Talk:Monty Hall problem/Archive 37

This is an archive of past discussions. 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.

Archive 30

←

Archive 35

Selvins letter

Selvin is wrong too. His solution also doesn't fit to his problem statement. You can not excuse this error with the remark:"Selvin is clearly not a very good writer." --85.72.41.84 (talk) 18:35, 14 May 2013 (UTC)

I can and do. Do you understand his solutions? Do you agree that there is a good question inspired by the Monty Hall show which does have 2/3 as its solution? Richard Gill (talk) 19:18, 14 May 2013 (UTC)

I understand his solutions but I don't agree that his question does have 2/3 as its solution. --94.68.194.35 (talk) —Preceding undated comment added 18:56, 15 May 2013 (UTC)

94.68.194.35 what exactly is your demur? The 2/3 chance to win by switching is valid for the famous paradox only, but never outside the paradox in deviant scenarios, e.g. if some unknown host is perfectly "known to be acting biased" in case he got two goats to show, and his one-sided bias as well as its direction as its extent is already perfectly known. The odds of exactly 1:2 for staying:switching is valid for the famous paradox only. Any other value but 1:2 is outright inapplicable for the famous paradox. So what exactly is your objection regarding Selvin? Gerhardvalentin (talk) 20:48, 15 May 2013 (UTC)

I think the problem is that these problems are brain teasers, trick questions. A kind of joke. Only when you hear the punchline (the solution) do you fully realise what the problem is about. Moreover, Selvin's question did not stand alone but was just a re-telling of an old joke which had already been going the rounds in Selvin's circles (academic probability and statistics teaching) for a hundred years (Bertrand's box paradox, Gardner's three prisoners problem). With vos Savant a further mutation occurred and the problem jumped the species gap and entered into popular culture.

Selvin's intended readers understood exactly what the intended problem was. Selvin published a letter to the editor in "The American Statistician". He didn't write a wikipedia article. Richard Gill (talk) 08:03, 17 May 2013 (UTC)

I agree that both Selvin's and vos Savant's readers well understood the intended question but I think it is an incorrect characterisation to refer to it as a trick question or a joke. I trick question is one in which is intended that the reader will understand the question in a different way from that on which the questioner proposes to base the solution. I know of no reports or sources of anyone finding any humour in the MHP. Martin Hogbin (talk) 08:58, 17 May 2013 (UTC)

Kraus and Wang describe how the visual details which we provide (door numbers, goat...) cause the "victim" to fixate on a static visual image, from which all trace of the sequence of events is erased. It is like a joke: building up a false impression. And Kraus and Wang say so explicitly. Mlodinow says "the quizmaster is setting us up" (tricking us). A trick question is a question to which the obvious answer is actually wrong. Richard Gill (talk) 07:18, 18 May 2013 (UTC)

The MHP demonstrates a weakness in human intuition regarding probability. It does not depend on misunderstanding the question for its effect.

I never said so, Kraus and Wang do not say so. Kraus and Wang explain how the static visual framing of the question leads people to *see* a symmetry between doors 2 and 3 and hence deduce equally likely probabilities, while the symmetries which lead to correct probabilities and the correct answer are hidden in the asymmetric and dynamic history of events leading up to the static image of two doors closed and one open. Richard Gill (talk) 11:55, 20 May 2013 (UTC)

The term 'trick question' refers to a puzzle in which the audience are expected to interpret the question differently from the questioner, who normally chooses to interpret the question in some bizarre or unnatural way. You have argued many times that most people interpret the question exactly as it was intended by vS and Selvin (and I agree with you) yet most people still get the answer wrong. Martin Hogbin (talk) 16:21, 18 May 2013 (UTC)

Not "bizarre" or "unnatural". The characteristic of a trick question is that the "natural" (immedate, intuitive) answer is wrong, while the right answer appears initially ludicrous. In fancy words: it's a "veridical paradox" (this is the terminology of the philosopher Quine). Richard Gill (talk) 07:44, 19 May 2013 (UTC)

pardon for intruding. IMO both of you are right. What matters is that the tricky paradox fools just about everyone. And thank you both for your efforts, I feel the article is much better to digest, meanwhile. Nevertheless the evidently implied absolute certainty in one aspect, and the evidently implied absolute uncertainty in another aspect should still be emphasised, to help avoid and brand post-hoc fallacy. Gerhardvalentin (talk) 20:32, 19 May 2013 (UTC)

Post-hoc fallacy? Evidently implied absolute certainty / absolute uncertainty? What are you talking about? Richard Gill (talk) 12:00, 20 May 2013 (UTC)

Richard, I guess we have different understandings of the term 'trick question'. I take it to refer to questions that are deliberately misleading, such as those to be found here. Martin Hogbin (talk) 12:53, 20 May 2013 (UTC)

The wikipedia article is about a question whose answer is 2/3. Maybe 94.68.194.35 can help us by proposing a good formulation to such a question, along with references to reliable sources. Richard Gill (talk) 08:07, 17 May 2013 (UTC)

That would be very useful. Martin Hogbin (talk) 09:03, 17 May 2013 (UTC)

Many of Selvin's intended readers said that he was wrong. His question describes a game with an free host, independent of any additional rules. If the letters he got from his readers would be published, we could detect what was their critic.

Selviin's readers (professional statisticians) understood his question as intended, but did not like his approach to the solution 2/3. He had to re-do the solution using conditional probability. Savant's readers first disagreed with her, but after doing the simulation experiment themselves they came to see she was right. Richard Gill (talk) 12:06, 20 May 2013 (UTC)

Many of Savant's readers said that she was wrong, too. Her question including her answer published together with the question doesn't give any hint to additional rules of the game. Her readers were knowing gameshows of this kind with an free and independent host, and they couldn't assume that the host in her question is intended to be only an "agent of chance". In later comments surprisingly she introduced additional rules referring to the three cups problem. So it is right what Gardner said:"The problem is not well formed!", what should be explained in the lead and in the following sections. If Savant would publish all the letters from her readers which criticized her solution we could find out what was their comprehension of her problem's wording. I don't see any technical problem to publish the readers' letters on the internet, for example as text files. Or wishes she to hide something? Till now she only published a few letters which don't refer to any additional rules.

I think an wikipedia article should report the human knowledge. And we know that Savant's answer doesn't fit to her original question. So what is the problem to say this in the article? --85.75.48.45 (talk) 10:59, 18 May 2013 (UTC)

Thank you. Misunderstandings are inevitable. Savant's intention was to present a paradox that already was known to Mathematicians, and in reading her question and in regarding her answer, this paradox can clearly be recognized – of course it excludes any host who could have revealed the car instead of a loser. So the clear premise is that the host knows which door hides the car and will always reveal a goat but never the car. Otherwise the paradox cannot arise. Some even presented the idea that any "closer" odds than 1:2 for "staying:switching" could be obtained for any "actual" game, a tail-biting serpent. But they did realize the paradox, indeed. In the first line, the article should help the reader to realize it, too. Gerhardvalentin (talk) 12:10, 18 May 2013 (UTC)

Selvin's and Savant's intentions are one, their problem statements are another thing. Obviously they have failed and the article should explain their error. Nothing else can be the aim of an wikipedia article than to report the human knowledge! --94.68.137.143 (talk) 12:42, 18 May 2013 (UTC)

@85.72.41.84: Take it easy! --Albtal (talk) 15:57, 18 May 2013 (UTC)

85.72.41.84/94.68.137.143 If you are the same person, why not register? It is free, easy, and anonymous.

The current article does report human knowledge. We know the question that vos Savant and Selvin intended and we know that that is exactly how the problem is interpreted by most people. We know this from reliable sources such as vos Savant, Selvin, and Krauss and Wang.

Misinterpretations of the problem do exist but most people interpret the problem correctly but still get the wrong answer. That is principally what the MHP is all about. Martin Hogbin (talk) 16:13, 18 May 2013 (UTC)

Gillman (1991), missing reference

The article cites Gillman (1992) in support of the Morgan et al. solution. Interestingly Gillman does not refer to those authors at all, but instead to a note by himself one year earlier "The car and goats fiasco", Focus (newsletter of the Mathematical Association of America, of which Gillman was a past president), volume (or number) 11, June 2011, p.8. He also reports on correspondence with Martin Gardner. He was a pure mathematician, working in topology and set theory, but interested in probability paradoxes. Does anyone have a copy of the 1991 note? He is careful with acknowledging sources so I am pretty sure he completely independently invented the Morgan et al. solution, even "inventing" the same notation "q" for the host bias.

This supports my opinion that the Morgan et al. solution is the perfectly natural mathematician's approach to this problem building on a long tradition (Bertrand box paradox, Gardner three prisoners problem). Notice that Nalebuff (1987) also connects the problem immediately with Bayes' theorem, and connects it to the probability puzzles tradition. Richard Gill (talk) 09:04, 13 May 2013 (UTC)

Richard you seem to want to talk about how natural the Morgan solution is. I would be happy to go through the actual facts with you on the arguments page. Martin Hogbin (talk) 12:56, 13 May 2013 (UTC)

No I don't want to go through "facts". I merely give evidence supporting my claim that the Morgan et al. solution is the straightforward solution automatically invented by any sensible (but unimaginative) mathematician with some grounding in contemporary mathematical culture, hence awareness of the prehistory (Bertrand, Gardner) of this kind of probability puzzle. In contradiction to Martin's claim that Morgan et al. somehow perverted the course of history by coming up with some crazy diversion for purely unscientific motives. Which is his personal point of view, so far unsupported by any historical or scientific-cultural facts at all.

Richard, where can I find Bertrand and Gardner's solutions to the related problems? Martin Hogbin (talk) 20:32, 15 May 2013 (UTC)

Wikipedia articles on Bertrand's box paradox and Three prisoners problem. Richard Gill (talk) 07:51, 17 May 2013 (UTC)

Bertrand and Gardner's solutions do not seem to be available there. I am looking for the original solutions by them. Martin Hogbin (talk) 08:49, 17 May 2013 (UTC)

That would be interesting. Bertrand's book is freely available on internet. The Bertrand's box paradox article included a section with a short and intuitive solution by symmetry. It said that it was difficult to give that argument in rigorous mathematical terms. I think it is easy to do so, and edited the article accordingly, adding some "own research", so far unchallenged. Richard Gill (talk) 07:22, 18 May 2013 (UTC)

Gardner's 1982 book "Aha Gotcha" contains a probability problem called "Three shell game"! Marilyn must have got the shell/cup game from there. Richard Gill (talk) 07:41, 18 May 2013 (UTC)

"The Second Scientific American Book of Mathematical Puzzles and Diversions" contains the "Prisoners and warden" problem. Richard Gill (talk) 11:02, 18 May 2013 (UTC)

And I ask if anyone can inform us as to the content of Gillman (1991). Richard Gill (talk) 19:30, 14 May 2013 (UTC)

Here is a less ambiguous citation for Gillman (1991):

Leonard Gillman, "The car-and-goats fiasco", Focus: The Newsletter of the Mathematical Association of America, Vol. 11, No. 3 (June–July 1991), p. 8.

Unfortunately, MAA's online archive of past issues of Focus only extends back to 2001.[1] For what it's worth, there is hardcopy in Leonard Gillman Papers, 1922-1924, 1945-2004,[2] Archives of American Mathematics, Dolph Briscoe Center for American History, The University of Texas at Austin. (Request box 2013-073/2 from remote storage.)

I found one secondary source that describes the content of Gillman (1991) in the following passage:

"Some of our colleagues were so moved by an ill-tempered response to Marilyn that [...] another took a page of MAA Focus to castigate ungallant and unprofessional behaviour."
— Ed Barbeau, "Fallacies, Flaws, and Flimflam", The College Mathematics Journal, Vol. 24, No. 2 (March 1993), p. 149.

I do not know who Gillman was calling out (i.e. whether it was just the 50/50 boneheads or included semantic nitpickers and other quibblers), nor whether he discussed the math at any length or just focused on the problem of demeanor. ~ Ningauble (talk) 21:07, 31 May 2013 (UTC)

Bertrand's problem and solution

Bertrand's 1889 paradox and solution (rough translation from the French by R D Gill): Three apparently identical cabinets each contain two drawers. There is a silver or a gold medallion in each drawer. One cabinet contains two gold medallions, one contains two silver medallions, one contains one of each.

A cabinet is picked at random. The chance it contains two different medallions is 1 in 3 (three equally likely possibilities, one favourable).

Now we open one drawer at random and see a gold medallion, say. The the other medallion is gold or silver. Two equally likely possibilities. Hence the chance that our cabinet is the one with two different medallions in 1 in 2.

