namely, to choose whether to switch or stay from the door initially chosen to the door left closed by the quizmaster). This is why Persi Diaconis fully supported Marilyn vos Savant's solution.
As I wrote, various textbook authors (highly reliable teachers of mathematical statistics and probability) solve MHP in exactly this way! One can completely take the soul out of a fun problem by being overly pedantic. In some contexts it is important to be as pedantic as possible. Wikipedia articles are for the general public, not for aspiring maths PhD's.
Please note: symmetry in MHP is completely natural if you have a Bayesian inclination. Symmetry in MHP is not completely natural if you have a frequentist inclination. A frequentist needs to be told explicilty that the quizmaster makes his choice by tossing a fair coin in order to be justified in using the symmetric solution. A Bayesian has no idea how the quizmaster makes his choice so for him, in one run of the game, either door would be equally likely opened if the quizmaster had a choice.
That's another good reason why it is unwise to write out all possible mathematical assumptions at the start. Not everyone thinks about probability in frequentist terms, and not everyone thinks about it in subjectivist terms. So the "usual" full list of assumptions make a lot of sense to some people, and no sense at all to others. Still, both frequentists and Bayesians like to try to solve MHP. Richard Gill (talk) 16:11, 13 September 2012 (UTC)
Richard, in my comment above you can see that I widely support your argumentation. But although "the topic which is going on here" has been very time-consuming over years it is a small side issue of the MHP. The main problem is that the crucial rule which leads to a 2/3-solution was missing in the version of Marilyn vos Savant: This rule is equivalent to: The contestant has to determine two doors, of which the host has to open one with a goat. And in the New York Times article of Juli 21, 1991, by John Tierney (see References) not only Martin Gardner and Monty Hall himself, but also Persi Diaconis said: The strict argument would be that the question cannot be answered without knowing the motivation of the host. And in the task setting of Steve Selvin I cannot find this rule at all; and Monty Hall never had offered a switch. (See my other comments for further information; see also the comments of Trovatore)--Albtal (talk) 17:07, 13 September 2012 (UTC)
There are two aspects of host behaviour which were not initially specified explicitly by Vos Savant. One was important. It was the fact (only implicit in her statement) that Monty always opens a door revealing a goat. That's what Persi meant by saying that we need to know the motivation of the host. He meant, we need to know why Monty opens a door. The other aspect is how he chooses a door to open when he has a choice. Since we are not told anything, for a subjectivist both his choices are equally likely. For a frequentist, we have to admit we don't know, and weren't told anything. Advice to a freqentist: pick your initial door at random and always switch. You'll get the car 2/3 of the time. You can't do better in a minimax sense: whatever strategy you follow, the game organizers can always stop you winning more than 2/3 of the time. Richard Gill (talk) 10:18, 14 September 2012 (UTC)
One was important. Yes, Persi Diaconis, Martin Gardner and Monty Hall meant that this condition was missing in the task setting. And I can't see another reason for the furor around the MHP. And it is my opinion that this furor wouldn't exist if Marilyn had published one of "the very small percentage of letters" whose writers where convinced that the furor resulted from the fact that this rule was missing; and if she had formulated this condition not only in a later explanation, but in a clear new task set. So the wrong formulation together with the claim of the 2/3-solution was going around the world, and the "Monty Hall Paradox" was created. And many proponents of the 2/3-solution up to today are convinced that the blank fact that the host opens a not chosen door with a goat leads to this solution. How is this possible? - I think the key is the fact that it seems to be a good explanation to "simulate" the situation. But they don't realise that the rule which is not part of the problem setting implicitly slips into the "proof". So they are able to formulate a problem which "intuitively" has a fifty-fifty-solution (which really is the best solution) but paradoxically leads to a 2/3-solution. I know that it is not the time here to point to this aspect. But there are thousands of sources which confirm this thesis. And this issue is not done by replacing the original by a "standard problem" - if possible within a footnote.--Albtal (talk) 13:10, 14 September 2012 (UTC)
I carefully explained why a Bayesian with no prior knowledge has symmetry (symmetry coming from his ignorance of the host's bias, not coming from his kniwledge that the host uses an unbiased randomizer) and hence why he may reduce the oroblem by symmetry before writing down what his prior is over everything. I think that many non-professional mathematicans who favour the simple solution are thinking in this way. "The door numbers are irrelebvant because I wasn't told otherwise". Ignorance! Subjective probability! One cannot discuss formulations and solutions to MHP without agreeing what probability means, for the present purposes. Different sources very clearly (to me) are working withins different paradigms, taking it for granted that everyone thinks the same. Different editors here, too. Richard Gill (talk) 06:35, 14 September 2012 (UTC)
If the contestant who only knows the basic rule of the game asks us (before the game starts or just before his final choice) for his chance if switching, I think it is completely true if we say: If now there would be 3000 shows in the world with contestants in the same situation as you, about 2000 will win if they switch. (To be on the safe side we should add here that the contestants should make their first choice (uniformly) randomly.) The contestant now says: Well, so I have a 2/3-chance with switching. I'll take the switch. That is he takes the "average probability" for the probability in his single situation revealing a special understanding of probability. Isn't it game theory which confirms his view?--Albtal (talk) 08:07, 14 September 2012 (UTC)
Not everyone who couches things in frequentist terms is a frequentist! There is no problem with the Krauss and Wang clarification from a Bayesian POV. You need a bit of justification for getting there from a more ambiguous description, but you do get there. Explaining that in a later section would be just fine.