How can the mere fact of opening a drawer raise the chance that the cabinet is the "mixed" cabinet from 1 in 3 to 1 in 2?

Solution. Imagine three hundred instead of three cabinets, one hundred of each type. Open one drawer at random of each cabinet. One expects to see about 150 gold and 150 silver medallions. Exactly 100 gold and 100 silver each from the "unmixed" cabinets, about 50 gold and 50 silver from the mixed cabinets. Given we see a gold medallion, it is twice as likely (100 : 50) from an unmixed as from a mixed cabinet. The chance we have a "mixed" cabinet has remained 1 in 3. (End of translation).

Bertrand's 1889 resolution of his paradox is therefore effectively by a calculation of a conditional probability. However at this stage of his book (we are right at the beginning with a list of thought provoking paradoxes) we haven't yet been introduced to the law of large numbers or to Bayes theorem so the solutions are purely intuitive. He doesn't actually divide the 100 mixed into roughly 50 and 50 but just says: "much less than 100 gold from the mixed boxes" hence the chance of an unmixed box is larger (100 : much less than 100) than the chance of a mixed box. Richard Gill (talk) 14:52, 21 May 2013 (UTC)

in the beginning he presented a completely false conclusion, related to the wrong "2 envelopes" conclusion. Gerhardvalentin (talk) 19:56, 21 May 2013 (UTC)

What do you mean, presented a completely false conclusion? He presents two arguments which lead to mutually contradictory results. In other words, he presents his paradox. After that he gives his solution, which shows that the second argument is wrong.

Like the two envelopes paradox, and like Monty Hall problem, Bertrand's paradox shows that we should be careful with saying "because there are just two possibilities, they must be equally likely". Richard Gill (talk) 12:49, 22 May 2013 (UTC)

Is his solution available online anywhere? Martin Hogbin (talk) 15:18, 20 May 2013 (UTC)

I wrote out my rough English translation of Bertrand's problem statement and accompanying solution. His book in available on internet as free eBook e.g. from [3] but the ePub version it is hard to read because it is the result of applying OCR to a scanned original. Probably the pdf (scanned original) is better, but the pdf is probably not searchable. Richard Gill (talk) 05:50, 21 May 2013 (UTC)

Where can I find your translation? Martin Hogbin (talk) 14:04, 22 May 2013 (UTC)

Right here in front of you. Richard Gill (talk) 06:30, 23 May 2013 (UTC)

Your use of the word 'conditional' somewhat vague and deceptive. To say that a solution is conditional means nothing unless you make clear exactly what events you are conditioning on. Martin Hogbin (talk) 10:35, 23 May 2013 (UTC)

Sorry. It is obvious to anyone familiar with elementary probability calculus, so, it seems to me, it should be obvious to anyone familiar with the elementary mathematical literature on the Monty Hall probem. Bertand calculates effectively the probability that the the other coin is silver given that the observed coin is gold and shows that the answer is 1/3, the same as the (unconditional) probability that the box initially chosen is mixed. Whether or not the chosen box is mixed is independent of whether the chosen coin is gold or silver, by symmetry. Richard Gill (talk) 14:55, 23 May 2013 (UTC)

It is indeed quite obvious what Bertrand is doing, it is your use of the word 'conditional' that confuses me. How would the problem be solved unconditionally? Martin Hogbin (talk) 15:53, 23 May 2013 (UTC)

I said, Bertrand's solution is a nice illustration of the use of conditional probability. The (unconditional) chance that the other coin is silver is different from the (conditional) chance that the other coin is silver given the chosen coin is gold. The word "conditional" to be understood in its technical meaning as in the phrase "conditional probability", get it? Richard Gill (talk) 16:35, 23 May 2013 (UTC)

As I wrote in Bertrand's paradox talk there is a very simple solution: There are two cases that the other coin is a gold coin, and one case that it is a silver coin.

No, I do not 'get it'. What is the unconditional chance that the other coin is silver? Martin Hogbin (talk) 17:35, 23 May 2013 (UTC)

P(other coin is silver) = 1/2.

That is not an answer to the question asked. Martin Hogbin (talk) 09:26, 24 May 2013 (UTC)

P(other coin is silver | your coin is gold) = 1/3. Richard Gill (talk) 06:20, 24 May 2013 (UTC)

Exactly, the answer to the question asked is by definition conditional. Martin Hogbin (talk) 09:26, 24 May 2013 (UTC)

It is not the question nor the answer which is conditional. The question is about probabilities, and it is solved by realizing we must distinguish between conditional and unconditional probabilities! MHP is about probabilities, and about being careful with conditioning. In MHP, Prob(the car is behind door 2 | a goat is behind door 3) = 1/2, but Prob(the car is behind door 2 | the player selected door 1 and the host revealed a goat is behind door 3) = 2/3. So resolving the paradox of MHP is about distinguishing between conditional probabilities conditional on different events. If the process which delivers you some information is itself influenced by that information, you have to take it into account. Like in the Boy or girl paradox.

One can answer Whitaker's question correctly without apparently using conditional probability (use symmetry and independence instead), but understanding the Monty Hall paradox is all about deepening one's understanding of conditional probability.

To say the the same thing without even using the word "probability": When you receive a message informing you of some facts, you should not not just think about those facts, but also about how/why the message got to be delivered to you. Richard Gill (talk) 05:09, 25 May 2013 (UTC)

There are three gold coins. For two of them, say G1 and G2 (which are in the same cabinet), there is a gold coin in the other drawer of the cabinet: For G1 this is G2, and for G2, this is G1. Only for one of the gold coins, G3, there is a silver coin in the other drawer. So by simply counting the cases we get the (unconditional) chances 2/3 that the other coin is a gold coin, and 1/3, that it is a silver coin. I can't see any reason for a more complicated proof. (My remarks below illustrate another proof without conditional probabilities.)--Albtal (talk) 19:49, 23 May 2013 (UTC)

A paradox arises if we consider a chance other than 2/3 as the intuitive solution (and the chance for a mixed cabinet other than 1/3). For we know at the beginning of the game that the chance to get a mixed cabinet is 1/3; and that we will with certainty get a drawer with either a gold or a silver coin. And if the event of seeing a gold coin (or a silver coin) would change the chance of 2/3 (1/3) we would know this chance before opening a drawer which contradicts the real chance of 2/3 (1/3) at the beginning.

As we know by former comments on this talk page an analogous contradiction would arise in (the well formulated) MHP if we would assume a chance other than 2/3 for switching; and if we assume that it doesn't matter if the host opens door 2 or door 3.--Albtal (talk) 10:41, 23 May 2013 (UTC)

Bertrand's box 'paradox' has been called Bertrand's box 'paradox' for more than a hundred years now. And the reason it is so-called is that though there are two choices for the other coin, they are not actually equally likely. Whether modern readers would find the problem paradoxical or not, is another matter. In my opinion it is amusing that though Bertrand's box problem and the Monty Hall problem are considered to be mathematically equivalent, people tend to have a very strong, very wrong, intuition about MHP, but I doubt it is the same with Bertrand. I think that most people will quickly realise what is the right answer for Bertrand. I think that people also more easily see the right answer for the three shells problem than for the Monty Hall problem. And Monty Hall interests almost everyone, but only few people find Bertrand's boxes interesting or amusing. Selvin's shift of context was a brilliant invention. After that we just had to wait for someone to pick it up and introduce it to the world (vos Savant). Richard Gill (talk) 15:11, 23 May 2013 (UTC)

For Martin. You pick a box at random and then a drawer at random. You see a gold or silver coin in that drawer. The other drawer has another coin which is also gold or silver. The box can have two different type coins or two same type coins. Mixed boxes and unmixed boxes.

Paradox (?):

P(other coin is gold) = 1/2

P(other coin is gold | your coin is gold) = 2/3

A solution (there are many more):

P(box is mixed) =1/3

Whether or not your coin is gold is statistically independent of whether or not your box is mixed, by symmetry.

Therefore P(box is mixed) =1/3 = P(box is mixed | your coin is gold) = P(other coin is silver | your coin is gold).

The type of your coin is statistically independent of the type of your box.

The type of the other coin is statistically independent of the type of your coin.

The type of your coin and of the other coin are statistically dependent.

I use the standard fact A and B are independent if and only if P(A|B) = P(A) (provided P(B)>0).

The conditional probability of A given B, P(A|B), equals (by definition) P(A and B)/P(B) ) (provided P(B)>0).

A and B are called statistically independent if and only if P(A and B)=P(A)P(B) Richard Gill (talk) 06:16, 24 May 2013 (UTC)

I have no argument with any of that but you describe the solution as using 'conditional probability'. The word 'conditional' seems somewhat redundant to me as you cannot provide any solution that is not conditional.

I am glad with your observation, Martin. Now, many reliable sources (e.g., Ruma Falk, 1992: a maths educationalist, cognitive scentist, psychologist) claim the three boxes problem is equivalent to the three prisoners problem and to the three doors problem. If they are right, you cannot solve MHP without using conditionsl probability either! At best you can hide that the problem is really about conditional probability by hiding it and using instead symmetry and independence. I think that Martin Gardner's three shells problem is the most succesful version of this disappearance trick. By replacing the doors by apparently indistinguishable shells, and physically shuffling them as well, the problem is reduced from a situation with naturally six sample outcomes (each of probability 1/6) and the obvious need to condition on three of them, to a problem with two outcomes (probabilities 1/3, 2/3) and no conditional probability at all. The so-called "combined doors solution" to MHP is essentially the realization that by symmetry (which generates independence, which gives irrelevance) we might as well pretend that the MHP is the three shells problem.

three boxes needs conditioning, three shells doesn't (MHP in the middle), yet they are "equivalent"! Richard Gill (talk) 04:42, 25 May 2013 (UTC)

Three card Monte

It seems to me that there are links to the street confidence tricks Three Shell and a Pea and Three-card Monte in the history of MHP. Note that Selvin's letter calls the host "Monte" not "Monty". Deliberate? Gardner (1988) invented the three shells and a pea problem, obviously refering to the three shells and a pea confidence trick, and Marilyn vos Savant clearly knew about Gardner's problem. Richard Gill (talk) 13:38, 20 May 2013 (UTC)

Player who chooses uniformly between sticking and switching.

In the article we currently have, Indeed, if a player believes that sticking and switching are equally successful and therefore equally often decides to switch as to stay, they will win 50% of the time, reinforcing their original belief. An editor has suggested that this is illogical but, I suspect, has actually misunderstood what is being said. Martin Hogbin (talk) 08:03, 23 May 2013 (UTC)

This strange fictive player indeed has a strategy which is always right: If one of two cases has the probability p, the other 1 - p, he will win 50% of the time if he chooses each possibility with q = 1/2: (1/2)*p + (1/2)*(1-p) = 1/2. So this player would reinforce their original belief that p = 1/2 even if p would be 1/1000000 ... The UFO argument of Marilyn vos Savant (The Power of Logical Thinking, p. 10) is similarly vacuous. --Albtal (talk) 10:09, 23 May 2013 (UTC)

The UFO argument is not vacuous. It underlines the point that information about how the present situation has arisen is crucial. Using it, you can improve your chance of winning. You know more than just "there are two closed doors and the car is behind one of them". You know that one of those doors was selected by yourself, the other was left closed by the host, who may well have used his own information as to the location of the car in order to do this. Richard Gill (talk) 15:19, 23 May 2013 (UTC)

Thank you Martin, for posting this here, too. It's exactly the same as the little green woman, and I repeat here what I just said to user Yabti. "The paradox":

Yes, two still closed doors, one hiding the car for sure, the other one hiding a goat for sure, but their chances are not 1/2 : 1/2, but their chances are 1/3 : 2/3 (chance of door first selected by the player : door offered by the host to switch on).

By randomly choosing one of those two doors: (1/2 * 1/3) + (1/2 * 2/3) the chance to win the car = exactly 1/2.

We indeed should say this explicitly in the article also, maybe even twice, as some readers still seem to look blank. Gerhardvalentin (talk) 13:52, 23 May 2013 (UTC)

I just tried to add this to the article and removed the citation tag. Please make it better if you like. Gerhardvalentin (talk) 17:27, 23 May 2013 (UTC)

Adding an explanation isn't usually removing the need for a citation. However Steinbach's article might work as a source: [4]--Kmhkmh (talk) 17:35, 23 May 2013 (UTC)

But please not as a source for this strange fictive player ...--Albtal (talk) 18:05, 23 May 2013 (UTC)

A rephrasing of an explanation for why the uniform selection has the results it does: if you always flip a coin before making any bet between two alternatives (where no possibilities exist outside those alternatives), you will win around half the time no matter what the relative probabilities are. For one thing, because flipping a coin can have no effect on the outcome, it doesn't matter when it happens. I could flip first then watch the event being anticipated, or I could watch the event and then flip a coin to see which outcome I "had bet" on. In the latter case, of course, it's easy to see that there's a 50% chance my coin will corresponded outcome that already happened. If flipping beforehand made a difference, then the coin would somehow be effecting the event. ± Lenoxus (" *** ") 18:09, 23 May 2013 (UTC)

Just tried to add Steinbah as a ref. Btw: the effect is simple and common in a variety of ways. Gerhardvalentin (talk) 18:25, 23 May 2013 (UTC)

Do we actually need a source? I think this might be properly called a routine calculation. Martin Hogbin (talk) 21:56, 23 May 2013 (UTC)

Basically, I'm with you, Martin. On the other hand: We have Steinbach as an excellent source on this very matter, from a very categorical point of view. So why not citing him? Gerhardvalentin (talk) 22:02, 23 May 2013 (UTC)

Strange things happen here.--Albtal (talk) 21:02, 23 May 2013 (UTC)

Steinbach says:

Der Kandidat hat keinerlei Anhaltspunkt, welches Spiel mit ihm gespielt wird! Wie also soll er sich entscheiden? Tür eins oder zwei? Interessanterweise darf er sich weder für die eine noch für die andere Tür (systematisch) entscheiden; vielmehr muß er zufällig und mit gleicher Wahrscheinlichkeit eine der beiden Türen wählen, also z.B. durch Münzwurf. Nur so kann er sicher sein, eine Gewinnchance von 1/2 zu erreichen!

If the contestant has no clue in which game he is in, how should he find the correct decision? Door #1 or #2? – Interestingly, he should on no account (systematically) select either door, but in fact he should choose completely at random, with equal probability, e.g. by tossing a coin. Only in that case he can be sure to achieve a chance to win of 1/2 !

You understand German. Do you see what Steinbach tells us concerning "choosing at random"? Regards, Gerhardvalentin (talk) 21:50, 23 May 2013 (UTC)

Steinbach tells us here, that the contestant who does not know the motivation of the host has no reason to prefer a door; but in order to get a chance of exactly 1/2 he should not decide himself systematically for one door, but with probability 1/2, for example by throwing a coin. Surely (1/2)*p + (1/2)*(1-p) = 1/2 for every p. But it would be a very stupid player who concludes that therefore p = 1/2 reinforcing their original belief. Strange things happen here.--Albtal (talk) 22:41, 23 May 2013 (UTC)

Steinbach is giving the game-theoretic mimimax solution for the non-standard problem in which the host doesn't always open a goat door and offer a switch. He may or may not do it, as he likes, depending on your choice and where the car is; and you don't know what he's up to. This nothing to do with Marilyn's UFO story. Of course it is a routine calculation that if the car is certainly behind one of two doors and you select one of them at random, you get the car with probability 1/2. But it's not entirely trivial to show in Steinbach's game that there's no way to be sure of doing better still, and moreover, with any other strategy, the host could make sure that you do worse. Richard Gill (talk) 05:58, 24 May 2013 (UTC)

Albtal, you say, '..it would be a very stupid player who concludes that therefore p = 1/2 reinforcing their original belief'. People really are like that, they are wrong as you point out, but I think you would be surprised how many people would make the incorrect deduction. Martin Hogbin (talk) 09:29, 24 May 2013 (UTC)

Strange things happened there.--Albtal (talk) 11:11, 24 May 2013 (UTC)

There's nowt so queer as folk. Martin Hogbin (talk) 12:39, 24 May 2013 (UTC)

Jeff Stibel says:

"The solution to this problem is not intuitive. Contrary to common sense, people should switch to the remaining door, as it will increase their probability of winning. Yet people consistently choose to stay with their first choice and report with high levels of certainty that the odds are 50:50 that the prize is behind either of the remaining doors (Falk, 1992; Gilovich et al., 1995; Shimojo and Ichikawa, 1989). A thought experiment helps most people understand the true probabilities of this problem. Imagine that you originally chose door 1. If the host now asked you to choose between your current choice or switching to both doors 2 and 3, you would switch. The Monty Hall dilemma is no different except that the host is giving you additional information (that the prize is specifically not behind one of the other doors). The key to the problem is to realize that, because the host’s choice is not random and is constrained (he will never open the door with the prize), the probability of the prize being behind the chosen door remains one out of three."

As to the intuitive but wrong 50:50 chance, correct only in "choosing randomly" whether to stay or to switch, above I said "We indeed should say this explicitly in the article also, maybe even twice." Gerhardvalentin (talk) 14:09, 24 May 2013 (UTC)

Jeff Stibel misses one step in his argument: it is careless/deceptive. He is also assuming that when the host must choose between two goat doors to open, his choice is random. Otherwise the probability the prize is between the chosen door could change.

About 50/50. Notice that 50/50 is the right answer to the question "what is the chance the car is behind door 1, given there's a goat behind door 3?" The paradox of MHP is that that question is the wrong question. One must take account of the process behind door 3 getting opened. It involves also the player's choice (door 1), and after that, the host's choice is constrained by the player's choice and by the location of the car.

50/50 is the right answer when the host opens door 3 anyway (e.g. by accident) and it just happens not to hide the car. Richard Gill (talk) 05:27, 25 May 2013 (UTC)

Here you obviously don't address Jeffrey M. Stibel et al., who are neither careless nor deceptive, but probably you are addressing Marc C. Steinbach. But even Steinbach is neither careless nor deceptive, but explicitly comes from the imaginable assumption "no rules, anything possible", whereat the contestant has no clue in which game he is in. You didn't read Steinbach? He does not address the "paradox", but he explicitly surveys the case "no rules, anything possible" and therefore has to allow for a host who could choose even at random, also. Even if in that case the player makes his decision to stick or to swap by tossing a coin, the chance to win is 1/2.

Or did you try to say that Stibel, although assuming that when the host can choose between to goats to show, his choice is random, but did not say so, explicitly? But IMHO this is quite another issue, as I am reading Falk in quite a different way. Gerhardvalentin (talk) 07:07, 25 May 2013 (UTC)

Gerhard, you appear to attribute some words to Stibel. Unfortunately, they contain a logical error. I don't know why you introduce this off-topic deception into the present discussion. Richard Gill (talk) 16:24, 25 May 2013 (UTC)

Richard, above you said "Jeff Stibel misses one step in his argument: it is careless/deceptive. He is also assuming that when the host must choose between two goat doors to open, his choice is random. Otherwise the probability the prize is between the chosen door could change." – And I asked you how to "understand" your fact-finding. Instead of answering my question, you blame me for "introducing an off-topic deception". What about just answering my question? What exactly do you mean when you say that the argument of Stibel "is careless/deceptive." ? Is it what I guessed ("did not explicitly say"), or not? What did you address? I still don't know it. Presently, inapprehensible to me. Gerhardvalentin (talk) 17:42, 25 May 2013 (UTC)

I agree with Gerhard. No one has yet produced a logical argument, as opposed to just an assertion that the door opened by the host is any more significant that the goat revealed by the host. Of course, from a Bayesian perspective we can ignore both since, as we have no information in the matter, we must assume probability is uniformly distributed between the options.Martin Hogbin (talk) 09:08, 25 May 2013 (UTC)

Some authors clearly state their assumptions (whether or not you like them is not the issue) and then make logically correct deductions from those assumptions. Some e.g. Stibel don't. Richard Gill (talk) 16:25, 25 May 2013 (UTC)

As to the famous "paradox", one must assume that the host, if he isn't slave to the act in 2/3, in the remaining 1/3 chooses randomly between his two goats, in order to show one of them. I am talking about the famous "paradox" (unchangeably 1/3 : 2/3), that arises in a special given scenario. Applying any one-sided bias for the "actual game the contestant is in" will lead to an incorrect, say utterly wrong value of probability. Any other value than "1/3 : 2/3" inevitably is wrong for any "actual game". See Falk: Only if one "knows", this is allowable. Otherwise false. Gerhardvalentin (talk) 10:16, 25 May 2013 (UTC)

Gerhard, I agree with you entirely what assumptions one ought to make. That is not the point under discussion. Richard Gill (talk) 16:24, 25 May 2013 (UTC)

Then why do you say, 'He is also assuming that when the host must choose between two goat doors to open, his choice is random. Otherwise the probability the prize is between the chosen door could change'? You agree that we ought to take the host's choice as random so the probability the prize is behind the chosen door cannot change. Nothing is missing and no one has been careless. Martin Hogbin (talk) 17:10, 25 May 2013 (UTC)

You just copied out, yourself, what is missing. It was missing, and it was careless to leave it out. He suppressed making explicit an assumption which so far was not stated in the quoted passage, and a deduction step which crucially needs that assumption. He is misleading his readers by appearing to present a logical deduction, while he actually makes no attempt to justify a non-trivial step in his chain of inferences. It uses a new modelling assumption which so far he did not put on the table. Instead, he makes an intuitive jump which amateurs do not notice, but he could just as well use such a jump in order to lead you to a contradiction. Professionals can see at this point that he is not doing his job well. Careless, shaky, incomplete...

His story certainly contains some useful intuitive ideas, some ingredients of a careful solution, but it misleads by missing one step. The author is a mathematician but by standard mathematicians' standards his story is sloppy.

It doesn't matter, it is all carefully explained in the second part of the article, which draw on the published sources concerning exactly this tricky step. Many readers won't care and won't appreciate the subtleties. If you just want to get an intuitive idea why the answer is 2/3, not 1/2, this is arguably enough. That's why Devlin's combined doors argument is in the first part of the article. If however you want to avoid making mistakes when looking at new problems in the future, it could well pay off to understand what is going on at a deeper level.

BTW, at the beginning of her paper on the Three Prisoners and related problems, Falk (1992) outlines "the" standard systematic mathematician's approach to solving not just MHP but also its precursors and variants. It's exactly the careful step by step analysis of actions and information flow which I described earlier, culminating in computation of the relevant conditional probability (and one also deduces in this way which probability to compute, as well as how to compute it). Her paper is pre-Morgan. Richard Gill (talk) 05:32, 26 May 2013 (UTC)

The sloppiness you refer to has nothing to do with mathematics or mathematicians, it has to do with the stage before mathematics becomes involved, where the problem is put into mathematical language. There simply is no standard, mathematical, or algorithmic way to turn a natural language problem statement into a precise mathematical formulation (I trust you agree with this). The problem solver has to decide what is important; what needs further explanation and what is obvious. The step you call non-trivial is as obvious to most people as is the unimportance of the players original door choice. Martin Hogbin (talk) 09:43, 26 May 2013 (UTC)

This now rehashing a (moot) point that has been made for years here. There is no point in speculating what most people might consider be "obvious" (while actually really understanding it rather than overlooking it), neither what one personally might consider as obvious. As Richard has correctly pointed out there is enough reputable literature that does not treat it as obvious and zooms in on it. Not all sources are doing that, but there are enough to warrant a mentioning of it in the WP article. Also I agree with Richard that you can call it sloppy, at least any person educated on the MHP knows that the "obvious" requirement is a thorny issue and has been considered as problematic by some, so simply completely ignoring it, could be regarded as sloppy or incomplete.--Kmhkmh (talk) 18:32, 26 May 2013 (UTC)

Yes, I know how the sources treat this. This is really a discussion for the arguments page or for userspace. Martin Hogbin (talk)

Agreed. Now Martin says "it has to do with the stage before mathematics becomes involved, where the problem is put into mathematical language". If only that were the case, I would not bring it up this question here yet again. I think most people agree that it is useful to separate these two phases: translate from natural to mathematical language; then solve within a formal framework. The first phase depends on personal taste, culture, experience, tradition. It's an art. The second phase is a matter of logical deduction. However, the "solution" we are talking about is sloppy because it does part of the translation, gives the appearance that a shift has been made to a formal mathematical analysis, but then makes an intuitive jump which in fact depends on a silently made further translation step. If Monty Hall paradox is about being careful with probabilistic reasoning, then this particular solution is a bad solution since it is careless with probabilistic reasoning. It counts on the reader being convinced by the authority of the writer together with some intutive gut feelings, not by thought.