To the stereotypical subjectivist Bayesian, K-W is not a description of reality, but of his Monte Carlo simulation of the posterior of whatever quantity is being computed. To some Bayesians, such as myself, it also describes the situation where the prior is "correct", or to use a less loaded term, (well-)calibrated. It is a privileged situation, and worth special consideration because in it the argument has more force than is obtainable on subjectivist grounds alone (not only "if you are rational as I define it, you must reason this way", but also "if you reason this way, it almost surely gives the best results in the imaginary infinite long run, even though I can guarantee nothing about the situation you are actually in".. it's complementary, really). The simulation is also useful for considering other situations, to see how sensitive the final decision (switch or not?) is to the choice of prior (answer: as insensitive as you could possibly hope for). -- Coffee2theorems (talk) 08:59, 14 September 2012 (UTC)
@Richard: I have no objection to your clever Bayesian. As you said yourself, he will reduce a problem by symmetry. And in presenting his solution he will show how he reduced the problem. But if he leaves this out, his solution will not be complete. Nijdam (talk) 21:21, 13 September 2012 (UTC)
There is a problem with taking such a shortcut instead of modeling the problem in the most direct way: it limits what you can say about the situation to what is expressible in your impoverished model. Without the host explicitly choosing the door with some probability, you cannot consider a host who does otherwise, and that is a problem because then you cannot talk about some of the cognitive psychology material, which explains why people have problems with the MHP. That "why" is a central topic, and the readers should be able to understand that discussion.
Consider the current "Sources of confusion section", particularly the material referenced to Falk. When you get to analysis at that level, I don't see how you can possibly make do without a full probability model, or why you should want to. Another problem with such simplified models is that you then have to explain how it corresponds with the original problem (as Nijdam says above), whereas the same is dead obvious with a full model. -- Coffee2theorems (talk) 06:00, 14 September 2012 (UTC)
Sooner or later we will want to discuss full probability models, I agree. But so what? The question is whether the whole article is to be dominated by the idea that there is only one proper complete mathematical specification of MHP and only one proper complete mathematical solution. This is a point of view encouraged by the way it is treated in many introductory probability textbooks but there are plenty of reliable sources who take other approaches. (I recall Nijdam saying that he insists that MHP is done "his way" on wikipedia because he wants to prevent his students in his introductory probability course from finding a "wrong solution" on wikipedia. I don't think Wikipedia has to suffer because he couldn't explain to his students *why* he formalizes the problem in a particular way and *why* he solves it in a particular way.)
Anyway, whatever model you take you have to explain why it corresponds to "the original problem". For the purposes of Wikipedia, "the original problem" (in my opinion) is the verbal question posed by Marilyn Vos Savant ("would you switch"?) with the context given by her together with the implicit understanding that the host always opens a door revealing a goat. Thus for any mathematization you have to explain why it corresponds to the original problem.
There is not one correct formalization of Vos Savant's problem. It's a fact that different authors make different assumptions. Wikipedia editors are going to have to live with this. Most authors make no attempt whatever to motivate their assumptions. Some authors just write down assumptions so that the problem can be solved in the way they want to solve it. Certain assumption sets have become traditional in introductory probability courses.
The more assumptions you write down, the more work you have to do to motivate them, and the more restricted is your solution to a special case. On the other hand, the more assumptions you write down, the stronger the conclusions that you can draw.