Martin also wrote "The step you call non-trivial is as obvious to most people as is ...". It's completely obvious to most people that the answer to the Monty Hall problem is don't switch because the chance is 50/50... Completely obvious and wrong. The history of statistics is full of terrible disasters because steps which statisticians called non-trivial were completely obvious to most people ... and wrong. Eg. Sally Clark, Lucia de Berk, ... The moral is: be careful with probabilistic reasoning. Sloppy reasoning costs lives. Richard Gill (talk) 07:20, 27 May 2013 (UTC)

Richard, if you are going to continue to argue your point can we do it on the Arguments page page please. Martin Hogbin (talk) 08:30, 27 May 2013 (UTC)

I don't want to argue: I want to present a point of view. But OK we can continue on the arguments page. Richard Gill (talk) 17:38, 28 May 2013 (UTC)

The key to the paradox: It's not random

The section The Paradox should read:

Leonard Mlodinow says, "The Monty Hall problem is hard to grasp, because unless you think about it carefully, the role of the host goes unappreciated." (Mlodinow 2008) The correct answer, that players who swap have a 2/3 chance of winning the car and players who stick have a 1/3 chance of winning the car, is based on the premise that the host knows which door hides the car and will always reveal a goat but never the car.

If the player initially selected the door that hides the car (chance 1-in-3 only to be in this lucky guess scenario"), then both remaining doors hide goats, the host may choose either door, and switching doors loses.

But if the player initially selected a door that hides a goat (chance 2-in-3 to be in a "wrong guess scenario", having picked either goat A or goat B), then the host's action is no longer random, because he has no choice but to show the second goat. In that 2-in-3 switching door wins for sure.

If the host however, contradictory to the premise of the standard paradox, had opened a door at random and revealed a goat simply by chance, then the odds staying:switching would have been reduced from 1:2 as per the paradox to 1:1 only, as per the most common intuitive wrong answer, because half of the potential winning cases are wasted when the host can accidentally reveal also the car (Rosenthal, 2005a), (Rosenthal, 2005b).

Is this okay?

furthermore

And we also should show there, following the sources, that with regard to the lucky guess scenario of the standard paradox, it's completely futile to ponder whether the host "could be biased to prefer a special goat to show, or to prefer a special door to open" – or not, resp. "to have been biased", or not. – Because one would succumb to the hidden cryptic fallacy that any such futile inapplicable "assumptions" are completely inappropriate as long as one is only just assuming but not exactly "knowing" (see Ruma Falk).

Because, by just "assuming", one necessarily would be misguided and necessarily would come to actually wrong conclusions with regard to the actual game of the standard paradox. We should show, as per modern sources, that it's completely futile "to condition on wrong assumptions", with regard to the "actual game" of the standard paradox. Gerhardvalentin (talk) 14:57, 4 July 2013 (UTC)

Regarding the proposed text for The Paradox section, it was not immediately obvious what you were changing, so I compared your text to the corresponding two paragraphs in the current article. There are the main differences:

Break out two sentences as bullet points.
Replace "On the other hand," with "But" (at the second bullet).
Add "to be in this lucky guess scenario"" (in parentheses at the first bullet).
Add "to be in a 'wrong guess scenario', having picked either goat A or goat B" (in parentheses at the second bullet).
Change "other goat" to "second goat" (at the second bullet).
Change ", and switching" to "In that 2-in-3 switching" (concluding the second bullet).

I don't think these changes are ok. #1 is unnecessary, and just takes up space. #2 is not very wrong, but I prefer the previous style of exposition. #3 and #4 are largely superfluous, excess verbiage. Introducing goat names in #4 is an unnecessary complication. #5 is not very wrong, but it is not a matter of sequential goats, it is just the other one. #6 is unnecessary and grammatically awkward. On the whole, this takes a passage written in rather poor quality English prose and makes it worse.

Regarding your furthermore remarks, I don't agree about putting it in this section. Although I agree with your understanding of what the standard paradox is, this is not the place to introduce discussion of what it is not or why, from an appropriate prespective, host bias may be considered a red herring. Sections that treat host bias should be clear about what they are assuming, or taking as given, but this is not needed in setting out the basic paradox. ~ Ningauble (talk) 17:19, 4 July 2013 (UTC)

Thank you, esp. for your tip to English language. Can you help? My aim is to accentuate Mlodinow's evidence:

"lucky guess scenario" (contestant having picked the car right off, and we "at least are assuming" that the host didn't care which booby-price he shows, goat A or goat B),
"wrong guess scenario 1" (contestant having picked goat A), and
"wrong guess scenario 2" (contestant having picked goat B).

The contestant's rating, in which of those three equally likely scenarios "he actually is in" (1/3 : 1/3 : 1/3) can never be "corrected" nor "upgraded" in any way, this is his main statement. This should be spelled out.

Indeed, Morgan et. al themselves recently acknowledged that the "correct" answer for staying : switching (regarding the paradox) is 1/3 : 2/3. Period.

Their "mathematical joke" was aimed to lessons of probability theory, and nothing else, but completely unable to address the actual answer to the famous "paradox". Not every math teacher got it.

Please help the article to get more lucid. Imo, to get really "excellent", the article should explain the facts, not "obscure" the facts. A hard challenge. would be great if you can help to back it. Gerhardvalentin (talk) 18:51, 4 July 2013 (UTC)

It isn't hard to grasp at all, I would say the Monty Hall problem is the easiest "paradox"s to understand in fact. If you swap you will always get the opposite of what you choose, and you are twice as likely to choose a goat. It's that simple. Robo37 (talk) 22:02, 4 July 2013 (UTC)

Thank you Robo37, for this comment and for having said this also at the time where Martin was formulating the section "The paradox". And it is important to show that the host's action never is "fully at random", but in 2 out f 3 is extremely "biased". And to show as per modern sources that in 1 out of 3, when he "has" a choice, it is not admissible to "fix" any forever unknown bias, regarding the standard paradox. Doing so actually gives a "wrong result" for the standard paradox. Gerhardvalentin (talk) 06:55, 5 July 2013 (UTC)

I think the last sentence of this section "The Paradox" is quite a disaster now - such idiosyncratic use of English that I doubt many readers can make sense of it. I'm not happy with attributing it to Rosenthal (article, book). Perhaps the editor who is responsible can tell us exactly what sentences of Rosenthal's works he is trying to paraphrase. And by the way, I do not find Mlodinow a good reference at all. He slickly presents a superficially attractive solution which however is logically flawed, as the article explains later (criticism and fix of Devlin's solution). Rosenthal on the other hand is careful and knows Monty Hall problem through and through. Richard Gill (talk) 17:33, 10 July 2013 (UTC)

For years and years the article did not – demonstrative enough for newcomers (including Paul Erdős) – point out the "key" to the famous paradox: that the host's action is not fully at random.

That in 2 out of 3 cases, if the contestant first selected goat A or goat B, the host then is forced to show the other goat. He then is forced to show the second goat only, but may not show the car.

If however, outside the rules of the "paradox", he opened a door at random and just by chance "happened" not to have shown the car, but just "happened" to show a goat, then the paradox will not arise. See uscd, Monty Knows Version and Monty Does Not Know Version, An Explanation of the Game.

Jeffrey S. Rosenthal says: suppose you select Door #1, and the host then just by chanc opened Door #3. The probabilities that Door #3 happens not to contain a car, if the car is behind Door #1, #2, and #3, are respectively 1, 1, and 0. Hence, the probabilities that the car is actually behind each of these three doors are respectively 1/2, 1/2, and 0. So, your probability of winning is the same whether you stick or switch.

We shouldn't disregard that not every new reader already "knows Monty Hall problem through and through", like you and like Rosenthal. Gerhardvalentin (talk) 08:09, 11 July 2013 (UTC)

One thing we should all learn from this article and discussions on the subject is that there is no key to the MHP. We of course need to make clear exactly how the problem setup is generally understood to be but there is clearly no key to the problem in the sense that there is a simple fact or explanation that will quickly enable everybody to understand the correct solution. I think to look for one is to look for the Philosopher's stone. We here have all long forgotten the times when we first met the problem so probably cannot even remember what it was that helped us get to grips with it. I think we need to recognise that different people will be helped in different ways. Martin Hogbin (talk) 08:51, 11 July 2013 (UTC)

Philosopher's stone?
Without a doubt the famous paradox (1:2 and not 1:1) does arise in the fixed scenario where the host's action as per MvS is specified to be restricted:

— in 2 out of 3 cases, if the contestant first selected the door hiding goat A or the door hiding goat B,
then the host may not open one of the two non-selected doors at random,
but is extremely biased (forced) to show the second goat only, but never the car,
so in that 2 out of 3 he clearly is never equally likely to show goat or car,
— and in the remaining 1 out of 3 cases, if the contestant first selected the door hiding the car,
we know "absolutely nothing" about any host's given bias "which one" of his two goats he will show / he did show,
for the contestant, and for us too, he is "equally likely" to show just one / any of them.

I call the well defined restricted role of the host to be the "key for the paradox to arise", and I call this the "key" to understand the arising paradox (1:2 and not 1:1).

And helping the reader to understand, by showing that with a host who "always" opens one of the two unselected doors just at random, the paradox will not arise is important in the very beginning. Please notice that in other scenarios the famous paradox will not be given to arise. We shouldn't obfuscate any more. But treating quite other scenarios in scientific papers gave birth to the Monty Hall Problem-"problem-problem." That is shown in the article, well enough. Gerhardvalentin (talk) 11:39, 11 July 2013 (UTC)

My points were (1) your English is rather difficult, Gerhard; and (2) please can we avoid pushing a personal point of view / own research? Please can we remember that our primary task is to survey the literature on Monty Hall problem in a balanced way, *not* to explain the Monty Hall problem to the world? Please take a look at the Wikipedia policy pages, especially the article "what Wikipedia is not", WP:NOT, and from there, especially WP:NOTGUIDE. I agree with Martin's remarks. Especially the point that different people will be helped in different ways. Richard Gill (talk) 11:54, 16 July 2013 (UTC)

My point: please read the rules. Calling reliable academic sources to be OR, in explicitly disliking their illustrative view, isn't constructive and is of no advantage in WP. Gerhardvalentin (talk) 12:54, 16 July 2013 (UTC)

I think you misunderstand me. Never mind. Richard Gill (talk) 11:54, 17 July 2013 (UTC)

Citation style

Since recently, the article does many references like this: "This is a statement which needs a reference. (Selvin, 1973) This is another which needs a different reference. (Selvin, 1974)" However I think this should be written with the reference (which is in brackets) inside the sentence to which it belongs, ie before the full stop at the end of the sentence, not after. So you should have: "This is a statement which needs a reference (Selvin, 1973). This is another which needs a different reference (Selvin, 1974)." Just move the full stops along a little.

What are Wikipedia rules on this? Richard Gill (talk) 13:36, 14 August 2013 (UTC)

I don't know if there are WP rules on this issue. The {{harv}} recommendation follows your before rule, but the parenthetical rule I learned (in legal pleadings) is after punctuation. (There are 'bots that enforce the footnote-after-punctuation.) I'd leave it alone for now -- at least until the harv references are actually converted to templates. Glrx (talk) 20:52, 15 August 2013 (UTC)

Trust the lawyers... But I will also leave it alone for now! But notice: when you *read* "(Selvin, 1974)" you read a parenthetical phrase, not a footnote. You also know that you can look up Selvin (1974) in the literature list at the end of the article, if you want a proper literature reference. Well and nowadays there is a hyperlink to make this easy... Richard Gill (talk) 14:47, 16 August 2013 (UTC)

Example Proof of the Monty Hall Problem, using Bayes Theorem

I think the subsection "Example Proof of the Monty Hall Problem, using Bayes Theorem" should go. Anyone who has some mathematical training can rewrite the verbal solutions and the arithmetic solutions which the article presents as conditional probability solutions, using math formulas. Such "proofs", by introducing notation, converting problem input data into formulas, and doing some algebra, are found in probability textbooks and many are cited in the article. The "example proof" tells us nothing about Monty Hall problem which wasn't said earlier in the article and in ways which are more accessible to more people. Richard Gill (talk) 12:00, 23 August 2013 (UTC)

I moved it to the arguments page. Richard Gill (talk) 12:12, 23 August 2013 (UTC)

Tree-Diagrams for multi-stage experiments

When this game is played under the vos Savant assumptions, there are 12 elementary outcomes, as demonstrated in the tree-diagram posted at

http://mathforum.org/kb/plaintext.jspa?messageID=723345

In order to analyze a multi-stage experiment, one must state a precise description of exactly what activity is occurring at each stage. Once this is done, one of the best ways to calculate posterior conditional probabilities, using Bayes' Theorem, is to first draw the corresponding tree-diagram.Italus (talk) 15:17, 20 August 2013 (UTC)

Agreed. And the article indeed contains a tree diagram: a rather smaller one, based on the idea that though actually the car is hidden first and then the player chooses his door, one might just as well imagine these steps done in reverse order, and fix the choice of door by the player as door 1 from the start. Since the choice of door by the player is known by the player and the player is going to condition on their own knowledge when they decide whether to switch or stay, the smaller tree diagram is "fit for purpose" (and of course gives the same answer).

The smaller diagram has 4 elementary outcomes. 12 equals 3 times 4. The reason that the player's initial choice and the car hiding operation can be thought of in reversed order is because of the assumption of statistical independence of these two choices. The smaller diagram actually gives a more general solution: it doesn't matter with what probability the player chooses each door. We condition on his actual choice anyway. Richard Gill (talk) 12:28, 23 August 2013 (UTC)

In this particular problem, it is true that "it doesn't matter with what probability the player chooses each door." However, in general, one must state a precise description of exactly what activity is occurring at each stage. The other issue that I have with the convoluted articles on this and other "probability paradoxes" is that the articles do not emphasize the fact that they involve posterior conditional probabilities--you are dealing with a multi-stage experiment; you are given the result of a FUTURE stage; and you are asked for the posterior probability that you started at a particular PREVIOUS stage.Italus (talk) 20:52, 23 August 2013 (UTC)

I disagree with "must". Certainly it is a wise model-building strategy to describe stage by stage (in time) what is going on. But it isn't obligatory. Secondly I agree that followong this strategy, it is clear that you want to know the posterior probability of what happened at earlier steps (but was hidden) given what you actually saw, at later steps. Unfortunately the vast literature on Monty Hall problem does not agree. A lot of sources present various "short-cut" solutions, and many of them moreover give nice insights into the problem. So that is why the article is how it presently is: very encyclopedic, rather long and clumsy. If you want a short sharp powerful analysis you have to look elsewhere. The article gives plenty of excellent references. Richard Gill (talk) 17:43, 28 August 2013 (UTC)

Formerly featured article

We read above that "Monty Hall problem is a former featured article. Please see the links under Article milestones below for its original nomination page (for older articles, check the nomination archive) and why it was removed." I can't find it in the archives referred to here. Does anyone have links / dates? Richard Gill (talk) 12:26, 23 August 2013 (UTC)

If you click the [show] button in the notice template, there are links to the archived discussion/process pages with dated links to article versions at the time of the indicated actions. If [show]/[hide] is not working for you (possibly due to script problems in your browser) you can find this information by viewing the wikimarkup source for this page at the point where template {{ArticleHistory}} is invoked. ~ Ningauble (talk) 17:10, 23 August 2013 (UTC)

Thanks! Hadn't noticed the show button. Richard Gill (talk) 17:32, 28 August 2013 (UTC)

Falsifiable

(This paragraph has been moved to the Arguments page.) Gerhardvalentin (talk) 08:20, 7 September 2013 (UTC)

Monty Hall in popular culture

The Monty Hall problem features in Ian McEwan's latest novel "sweet tooth". See book review in "Significance". http://www.significancemagazine.org/details/webexclusive/4207941/Ian-McEwan-sinks-his-Sweet-Tooth-into-The-Monty-Hall-Problem.html Richard Gill (talk) 12:50, 15 September 2013 (UTC)

And in art: Stephan Huber's "Saussure's Heart". http://wissen.spiegel.de/wissen/image/show.html?did=18924671&aref=image025/E0115/SCSP200101502400242.pdf&thumb=false Richard Gill (talk) 13:36, 15 September 2013 (UTC)

"...at first sight, ludicrous"

I have to take issue with this phrase in the lede. "At first sight, ludicrous" assumes a great deal. Some people may certainly realize the correct answer at first sight, and to those who don't, the correct answer may not necessarily appear "ludicrous". Perhaps it should be changed to something like "at first sight may appear incorrect"? Joefromrandb (talk) 03:15, 10 October 2013 (UTC)

Or, alternatively, that it is simply "not intuitive"? I, JethroBT ^{drop me a line} 06:12, 10 October 2013 (UTC)

Yes, not intuitive, but moreover rather ludicrous, in just only considering that one of both still closed boxes definitely must hide the prize and the other one being empty for sure.

But imo it's a fault that the lede, even before clearly presenting the intended famous paradox, rashly deals with aberrant confusing variants, e.g. if Monty only offers a switch when the guest first selected the car ....

The article still avoids to clearly distinguish between the arising of the famous "paradox" and aberrant scenarios where the paradox may never arise. To the detriment of the reader who never will know what the lengthy verbose article is just talking about. Gerhardvalentin (talk) 08:49, 10 October 2013 (UTC)

Gerhardvalentin, your comment is nearly indecipherable. Could you please simplify and restate? — Preceding unsigned comment added by 72.95.199.241 (talk) 21:44, 21 October 2013 (UTC)

@72.95.199.241: Thank you. Did you read the article about the famous paradox? It's about two still closed doors/boxes, one of them holding a chance of 2/3 indeed to hide the prize, whereas the other one is holding a chance of 1/3 only. This is not really intuitive and moreover seems rather ludicrous, in narrowly (!) just only considering that one of both still closed boxes definitely must hide the prize and the other one being empty for sure. The article still is not clearly arranged to help the reader to understand what it is talking about. Obviously that famous paradox will arise only due to one special precedent scenario, see "The paradox". In not recognizing this fact and, without explicitly saying so, in mingling quite other scenarios, the paradox may not appear. The article does not distinguish clearly enough between the necessary precedent scenario and quite other aberrant scenarios where that paradox may never arise. The article still is an unclear mishmash, and instead of being crystal clear, imo is still rather confusing. Please tell us your opinion in this respect. Thanks, Gerhardvalentin (talk) 22:04, 22 October 2013 (UTC)

Gerhard, the subject of this thread is about the appropriateness of using the word "ludicrous" to describe how the correct solution appears at first sight. Your theories about why people often doubt the correct answer do not justify using this sort of subjective language in an encyclopedic article. I agree with Joefromrandb that it is inappropriate, and I endorse the change that has already been made in the article. ~ Ningauble (talk) 14:16, 24 October 2013 (UTC)

Actually I would like to put in a plea to go back to "ludicrous", or if you prefer "ridiculous". Experience shows, and the literature abundantly confirms, that 99.99% of people first hearing the problem *know* that the answer is "it doesn't make a difference", and they are incredulous about any claims that the answer is "you should switch". Probably it is a good idea to find some good literature references and if they can be found, go back to a stronger statement. Richard Gill (talk) 13:54, 6 November 2013 (UTC)

@Richard Gill You said "Probably it is a good idea to find some good literature references and if they can be found, go back to a stronger statement." If I included the Monty Hall Problem from the book This Book Does Not Exist, would that be helpful? The author notes that even some scientists get this wrong, so we as editors don't have to say it. The book also briefly describes its history, solution, why people tend to get it wrong, etc. meteor_sandwich_yum (talk) 08:28, 7 November 2013 (UTC)

@ Meteor sandwish yum I don't know that particular book. I have read just about every book I could find on MHP. In particular Jason Rosenhouse wrote a complete book just on MHP ... and everything we need can be found there too. Richard Gill (talk) 17:14, 16 November 2013 (UTC)

@Richard Gill Hey, sorry it took me so long to get that book. You guys probably don't even really need it now. I had intended to contribute more, but just sort of lost interest in the topic, as of late. I put a reference to the book with a lengthy quote and ISBN in the ref's section. If you guys can find a way to incorporate (part of) it into the article, great; otherwise, no big loss, I suppose. At any rate, hope what I could do helped. meteor_sandwich_yum (talk) 09:24, 22 November 2013 (UTC)

I agree with Richard. This is a very persuasive paradox.

We already have some relevant quotation in the article, vos Savant said 'Even when given explanations, simulations, and formal mathematical proofs, many people still do not accept that switching is the best strategy '.

Massimo Piattelli-Palmarini said '... no other statistical puzzle comes so close to fooling all the people all the time' and 'that even Nobel physicists systematically give the wrong answer, and that they insist on it, and they are ready to berate in print those who propose the right answer'.

I do not think this aspect of the puzzle can be over emphasised so if you have another source, Meteor sandwich yum, that would be most welcome. Martin Hogbin (talk) 09:27, 7 November 2013 (UTC)

Something stronger than "appear incorrect" or "not intuitive" may indeed be appropriate to describe the utter disbelief many people experience. However, "ludicrous" and "ridiculous" describe an attitude of derision and, although some of vos Savant's correspondents did express ridicule, I don't think this has been shown to be the typical reaction. I would suggest replacing the current clause "may at first sight appear incorrect" with something like this:

"is so counterintuitive it can seem absurd"

I think "counterintuitive" expresses the essence of what is meant by a "veridical" paradox, and "absurd" fairly characterizes the strength of incredulity evoked by the answer. ("Absurd" is also the term used in the article that explains veridical paradox.) ~ Ningauble (talk) 13:46, 7 November 2013 (UTC)

That would be fine by me. Richard Gill (talk) 15:12, 11 November 2013 (UTC)

Fine with me, also. – Btw my ceterum censeo: IMHO of much more importance, for those variants that slightly differ from the standard paradox, would be to explicitly indicate such specific differences. Imo such clear hints are of prime importance, and helpful for the reader. Gerhardvalentin (talk) 13:35, 13 November 2013 (UTC)

Okay, I updated the last paragraph of the lede section accordingly. Thanks. ~ Ningauble (talk) 15:43, 13 November 2013 (UTC)

"The Paradox" section

In order to help the reader to clearly distinguish the standard paradox's premise and quite other problems, I propose to say in section "The Paradox" as follows:

"The correct answer, that swapping doors doubles the chance on the prize because players who swap have 2/3 chance of winning the car whereas players who stick only have 1/3 chance, is based on the premise that the host knows which door hides the car and, after the player's initial selection, intentionally reveals a goat from behind one of the two remaining unselected doors. If the player by luck selected the door with the car (chance to be in that lucky-guess-scenario irrevocably is 1/3 only), then both remaining doors hide goats and the host may choose either door at random, and switching doors loses. But if the player initially selected one of the two doors each hiding a goat (irrevocably 2/3 chance to be in such wrong-guess-scenario), then the host's choice is no longer at random, as he has no options but to show his only goat, the second goat, and is forced to offer his still closed door that hides the prize. So in both wrong-guess-scenarios, switching doors wins the car for sure (Mlodinow 2008).

Consider a completely different problem that contradicts the assumptions of the standard paradox: the host, instead of intentionally revealing a goat, reveals one of the remaining two doors at random. If the host opens a door at random and reveals a goat simply by chance, then the odds are reduced from the standard paradox's 2:1 in favour of switching to 1:1 only. The latter odds track the common intuitive wrong answer because half of the potential winning cases are wasted when the host accidentally reveals the car, and by that discards such plain winning event (Rosenthal2005a) (Rosenthal, 2005b)."

Would be great if you could help to say that crystal clear in plain English. Gerhardvalentin (talk) 21:48, 23 November 2013 (UTC)

Two points of clarification:

I have placed the above post under a new heading because it appears to be a separate issue, unrelated to use of the word "ludicrous" in the lede section, which was the subject of the thread above.
Much of this language already appears in the abovementioned "The Paradox" section. It would be helpful to state specifically what you propose to change, and why you think it would be better.

(Cf. a similar proposal a few months ago.) Thanks. ~ Ningauble (talk) 14:54, 24 November 2013 (UTC)

?? – Clarification: Please read the wording of my above proposal, that's exactly what I propose to change. IMO this is better for the reason I also called above:

In order to help the reader to clearly distinguish the standard paradox's premise and quite other problems.

And there are lots of reasons, see the sources:

1. E.g. with one host and two players that simultaneously both do select different doors, say #1 and #2, and the host opens #3, the only unselected door. And there is quite a lot of other aberrant "problems". Since, you should preclude and obviate.

2. In first selecting a door, the player knows that in "exactly 1-out-of-3", he inevitably will be in the lucky guess scenario. And in "exactly 2-out-of-3" he inevitably will be in one of the two wrong guess scenarios. This rate of 1/3 : 2/3 can never be changed nor updated for the actual game being, this rate remains immutable. Please note that this is not only "an overall rate", no, this immutably is the actual rate of any single game. So we should preclude and obviate. Consider that even M et al's host can never change that rate, he just could be taken to give further advice as to the actual scenario the player actually is in, e.g. in disclosing that the player actually is in one "wrong guess scenario" for sure, and thus the "conditional probability" to win by switching approximates 1. But no-one, neither M et al's host, can change the rate of 1/3 : 2/3. And it is wise to help the reader to be aware of this fact, just from the beginning. No OR, see Mlodinow who has been tagged as careless. Once more: This ratio of 1/3 : 2/3 is invariable, for any single actual game.

3. And it is wise to help the reader to see just from the beginning that, in both wrong guess scenarios, the host never is "eqully likely", but is forced to offer his door with the car behind. Yes, "is forced". Despite I constantly heard that this was OR.

So please read again what I wrote above: In order to help the reader to clearly distinguish the standard paradox's premise and quite other problems.