A mathematician can analyse a not completely specified probability model just as well as a fully specified model. It just means that some parameters are treated as unknown variables instead of being assigned specific values. Richard Gill (talk) 09:42, 14 September 2012 (UTC)
To avoid confusion, I'll clarify what I meant by "the original problem": it's the pre-formal description of "the MHP", whatever that is; nothing is intended to be implied about history by the word "original". The version we should present most prominently is the one that most informed people today think the MHP is. Krauss and Wang claim that it is their version, and people do not cry foul when presented with simulations embodying those assumptions, so that sounds plausible enough.
Good terminology: (1) the original problem is the pre-formal version; and to be specific (I suggest): Vos Savant clarified with explicitly adding her implicit assumption that Monty Hall always opens a goat door (as he can, because he knows where the car is hidden). (2) The most common assumptions are the Krauss and Wang assumptions: car is hidden uniformly at random, Monty Hall chooses uniformly at random when he has a choice.
Your own suggestion that people think of the MHP in Bayesian terms (which is also plausible enough) also "obviously" implies the K-W assumptions for Monte Carlo simulations. Weren't you the one who has unbounded faith in people's intuitions being justified when they claim that something is obviously true and also happens to be true? If you don't accept it for the K-W assumptions but do accept it for the symmetry argument, you're being mightily inconsistent! (If anything, the latter is the one you should doubt, as people think the probability is 2/3 even when the symmetry is removed, so their reasoning actually doesn't depend on it and therefore in the K-W situation they have an unjustified true belief, not knowledge. My position is that we should strive to give people knowledge, not mere true beliefs.)
I don't suggest that people should think of MHP in subjective Bayesian terms. I simply guess that many people do. And we are talking about a popular brain teaser, so I am talking about ordinary folk without training in maths or probability. If we attack MHP from the subjective Bayesian point of view then the K-W assumptions are expressions of the complete absence of any knowledge concerning how the car is hidden and how the host chooses to open a door when he has a choice. I do not have unbounded belief in intuitions being true. Monty Hall itself shows that intuition often goes wrong. I'm saying that from the Bayesian point of view, ignorance implies symmetry. Symmetry implies that specific door numbers are irrelevant. Some people (some writers of text books) say so out loud, some people take this step silently. Why do the conditionalists only condition on initial choice and door opened? Why don't they also condition on the day of the week? Because in the pre-maths stage you choose, rationally, to leave the day of the week out of it. You can also choose, rationally, in the pre-maths stage to leave the door numbers out of it. There are text books which do it this way. The probability of winning by switching is 2/3, independently of the day or the week or the numbers on the particular doors initially chosen and opened. Richard Gill (talk) 12:51, 14 September 2012 (UTC)
I didn't mean to imply the "should" part. I guess "Your own suggestion that people are thinking of ..." would have been a better wording.
There's quite a difference between days of the week and door numbers. If you simulate the problem in the natural way, then your simulation will contain simulated doors with numbers (or equivalently, other unique identifiers), but it will not contain days of the week. Days of the week are not mentioned in the problem description and are clearly superfluous. You can't do away with the door numbers without losing the obvious correspondence between the simulation and the original problem (in the original you can distinguish between the doors; if you cannot do the same in the simulation, then obviously it's not a faithful simulation). You then condition on everything the simulated player knows at the decision making point, because that procedure is generally valid. If you do not condition on something, then you need to show why you can leave it out.
Door numbers are superfluous, that's part of the problem description; at least, as many people read it. Vos Savant said "say" door 1, "say" door 3. She told you that the door numbers are superfluous. She said: you pick a door, the host opens a different door. She gave the door which you chose the name "door 1". She gave the door which the host opened the name "door 3". She doesn't tell you whether door 1 is the left, middle, or right hand door, as seen from the audience! And she doesn't tell you whether door 3 is the left, middle, or right hand door, as seen from the audience! Richard Gill (talk) 19:10, 14 September 2012 (UTC)
You're missing the point. Distinguishability of the doors is implied by the problem statement, underlined by the explicit labeling of doors with distinct numbers. In a faithful simulation (≈ model), this distinguishability needs to be preserved, and that would remain true even if there were no door numbers in the problem statement. The simulated player knows which door he and the host picked because he can distinguish them. If he can't, then you've really got some explaining to do as to how your simulation corresponds to the MHP (do you assume a blindfolded player or marbles in a bag instead of doors or something? Isn't that a different problem?). Taking shortcuts based on untrustworthy intuition at the modeling stage just to make the proof marginally simpler is not a good practice!