I ask you to help to improve that important section, it should be as crystal clear as possible and, just from the beginning, help to avoid any possible ambiguity. Yes, we should preclude and obviate. Regards, Gerhardvalentin (talk) 22:22, 24 November 2013 (UTC)

I think that the present text in the article is perfectly adequate. It makes exactly the same points as you want to make, and it makes them more clearly (it is much shorter, and it is written in more or less decent English). Richard Gill (talk) 16:07, 26 November 2013 (UTC)

This probably seems so obvious to me. It is entirely due to how the host is forced to act in only one case:If you pick door 1, and it is door 2, the host is forced to pick 3, therefore you should always switch to door2 to take exploit the advantage of this..

(12/11/2013) — Preceding unsigned comment added by 99.232.160.130 (talk) 19:46, 11 December 2013 (UTC)

Yes, exactly. For this famous paradox it helps to say that there are exactly three possible scenarios: the one and only "lucky guess scenario", where the contestant by luck first selected the door with the prize behind (in only 1 out of 3 cases) and switching doors will lose the car. In this scenario both unselected doors hide goats and the host may randomly open any of them.

But in both "wrong guess scenarios", where the guest first selected one of those two goats, the host obviously is forced to show the other goat, and to offer a switch to his door that definitely hides the car.

And please consider that in the beginning, the contestant knows exactly that his first choice has 1/3 chance only. And as to that famous paradox, after the host has shown a goat and offered a switch, the contestant still knows that his first choice unchanged has only 1/3 chance, so switching doors doubles his chance to win the car. Gerhardvalentin (talk) 20:48, 14 December 2013 (UTC)

And all this is written in the article! Readers who read carefully can find everything they need. It does not need to be repeated again and again in different words. Wikipedia is not a text book, it is not a pedagogical guide, it's an encyclopedia. The purpose of wikipedia is not to teach but to help people find information. Richard Gill (talk) 20:52, 17 December 2013 (UTC)

Simpler more intuitive explanation

The contestant chooses a losing door in two thirds of all games played. The winning door is chosen in only one third of all games. Therefore, in only one third of all games is the winning door and the chosen door the same. Since the host never opens the chosen door nor the winning door, it is only in these one third of all games that the host has the option of opening either of the two remaining doors. Whereas, in two thirds of all games, the chosen door and the winning door are different doors leaving the host only one door to open. In these two thirds of all games, the door which the host leaves unopened is the winning door. In only one third of all games played, does switching lead to the wrong door. In two thirds of all games, switching leads to the winning door.

If others agree this is a more intuitive explanation which leads readers to the correct conclusion more quickly, please comment and replace the paragraph below which appears as the fifth paragraph in the article with the paragraph above.

"Contestants who switch have a 2/3 chance of winning the car, while contestants who stick have only a 1/3 chance. One way to see this is to notice that, 2/3 of the time, the initial choice of the player is a door hiding a goat. When that is the case, the host is forced to open the other goat door, and the remaining closed door hides the car. "Switching" only fails to give the car when the player picks the "right" door (the door hiding the car) to begin with. But, of course, that will only happen 1/3 of the time."

--75.199.38.85 (talk) 14:09, 20 December 2013 (UTC) --Jerrykeywest (talk) 14:16, 20 December 2013 (UTC)

Keep existing. I like the existing version; its topic sentence gives good guidance. The proffered replacement does not. Glrx (talk) 00:08, 21 December 2013 (UTC)
I too prefer the existing wording. Someone tried rewriting it somewhat like the above but only made it less clear and obvious, so I've restored the original wording.--JohnBlackburne^words_deeds 23:59, 21 December 2013 (UTC)

Sample space question

In this problem, is it correct to say that the sample space is this:

1. At 1st choice; 1 car, 2 goats, 3 doors.

2. At 2nd, choice; 1 car, 1 goat, 2 doors.

In other words, does the Sample Space shrink when the host removes a door from the available choices?

Tweedledee2011 (talk) 19:05, 22 December 2013 (UTC)

NO; read again or consider remedial mathematics. Cheers Crusoe8181 (talk) 09:40, 24 December 2013 (UTC)

Not a very helpful or polite comment. Martin Hogbin (talk) 00:26, 26 December 2013 (UTC)

I'm not sure you understand my question. A sample space is the set of all possible outcomes of an experiment. When a door is removed by the host, it ceases to be part of the sample space, because it's no longer choose-able and if it's no longer choose-able, it's not part of the second experiment. When that happens, you are left with 2 doors, 1 goat and 1 car. Therefore, if door #3 is removed, the remaining possibilities are D1G D1C D2G D2C. The sample space at the point of the second choice is two possible positions for either the goat or car. The first experiment results in us getting an unknown outcome from an event. We choose a subset of the sample space (a door), but we don't yet look behind it. Then comes another experiment. This is when we make another choice, and switch (or stay). This event results in a knowable outcome because as part of this event, we open the door. Suffice it to say, this question is dishonest because the first event does not come to fruition - the outcome does not become known as a result of the event. Rather, the outcome of the removal of the door is pre-determined by a prior event - the prior inspection of the door contents by the host to verify that only a goat door will be removed. What this problem actually does is inject a player into a partially completed series of experiments, and then illegitimately allows the player to calculate a choice based on the prior knowledge of the host - which was acquired externally to the experiments conducted by the player. In other words, the "winning" method requires reliance on information acquired outside of the boundaries of the player's experiments. Not only that, but by removing a door, that door can no longer be chosen, so it's not possible that the removed door has the car. As a result, to include any math related to the state of the removed door, is to calculate odds based on information which depends on a set that no longer exists - the set of three doors. At the point of the second choice, the removed door is no longer part of the set which available to choose from, so it's not part of the sample space. Hence the sample space of the second choice must only be D1G D1C D2G D2C. That's two doors, and the possibility that either goat or car may be behind either of them. The odds of the second choice, if calculated using only the possibilities available to the second event, are 50/50. Tweedledee2011 (talk) 20:37, 24 December 2013 (UTC)

Please read again section "The paradox". Your reasoning does not address the famous paradox, where the host opens one of the two REMAINING doors in order to show a goat, AFTER the guest already has made his first decision – with a chance on the car of only 1/3 (in that 1/3 staying wins), but with a chance on a goat 2/3 (in that 2/3 switching wins).

You refer to another variant (3 doors, originally 1/3:1/3:1/3) where a goat had been removed BEFORE the guest afterwards makes his first decision out of only 2 doors (1/2:1/2). In your aberrant variant, chances of sticking:switching indeed are 1/2:1/2. Please read the introduction. Gerhardvalentin (talk) 16:38, 25 December 2013 (UTC)

Perhaps I did not make clear what I am saying. Allow me to simplify: At the point in the game, where the door is removed, the removed door ceases to be part of the sample space of the game. That is what I am saying. Why do I say this? Because of the very definition of sample space which is "A sample space is the set of all possible outcomes of an experiment". This game is not one single experiment. It is a series of experiments. The knowledge acquired about the true door contents, is acquired by the host prior the beginning of the game, that's experiment #1 - this is when he familiarizing himself with all the doors/contents. The first choice of a door by a player, that's experiment #2. But experiment #2, unless the player sticks with that door, never comes to fruition, because it's not opened. After the player makes choice number one, the host removes a door - but that removal is NOT an experiment. The information required by the host so as to make a removal of a goat-door was acquired in experiment #1, prior to the start of the game. What I am claiming is that the goat-door removal by the host, reduces the size of the sample space because the player is NOT allowed to choose the removed door, therefore the removed door DOES NOT hold any possibility of having the car (or goat) for the player. To be part of the sample space, a door must have a bona fide possibility for both yes and no. The sample space at the player's first choice is D1G D1C D2G D2C D3G D3C in however a pattern you want to arrange it, provided that you limit the car to one unit and fill the other two with goats. Those are the total possibilities in the starting sample space (set of possibilities). However, when a door is removed, if the rules of the game are followed, the removed door CAN NOT have a C or G possibility for the player's next choice. Therefore, since the removed door is not available to the player, it's not possible that it's part of the game's sample space. The only doors which can be counted as part of the sample space of the game, are doors which have both POSSBILITIES to the player - which requires that they be available to be chosen. Any door which is not available to be chosen, fails the predicate definition of Sample Space and is therefore excluded from the game's sample space at the moment it fails the test. It's not the fact that removed door #3 has a goat which makes it fail. Rather, it's the fact that it's not choose-able which makes it fail. It fails the definition of possible, therefore it's not part of the player's sample space after it's removed. And to be extra clear, let's be sure to call each of the player's chocies what they are: Events. And since an event is a subset of a sample space, then if at the point of the second choice, the 3rd door is not part of the set from which the choice can be made; it's not, and can't be part of the event which is the second choice. My contention is that the removed door, because it's not available to the player to choose, is not part of the player's sample space. In other words, the size of the sample space available to the player changes when the door is removed. This is what I am saying. Tweedledee2011 (talk) 19:40, 25 December 2013 (UTC)

Tweddledee, I think you are possibly misunderstanding the concept of a sample space. A sample space is, 'the set of all possible outcomes or results of that experiment'. Follow the link to see what is meant by 'outcomes'. Martin Hogbin (talk) 00:31, 26 December 2013 (UTC)

Martin - the wiki article about 'outcomes' which you directed me to, says the following:

'In probability theory, an outcome is a possible result of an experiment'.

If that is an accurate statement, then it proves my point, which is this: The total possible outcomes from the second choice is less than the total possible outcomes of the first choice, because after the first choice and prior to the second choice, two of the possible outcomes are removed from the game. The removed outcomes are D3G and D3C. If D3 is removed from the game, it's not possible for the player to choose it, so anything it contains is not a possible outcome of the second choice. The player can not conduct an experiment on the 3rd door - because it's removed. If the player deduces information about the third door, that information was not arrived at as a consequence of the player's experiments. Rather, it's only if the player knows that the host will not remove the car door, that the player acquires information about the third door. That information, again, was not arrived at from the player's experiments. Rather, as I said above, it precedes the player's actions in the game. For this trick question to work, it must rely upon information which was developed before the start of the game, by the host's experiment (see above). Again, let's return to my question: Does the removal of the door reduce the size of the sample space for the player's 2nd choice? I say it does, and if it does, then playing this game honestly would require disregarding any information known or deduced about the contents of door #3. This is why if a new person comes in and chooses, without any information, there is no advantage to switching. The advantage to switching lies in the knowledge about the 3rd door. But to be intellectually honest, one has to admit that it's illicit to use that knowledge if door #3 is removed from the game - because door #3 is no longer on the playing field - and it's no longer part of the player's sample space. Tweedledee2011 (talk) 04:03, 26 December 2013 (UTC)

Tweedledee2011 (talk) 07:03, 26 December 2013 (UTC)

You are correct that the host opening a door reduces the sample space. Looking at this in more detail (assuming the player picks door 1) the sample space before the host opens a door is:

D1=C, D2=G, D3=G, HostOpens=D2

D1=C, D2=G, D3=G, HostOpens=D3

D1=G, D2=C, D3=G, HostOpens=D3

D1=G, D2=G, D3=C, HostOpens=D2

The probabilities of these are 1/6, 1/6, 1/3, 1/3 (in the case the car is behind door 1 the host can open either door 2 or door 3, assuming this is an equal choice the probability of either of these is 1/2 the probability the car is behind door 1). If the host then opens door 3 the remaining possibilities are

D1=C, D2=G, D3=G, HostOpens=D3

D1=G, D2=C, D3=G, HostOpens=D3

with original probabilities 1/6 and 1/3. Scaling these up as conditional probabilities (given the host has opened door 3) yields the typically stated 1/3 chance of winning if you stay (with door 1 given the host opens door 3) and 2/3 chance of winning by switching (to door 2 given the host opens door 3).

There being N possible outcomes does not mean the probability of each occurring must be 1/N. -- Rick Block (talk) 22:06, 29 December 2013 (UTC)

Here's my concern: The Monty Hall problem is a deceitful pastiche which does not teach either reasoning or probability very well. If one takes all the information into account, then yes, switching is the best thing. However, if just prior to choice #2, you ask the question "What are the odds of the car being behind either door?", the answer is 50/50. The odds of finding the car takes into account the information derived from the removed door. But the distribution of cars to doors at the point of choice #2 is clearly 1 in 2 - as my sample space diagram shows. The smugness with which the experts explain the "switch is better" option hides the fact that there are multiple sets of data involved here: The original sample space and the smaller, "2nd choice" sample space. This is why, if a new player enters at choice #2, his odds are 50/50 - because he has no access to the information derived from the larger sample space. Tweedledee2011 (talk) 18:32, 30 December 2013 (UTC)

In your question (just prior to choice #2) what do you mean by "what are the odds of the car being behind either door?" If you mean what is the limit of the fraction of times the car will be behind door 1 (or door 2) if you repeat this setup numerous times (I'll assert this is what most people think of as "probability"), the odds are 1/3 for door 1 and 2/3 for door 2. If you mean what is the conditional probability the car is behind door 1 (or door 2) given the player has initially picked door 1 and the host has opened door 3 (knowing it will reveal a goat and choosing door 3 randomly if the car is behind door 1), then the odds are also 1/3 for door 1 and 2/3 for door 2. If you mean what are the odds of selecting the car by random choice (as a new player who enters at this moment must), the odds are 50/50 (since there are two doors). However, this latter meaning is not what most people think of as "probability". The reason the random choice ends up 50/50 is because a random choice between any two alternatives always results in a 50/50 success rate. In particular, if the probability the car is behind door 1 is 1/3 and the probability it is behind door 2 is 2/3, then the overall success of a random choice is (1/2)*(1/3) + (1/2)*(2/3), which is 1/2! This is not saying anything about the "probability" (in either of the other two meanings I've suggested) of the car being behind each door - just that the chance of picking the car with a random choice is 1/2.

Lets say you are at a university where you know 90% of the students are male and 10% female. If you run into a random student what are the odds this student is male? Is it 50/50 because there are only two choices (male or female)? If you encounter 100 students and flip a coin to guess male or female (lets say heads means your guess is male), you'll be correct about 50% of the time. However, I think most people would say the odds are 9/10 male and 1/10 female (of the 100 students you encounter, about 90 will be male and about 10 will be female). The Monty Hall problem is similar. At choice #2 the car really is twice as likely to be behind door #2 than door #1! Try the card experiment - see for yourself. Take the ace of spades and 2 red 2s. Shuffle. Deal 3 cards. Look at the 2nd and 3rd (the player picked #1). Turn over a red 2 (flipping a coin to decide which one to turn over if both are red 2s). If you happen to have turned over the 3rd card (host opened door 3), make a note of which card is the ace (#1 or #2). Do this until you have about 30 samples (more is better). You'll see the ace will be card #1 about 1/3 of the time and #2 about 2/3 of the time (as you do more and more of these experiments the proportion will become more and more exact). Note that it is important always to look at both cards and actually flip a coin to pick which one to turn over if both are 2s (lets say heads means you flip over #2). If you just look at #3 and flip it over if it's a 2, the experiment won't work (you're not following the proper rules). You have to both flip over a card you know is a 2, and pick evenly between them if both are 2s. If you do this experiment, please report back what you find. -- Rick Block (talk) 21:16, 30 December 2013 (UTC)

Ahem. Paradoctor (talk) 22:01, 30 December 2013 (UTC)

The problem with this question, is how the vernacular trips people up. The odds of finding it are not the same for the original player as they would be if a new player came in just prior to choice #2. This is because a 2nd player coming it at that point would not have the information gleaned from the larger sample space which existed prior the door removal. And, even though if this series of events were not interrupted, then yes, switching would be better, even for a new player, the new player could not know that - because he lacks the information. What I am saying is that this is a trick question and this article is weak, because it omits a proper explanation of "sample space" and "experiment". If each choice is a separate experiment, then the reduction of the size of the sample space must be taken into account. This is so because in the real world, when calculating probabilities, there is no perfect knowledge of the sample space. For that reason, this question (and article) deludes people into misapplying the parameters of a true probability event. For something to be part of a sample space, it has to hold the possibility of being true - but the removed door fails this test. By restricting the removed door to only goat doors, it's not possible to say that the removed door is part of the sample space. And if it's not part of the sample space, then making calculations based on knowledge of it, is outside the scope of the second experiment. That is, if the 2nd experiment (2nd choice) is discrete. So then, perhaps that's the real question: From a standpoint of calculating the event of the 2nd choice, is the 2nd choice a discrete choice or is it part of a series of choices? If it's a discrete choice, then it can't use the information gleaned by the door removal as that act pre-dates the experiment which is the event of the second choice. By definition, an "event" is a subset of a "sample space" and at the point of the 2nd choice, the removed door - and all its possibilities are no longer part of the sample space. And as such, it's axiomatic that the removed door - and all its possibilities - must be excluded from the 2nd choice. This question isn't about the math, this question is about how to honestly explain what logic-rules should be used to apply the math, and how, at which stage in the process. Tweedledee2011 (talk) 23:25, 30 December 2013 (UTC)

Since this discussion is not apparently about the article but rather the math behind the problem, it is probably more appropriate for the /Arguments subpage. We could move it there if anyone here is unduly bothered. -- Rick Block (talk) 22:29, 30 December 2013 (UTC)

I'm not happy with the suggestion that I am "unduly" concerned.

If the discussion is on topic, it belongs here. If it's not, it doesn't belong anywhere on Wikipedia, per WP:NOTFORUM. Paradoctor (talk) 02:31, 31 December 2013 (UTC)

Please do not move this discussion. It's needed here to clarify whether or not the change in the sample space should be fully explained in the article. Tweedledee2011 (talk) 23:09, 30 December 2013 (UTC)

If we're talking about the article, and not the math behind the problem, then the first question has to be what references are you talking about? If there are no references supporting your point of view, then the discussion is over. If there are references, then the question becomes how (or possibly even whether) to include what they say into the article. -- Rick Block (talk) 23:54, 30 December 2013 (UTC)

I have linked to event and sample space - are those enough? Tweedledee2011 (talk) 00:05, 31 December 2013 (UTC)

No. You need reliable sources discussing sample space in the context of the Monty Hall problem, otherwise this is WP:OR. Paradoctor (talk) 01:27, 31 December 2013 (UTC)

I don't understand you saying that. It's axiomatic that sample space is an essential aspect of a probability calculation, so to omit any reference to sample space in an article about the most well-known probability brain teaser seems needlessly obtuse to me. Tweedledee2011 (talk) 01:39, 31 December 2013 (UTC)

I agree with Paradoctor. A general discussion of events and sample spaces as applied to the Monty Hall problem would be original research (OR) without a specific reference. Since there are hundreds of published articles and multiple books specifically about and covering every conceivable aspect of the Monty Hall problem, virtually any approach (or any thought about it) has already been published - and if it hasn't been published it's OR (pretty much by definition).

Another approach for you here would be to suggest a specific change - exactly what are you suggesting be changed and in exactly what section of the article? For example, are you suggesting expanding the section "The little green woman"? Perhaps I'm misunderstanding, but it sounds like you're suggesting adding something that says the probability is 50/50 after the host has opened door 3 (when the player is deciding to switch from door 1 to door 2) because there are only two possible locations for the car. There aren't any reliable sources that say this (it's not correct), so if this is what you're looking for I think you're going to be disappointed. If this is indeed what you're thinking I'd be happy to discuss the math with you on the /Arguments subpage if you'd like. Rick Block (talk) 01:47, 31 December 2013 (UTC)

No what I am saying is that this article, by omitting any explanation of 'experiment', 'sample space' or 'event' leaves novice readers at a disadvantage. This article, as a matter of editorial choice, ought to explain those concepts. As it stands, even some of the questions I asked here have not been answered. For example, if the sample space shrinks upon door removal, how can one honestly say that the 2nd choice can calculate in the information which the removed door provided? In other words, a door choice is an "event" and an event is a subset of a sample space. But the event of the 2nd choice is not a subset of the original sample space. Rather, it's only a subset of the reduced sample space. And the reduced subset, I contend, is a wholly distinct sample space from the original one, not a subset of it. The removal of the door from the original sample space destroys that sample space because by definition, a sample space is all the possible outcomes of an experiment (or trial), but when you open a door, you eliminate that door as a source of possible outcomes for the 2nd choice. This is why I'm saying the article doesn't explain things well. It might very well be a distinction without a difference, but I still think it should be explained how we are still able to calculate the odds, even though the sample space is modified. We are able to do so, essentially, by thinking "outside the box" at the point of choice #2. That's because it's literally true that the information derived from the removed door of the original sample space is from a transient superset which no longer exists (the original sample space). Tweedledee2011 (talk) 02:20, 31 December 2013 (UTC)

Please use proper WP:indentation.

"the reduced subset, I contend, is a wholly distinct sample space from the original one" (my emphasis) That is the problem. What you contend is not encyclopedic. Is there a reliable source saying this? Paradoctor (talk) 02:37, 31 December 2013 (UTC)

Paradoctor: I don't understand your point. Either you agree or disagree. If you agree, then help me bring my contention into line by finding a good source. The absence of a source is why it's on this page - to be discussed. But discussing it means just that - discuss it - not reject it out of hand merely because it's not sourced yet. If it's not a distinct sample space, then why not? And if is, then help me find a good citation. Tweedledee2011 (talk) 02:49, 31 December 2013 (UTC)

Quote from talk page guidelines: "Use indentation as shown in Help:Using talk pages#Indentation (or, more specifically, Wikipedia:Indentation)". Please note that a guideline is "a generally accepted standard that editors should attempt to follow".

"bring my contention into line by finding a good source" WP:BURDEN. Furthermore, I don't think there is such a source.

"Either you agree or disagree." My opinion on this matter is just as uninteresting as yours here. We report what the sources say, not our own musings. Paradoctor (talk) 03:41, 31 December 2013 (UTC)

(responding to Tweedledee) A full decision tree enumerating all events that might happen is presented in the Grinstead and Snell book (an elementary probability textbook in the article's references, available online). Although presented there as a decision tree, this tree is for all intents and purposes the same thing as a sample space. A similar tree (from both Carlton and Chun, also in the references) showing only the events following "player initially picks door 1" is presented in the article in the "Conditional probability by direct calculation" section. If you're having trouble understanding the text in this section or simply think it's unclear, it's fair to ask here for the text to be clarified. I hope you don't take offense, but it seems like you're having trouble understanding the basic concepts of sample spaces and how they change in response to events. This sort of discussion belongs on the /Arguments subpage as this article is not a tutorial on basic probability theory. And, perhaps a more direct response, the article does not mention sample spaces because that's not how most references approach the problem. -- Rick Block (talk) 06:09, 31 December 2013 (UTC)

(responding to Rick Block) I think you are mistaken that a sample space changes in response to an event. Think about the search for the SS Central America. The search method they used took into account not only the probabilities of the wreck being in various quadrants in the search grid, but also the likelihood of actually being able to successfully locate it in the grid, even if the correct grid was searched. Suffice it to say, unless and until the wreck was found, then in regards to any particular quadrant, unless the searchers were certain (after inspection) that the quadrant was empty, then it wasn't eliminated from the sample space. But what is it that enables quadrant elimination? Inspection, that's what. However, in this game it's not an inspection by the player which eliminates the door. Rather, it's the prior knowledge by the host which eliminates the door. The actual opening of the door is a perfunctory action to prove that removal is valid (i.e.; has no car). Suffice it to say, it was not the action of the host's inspection of the door which eliminated it. Rather, it was the action of the host to deem the door eliminated which does that. An "event" is not an "action" in this context of this vernacular. Rather, an "event" is a subset of a sample space. But the subsets are entirely arbitrary and do not eliminate themselves from further consideration. Rather, it is our confidence in the accuracy of the information yielded by the experiment (look and see) that leads us to eliminate things from further consideration. However, in the game, it's not the player who eliminates the door, it's the host. And if the player is not aware of the fact that he can use the derived information about what the removed door does not contain (the car) to best formulate his next choice, then he's not able to understand that his results can improve by switching. So without beating a dead horse, I still think that this article could benefit from explaining the changing composition of the sample space and how a wise player can, even if the samples space shrinks, use the information from the previously large sample space (and removed door). In other words, because this article is written by those well versed in the topic, it overlooks that novices might like to see information about sample space in this article. Tweedledee2011 (talk) 07:38, 31 December 2013 (UTC)

Four users have told you that you're wrong. Two have told you that this discussion doesn't belong here. You have been offered to continue the discussion at /Arguments. Further edits in violation of WP:TPG will be deleted. How much farther you escalate this is up to you now. Paradoctor (talk) 07:57, 31 December 2013 (UTC)

Tweedledee, although I do not agree with some other editors about absolutely everything, we all agree that, in the standard interpretation of the problem, the probability of winning by switching is 2/3 and not 1/2. This page is for improving the article but you have not found any valid reason to make changes to the article so further discussion here is pointless.

If you want to discuss the underlying theory of this problem you are welcome to continue this discussion at the /Arguments page, where I and others will be happy to discuss more general issues of probability with you. Martin Hogbin (talk) 09:56, 31 December 2013 (UTC)

2 small corrections?