You need to condition on all the knowledge of the simulated player if you're going to make an optimal decision relying on generally true principles only. If you fail to condition on something, you need to prove why it is allowed, because there is no general theorem you can point to. Note that I'm talking exclusively about simulations, because nothing can ever be proved about reality, only about models of reality, and the idea of "conditioning on everything you know" is only sensible in simulations (≈ models), not reality. -- Coffee2theorems (talk) 07:58, 15 September 2012 (UTC)
You cannot prove things about reality, only about models of reality. You have to take the correspondence of a model with reality as "intuitively obvious", preferably bolstered by a heavy dose of empirical success of the model (sadly, not available here, but is applicable to more important things like the proof of conservation of energy). Since the correspondence is not proved but largely taken on faith, it is best if it is made as clear as possible, especially in cases like MHP where intuition has been shown to fail in multiple ways. Once bitten, twice shy. -- Coffee2theorems (talk) 14:11, 14 September 2012 (UTC)
Exactly. Now in the case of MHP, maths supplies us with a range of assumption sets and corresponding range of conclusion sets. It will be a matter of taste for any particular reader ("consumer" of maths) which assumption set they are prepared to buy. Which one corresponds best to reality as they see it? In particular, different consumers make probability in maths correspond to probability in the real world in completely different ways, and that is their good right: to have their own "real wirld" concept of probability. In good mathematical midelling (correspondence building, bridge building) one does not only do what appears "intuitively obvious". One brings knowledge and experience to bear, one can use logical deduction and even mathematical analysis to the task of establishing sound bridges. Richard Gill (talk) 07:28, 15 September 2012 (UTC)
The problem with having multiple probability models is that then you have to explain multiple models, which takes more of the reader's time than necessary, and leaves doubt as to whether they actually describe the same thing. It's far more convincing to first define the object of study with crystal clarity and then proceed to look at it from all angles you ever want to, confident that no matter how you look at it it's the same object and that your inferences are coherent. It's not that there is only one way, it's just that it's better to pick one and stick to it so that you get a coherent, correct and complete story. That's what I was going for in the "conditional solution by simulation" section, by the way. -- Coffee2theorems (talk) 11:18, 14 September 2012 (UTC)
If you restrict attention with crystal clarity to the K&W situation then you may get a coherent and correct story but you certainly don't get a complete story. You also artificially create the impression that all the other solutions and all the other stories are somehow inferior or even wrong. Sorry. Wikipedia has to reflect the reliable sources in an unbiased way, not to present a biased personal point of view which has some pedagogical advantage to mathematical purists. Anyway: MHP is about a popular brain teaser. It involves translating an informal real world problem into a formal mathematical problem with a clean solution. That transition is not unique. It's a matter of taste, a matter of compromise. That's a fact of life. And it's a fact of the life of the MHP. Richard Gill (talk) 12:59, 14 September 2012 (UTC)
Of course you can use a more general model than the K&W one, like Morgan et al. do. I wouldn't really have a problem with using a formal model that subsumes the K&W one throughout the article, even one with a fully general joint probability distribution for car placement, player choice and host choice.
I don't for a moment imagine that people here could possibly agree to doing something like that, though. I'm even wondering now whether I should've used some safer way of referring to Morgan than the real name. Perhaps "You-Know-Who" or something. People really hate generalizations of the K&W description here. This being so, I'd much prefer to have the K&W version explained clearly, followed by rest of the stuff in a "Variants" section, compared to a complete lack of an overarching model. (and when I say "model", it is enough for me if things have somehow been specified clearly and unambiguously enough for computer simulation, no need for any scary mathy words) -- Coffee2theorems (talk) 14:34, 14 September 2012 (UTC)
There are four steps in a simulation of MHP. (1) The host decides how to hide the car. (2) The contestant decides, independently, which door to pick. (3) The host decides which door to open, dependent on where the car is and which door the contestant chose. (4) The contestant decides to switch or not to switch, dependent on which door the host opened. Specify these four steps fully, and you have fully specified a computer simulation model. The K&W specification tells us how the host performs his two steps (both uniformly at random). Of course it could help a contestant to know the host's strategy (choice of these two steps). But MHP can be "solved" without knowing the host's strategy, in the sense that we can give the contestant good advice what to do. My advice to the contestant who has no knowledge of the host's strategy would be (1) choose your door initially completely at random and then (2) switch, whichever door is opened by the host, and whichever door you initially chose. This way the contestant is guaranteed to win the car 2/3 of the times, and that's the best the contestant can hope for. Give me more info about the host's behaviour and I'll be able to tell you more nice things about this solution for the contestant, but my advice to the contestant stays the same: SWITCH. If you ask me advice later in the day (after you made your initial choice) then you are too late for some of my good advice. But I can still tell you: (1) if your initial choice had 1/3 chance of being right, then by switching you increase your chance of success to 2/3. (2) if any initial choice had 1/3 chance of being right, then by switching you increase your chance of success to 2/3, and that is the best you can do. (3) If the three doors are initially equally likely to hide the car, and if the host is equally likely to open either door when he has a choice, then the chance of winning by switching given your choice and given the host's choice is 2/3, whatever those choices happened to be. These various combinations of assumptions and solutions cover encyclopaediacly everything that is in the literature. And all combinations I have mentioned are favoured by different reliable sources. So the only question to wikipedians is how to survey all these assumption-sets and solution-sets in a reader-friendly way. Richard Gill (talk) 18:02, 14 September 2012 (UTC)
Re: "If you want to simulate MHP you are going to have to simulate the quizmaster's behaviour and the player's behaviour", no, you don't. In fact, one of the best simulations for convincing those who insist that there is no benefit to switching is to simulate the game with plastic cups and small objects, with "Monty" following an instruction sheet that conforms to the Kraus and Wang description (every game involves the host rolling a die twice to get the 3:1 and 2:1 randomization, even if the 2:1 roll is not needed) and a human player. --Guy Macon (talk) 09:17, 14 September 2012 (UTC)
Guy, that's exactly what I said. The instruction sheet to Monty is the specification of his behaviour, which you then procede to simulate! OK, the player just plays. He has to make initial choices, and he has to decide to switch or not to switch. Because of your instruction sheet, it doesn't matter how he makes his initial choice. The player will then try out repeatedly switching, and repeatedly staying. He'll simulate two possible player strategies. Richard Gill (talk) 09:38, 14 September 2012 (UTC)
PS I'm interested to know, when you have a friend doing this simulation, do you ask your friend to take careful note of how often they win for each particular initial door they chose and which particular door was opened by the host? So in principle, six different succes rates, not one. I suspect that all those school teachers who had their pupils do the experiment only looked at overall success rate. So (to Rick Bloch, Nijdam), were all those school teachers and their classes answering the wrong question? Richard Gill (talk) 09:58, 14 September 2012 (UTC)
Yes, plastic cups and small objects, or simply three playing cards. And simply formulate the rule correctly (equivalent to The contestant has to determine two doors, of which the host has to open one with a goat.). It is now easy to show before the game starts that the player will win by "switching" in two of three cases. And if you find someone who insists that there is no benefit of switching you may start the game, best with your partner as the "host". He will soon realize that you as the player win in two of three cases. But now comes the most interesting moment: A participant of the Wikipedia discussion enters the room and calls: That's all dead wrong! - And reputable sources say that most people insist on the wrong solution. - Can you name sources for what you are doing here? ...--Albtal (talk) 10:17, 14 September 2012 (UTC)
Some editors here will say not that this is all dead wrong, but that it is irrelevant. They'll say that you have to look at the six different success rates for each combination of initial choice of plastic cup by the player and plastic cup turned over by the host.
I would however point out that the overall success rate is equal to the average of the six conditional success rates (weighted by their relative frequency). If the overall success rate has been got to be equal to 2/3, and if no higher overall success rate is possible, then none of the six conditional success rates can be improved either. I think it's pretty unthinkable that one could beat overall 2/3. And in fact it's true: it can't be beaten, because whatever you do there's always a car-location such that you'ld miss that car.
Proof: suppose you initially choose door 1 and then always switch. You'll certainly miss a car behind door 1.
Suppose you initially choose door 1 and then always stay. You'll certainly miss a car behind door 2.
Suppose you initially choose door 1 and switch when the host opens 3, stay if the host opens 2. You'll certainly miss a car behind door 3.
All other cases just consist of permuting the numbers on the doors. I've shown you that if the car is initially equally likely to be behind any door, then whatever you do (and however the host opens a door if he has a choice), you run at least a 1/3 chance to finish with a goat. Thus you can't beat 2/3 overall success-rate. Since 2/3 is achievable by always switching, there's nothing better to do, nothing more to say.
If by any chance you are interested in conditional probability you can now deduce that the conditional probability that the car is behind door 2 given you chose door 1 and the host opened door 3 must be at least 1/2. Because otherwise, your overall success-rate of 2/3 got by always switching would be improved by staying in the just mentioned situation.
Note: I did not use any assumption that the host chooses completely at random when he has a choice of door to open. I completely solved the original problem and I only make one assumption. You can explain this solution to your grandmother. I did not use conditional probability. Richard Gill (talk) 13:10, 14 September 2012 (UTC)
If we leave it at this, we'll have had a long discussion with no conclusion. AFAIK this is one of the least contentious points, so getting a consensus here should be achievable. If we do get one, we can point at this discussion in the archives if the issue comes up again, and move forward. I propose we try.
(Clarification: "Krauss & Wang version/interpretation" here means anything with the same assumptions, not necessarily the literal K&W text. I chose to call it "K&W version" rather than "standard version" here because the former is more precise and neutral. The discussion above is also mostly about the assumptions, not the wording.)
For those who are for Krauss & Wang:
If you are still for keeping the K&W interpretation as the most prominent one in the article, please pipe in this thread. My impression before this discussion was that basically everyone is for K&W, because about two years ago almost everyone wanted to assume that the host makes his choice with 50:50 chances for either goat. Has that changed? The discussion above leaves the impression that it has, but I suspect it's just an illusion.
For those who are against Krauss & Wang:
If there are still many who want the K&W version to have the spotlight, then any consensus would have to involve letting it have it. Is it possible to do that while still being satisfactory from your POV? Would, for example, the following do?
In the lead, give the spotlight to a more general version (say, MvS).
In the first section, present K&W as the usual (or even "standard", as that appears to be the term in the sources) interpretation, but make it clear that there are others and that it is not in any way "the only correct" interpretation of the MHP. Might also say that the K&W assumptions have other justifications than plain old "I think (or 'almost everyone thinks') it's implied by the problem", and refer to the section(s) explaining the justifications.
Present a clear, coherent and correct story for the K&W version, making it clear that it solves the K&W version only.
Present other interpretations in a later section (say, "The general case" section?), and tie in to the K&W version when it arises as a special case (equilibrium solution in game theory, naturally induced assumptions from Bayesian analysis of the MvS version; these are the justifications I meant above).
Present the MvS-incompatible stuff (quantum version, many doors, ...) in a "Variants" section.
What I'm essentially asking is whether you could be satisfied with still presenting the K&W story first (after a more general lead), provided other interpretations are included later in the article, and proper care is taken with tone, correctness and contextualization. If you feel that the literal K&W text is too heavy, then you might argue for paraphrasing it or otherwise explaining the extra assumptions, instead of quoting verbatim, but that's something of a separate issue.
"A false dilemma is a type of logical fallacy that involves a situation in which only two alternatives are considered, when in fact there is at least one additional option. The options may be a position that is between the two extremes (such as when there are shades of grey) or may be a completely different alternative." --False dilemma
I am not so much pro-Krauss-&-Wang as I am anti-any-known-to-be-flawed-description. We know that vos Savant is flawed. That doesn't mean that K&W is the best alternative. Perhaps Gill's description would be better than either, avoiding the flaws in vS and avoiding the turgid prose of K&W? --Guy Macon (talk) 08:28, 19 September 2012 (UTC)
What I mean by "K&W version" is any version that embodies the same assumptions, not the verbatim text of theirs ("if you feel that the literal K&W text is too heavy, then you might argue for paraphrasing it or otherwise explaining the extra assumptions, instead of quoting verbatim, but that's something of a separate issue"). The assumptions are what matters to the structure of the article, not the wording we choose. What I call "K&W version" here others may call the "standard version", but I avoided that term because it is more ambiguous and might bias the discussion.
If, despite that clarification, you have a third option, then by all means present it. It would probably be something quite new to the discussion (AFAIK this has been mostly a "standard/K&W version" vs "variants" dilemma on these talk page, but alas I can't claim to have read all of the archives). -- Coffee2theorems (talk) 08:56, 19 September 2012 (UTC)