After rereading the current article I noticed 2 things that appear to be incorrect/incorrectly sourced.

In the lead there is currently the following line: "The argument relies on assumptions, explicit in extended solution descriptions given by Selvin (1975a) and by vos Savant (1991a)". However as I understand Selvin is giving the extended solution description, in particular the explicit mentioning of the assumptions, in his 2nd letter only. Hence the source for that sentence should be (1975b) rather than (1975a) or at least (1975a, 1975b).
The Recent discussion section lists several publications that have referenced the WP article on MHP. It states: "Nowadays, Wikipedia is frequently cited as a source for Monty Hall problem (it is cited, for instance, by Ruma Falk, Rosenhouse 2009, Gill 2012, Gnedin 2012).". The only publication by Falk given in the References section however is an article from 1992, which certainly doesn't reference WP (only being around since 2001). So if there is another (later) publication by her which indeed references WP, then it should be added to References section otherwise she should be removed from that enumeration.

--Kmhkmh (talk) 09:36, 31 December 2013 (UTC)

Done Just out of curiosity: Is there a specific reason you didn't do it yourself? Paradoctor (talk) 16:56, 31 December 2013 (UTC)

brain teaser

Please see my edit calling this a "brain teaser" - and take note of the accompanying edit summary to that edit. Tweedledee2011 (talk) 11:22, 1 January 2014 (UTC)

With no reference to brain teaser in the body of the article, why should it be in the lead? Flat Out let's discuss it 11:28, 1 January 2014 (UTC)

Not a brain teaser at all. It's just a trivial little probability exercise. The thing that is teasy about it, is that some people seem to find it difficult—or even important. Without a proper source the predicate brain teaser reeks a bit like original research, or as just an opinion at best. Let's strike. - DVdm (talk) 12:21, 1 January 2014 (UTC)

Read Brain teaser and tell me this problem doesn't meet that definition. Tweedledee2011 (talk) 13:10, 1 January 2014 (UTC)

The lead is a summary of the main points of the article and brain teaser isn't in the article.Flat Out let's discuss it 13:20, 1 January 2014 (UTC)

Agree to strike: "brain teaser probability puzzle" is completely redundant. Puzzle is a perfectly good word (used by several of the sources) and there is no need to repeat the concept with a crude, informal vernacularism. ~ Ningauble (talk) 13:41, 1 January 2014 (UTC)

Otoh, this. Sheesh :-) - DVdm (talk) 15:18, 1 January 2014 (UTC)

The article is about a puzzle that reliable sources call
"The World's 200 Hardest Brain Teasers". (Gruber, Gary 2010). Sourcebooks, Inc. p. 136. ISBN 978-1-4022-3857-4. and
"The Psychology of the Monty Hall Problem: Discovering Psychological Mechanisms for Solving a Tenacious Brain Teaser". Krauss, Stefan and Wang, X. T. (2003), Journal of Experimental Psychology: General 132(1), retrieved from http://www.usd.edu/~xtwang/Papers/MontyHallPaper.pdf March 30, 2008.

It's a puzzle, yes, but in addition to that, reliable sources emphatically call it "a Tenacious Brain Teaser", and "one of the world's 200 hardest brain teasers". Gerhardvalentin (talk) 15:40, 1 January 2014 (UTC)

Thank you, Gerhardvalentin Tweedledee2011 (talk) 17:57, 1 January 2014 (UTC)

Yes, a reliable source and probably end of discussion. But still, for such a no-brainer... sheesh :-) - DVdm (talk) 16:00, 1 January 2014 (UTC)

So now we have "a tenacious brain teaser probability puzzle". Sheesh, how much purple prose can we cram into the opening sentence?

Yes, a couple writers use the brain teaser idiom, but to the extent this is synonymous with puzzle it is redundant in the first sentence (and to the extent it is not considered synonymous it is not a well defined or informative distinction). Yes, one source refers to the problem as being tenacious, but there is no need to copy their proclivity for emphatically titled prose. As Flat Out notes, the lead should be a summary of the main points of the article. The article does not discuss brains, teasing, or tenacity.

It is not appropriate for the opening sentence to try to list every adjective or synonym that has ever been used to describe the problem. ~ Ningauble (talk) 17:05, 2 January 2014 (UTC)

reliable source from MIT

This paper [5] (from two professors) raises the exact issue that I have been raising, which is "Was Marilyn really right? Our analysis suggests she was. But a more accurate conclusion is that her answer is correct provided we accept her interpretation of the question...". The authors make this point on a different aspect of the question that I have been pointing to, but it's the same point I've been making: Are we presenting this question correctly? We are calling it a probability problem, but we are omitting any discussion of sample space - and whether (or how) the sample space changes during the game, and what effect that change has on the choosing process. As a result of this omission, the article is not as good as it could be. Tweedledee2011 (talk) 20:58, 1 January 2014 (UTC)

The mentioned paper uses the Monty Hall problem (MHP) to introduce probability theory. This is not a good idea, as in the MHP the probabilities of interest are conditional probabilities. The conclusion that vos Savant was right in her interpretation of the problem is disputable, as the posed question is not whether switching is better than staying in general, but whether switching is better than staying GIVEN door 1 has first been chosen AND door 3 has been opened. Only when you understand the difference you may understand the MHP. BTW: the assumption made in the paper that the first choice is done randomly, is superfluous. Nijdam (talk) 11:40, 4 January 2014 (UTC)

The underlying scenario of this world famous paradox is shown in the article. Leonard Mlodinow says "understanding the Monty Hall problem requires no mathematical training. But it does require some careful logical thought". For years the article unnecessarily had been a math textbook on Bayes. Please stop trying to teach the bascis of "sample space" where maths and sample space isn't the core. Gerhardvalentin (talk) 23:39, 1 January 2014 (UTC)

Tweedledee2011, there is nothing new in your source. They show that if you interpret the question differently you get a different answer; that is no surprise. Vos savant, and Krauss and Wang have both shown that most people interpret the question in the 'standard' way. Variants of the problem are covered later in the article. Martin Hogbin (talk) 23:58, 1 January 2014 (UTC)

Thank you both for your comments. It seems that the editors camped on this page have no interest in making this article available to a wider variety of readers, including those who rely on verbal reasoning and/or are only modestly skilled at probability calculations. Bearing that thought in mind, I doubt I will comment on this page again. Tweedledee2011 (talk) 01:22, 2 January 2014 (UTC)

That is not fair. I am very interested in making this page available to the widest variety of readers but I cannot see how what you are suggesting that will achieve this. The question (the Whitaker/vos Savant version) is badly stated and open to misinterpretation but the evidence we have is that nearly everybody does interpret the problem in the same (standard) way. Misinterpretation of the question is not the main reason that people get the answer wrong. There is ample evidence that even when the question is carefully and rigorously explained to people they still get it wrong. Any help in helping people to understand why the answer is 2/3 is most welcome. Martin Hogbin (talk) 10:23, 2 January 2014 (UTC)

Martin - if that's true, then I invite you to dialog with from the standpoint of trying to understand what I am saying, without thinking that I am trying to make you agree. (Redacted) In other words, I am pretty sure that with help from someone like you, I can articulate what I think the issues are - but I can't do it if I keep getting chopped off at the knees. Please see my post at [[6]] What I want to do next is establish an exactly correct visual representation of the sample space at Choice #1 and Choice #2. If (and only if) I am able to accomplish that do think I can make any progress here. Look above - this was my reason for posting here in the first place. (Redacted) But I first need group consensus on a block diagram (not a decision tree) of what the sample space looks like at choice #1 and Choice #2. I am persuaded that the secret to understanding the way people go astray here lies with some of the ideas I picked up from DeBono ISBN 978-0816031788, along some clear visuals and some very exact parsing of the verbal aspects of the principles of probability. The math only proves things the right way - it doesn't help with understanding how things are reasoned the wrong way. Trust me when I say this: (Redacted) if you help me, I can develop something which is a) useful to the article and b) supported by reliable sources Tweedledee2011 (talk) 08:10, 3 January 2014 (UTC)

A sample space illustration would be appropriate only if sample space is at least mentioned in the article, which it isn't. If you can find sources supporting your position, please do so, first. If you need help searching for sources, just ask.

"The math only proves things the right way - it doesn't help with understanding how things are reasoned the wrong way." That's a valuable insight into what it takes to solve a paradox. But do not forget this: Whatever argument is provided, it must be from reliable sources, not just your own reasoning. Paradoctor (talk) 13:38, 3 January 2014 (UTC)

It's (sample space) not mentioned in this article for the simple reason that the editors here seem hell-bent on keeping any discussion of it out of the article. This is one of (if not the most) well known probability problems and yet, there is staunch opposition on this page to introducing some information about the essential probability concept of "sample space" into the article. Why is that? Tweedledee2011 (talk) 01:17, 4 January 2014 (UTC)

Because the literature on the Monty Hall problem does not contain discussions of the sample space WRT to the Monty Hall problem. This means it won't be mentioned in the article either. Provide sources or drop your attempt at including discussion of the sample space into the article. Paradoctor (talk) 01:37, 4 January 2014 (UTC)

Here's one, at UMBC.edu [7]

Here's something else, from a well known author [8]

Here's a PPT at another university which mentions sample space and MHP [9]

Tweedledee2011 (talk) 02:09, 4 January 2014 (UTC)

The UMBC image is not a reliable source. Mlodinow gives the sample space on page 54 as {Maserati is behind door 1, Maserati is behind door 2, Maserati is behind door 3}. The lecture notes are not a reliable source, especially given the amount of existing literature, but we can include it here for the sake of discussion. The presentation gives the sample space on page 35 as {prize behind door 1, prize behind door 2, prize behind door 3}.

The sources merely declare the sample space, there is no discussion of the contents of the sample space, much less of any changes to it resulting from the player's choices. Can you provide sources doing that? Paradoctor (talk) 03:06, 4 January 2014 (UTC)

The block diagram I want to add is no different than saying 3+7=10. It's an empirical fact that the sample space is whatever it is. All I want to do is put in a block diagram of whatever it is. Why is that so hard for you to understand? Tweedledee2011 (talk) 03:27, 4 January 2014 (UTC)

nyet A block diagram for "{prize behind door 1, prize behind door 2, prize behind door 3}"? I don't think so. Of course, I'll defer to consensus. Show the diagram you want to include, state where it should be placed, and if you can get support for it from other contributors, then there is no problem. Paradoctor (talk) 03:44, 4 January 2014 (UTC)

Ok - see new section below. And take note, this is my initial proposal - I'm not sure if it's correct.Tweedledee2011 (talk) 04:01, 4 January 2014 (UTC)

I believe the large figure in the section "Conditional probability by direct calculation" effectively shows the sample space (even though it is sourced to the decision tree - note that these are completely analogous views). Assuming the player's initial pick is door 1, the sample space is {(3,2), (1,2), (1,3), (2,3)}, where (x,y) indicates the car is behind door x and the host opens door y. The probabilities of these outcomes are 1/3, 1/6, 1/6, and 1/3, respectively (indicated visually in this diagram by column width). If the host opens door 3, only the latter two (shown on the right) remain as possibilities with the probability the car is behind door 2 twice the probability it is behind door 1. I don't know what else would be shown following a strictly sample space approach. -- Rick Block (talk) 16:56, 3 January 2014 (UTC)

Initial sample space diagram

I've read plenty of reliable sources that say the sample space at the first choice has 12 possible outcomes. But I think this diagram might be accurate and if it is, there are 15 (if you distinguish between goats). I would like to reach consensus on the correct number, then redo the diagram to reflect that number. As best as I can tell, the correct number is either 9, 12 or 15. Tweedledee2011 (talk) 04:01, 4 January 2014 (UTC)

"I've read plenty of reliable sources" In that case, you should be able to cite them. "Attribute all quotations and any material challenged or likely to be challenged to a reliable, published source using an inline citation." WP:ONUS Paradoctor (talk) 04:24, 4 January 2014 (UTC)

Where should it be placed? Which argument/variant is it meant to illustrate? Paradoctor (talk) 04:28, 4 January 2014 (UTC)

The complete sample space is shown in the reference you provided [10], which exactly matches the tree diagram provided by Grinstead and Snell (in the article's references). Consensus here about the sample space is irrelevant. Consensus among reliable sources is completely relevant. I really don't have any idea where you think you're going with this, but I strongly suggest you start a thread on the /Arguments subpage. -- Rick Block (talk) 05:26, 4 January 2014 (UTC)

Rick, if you don't have any idea where I am going, then perhaps you have not been reading my posts. But just in case, here it is again:

1) Not everyone who comes here will understand this material as well as you.

2) I think the article would be more accessible, if we do the following:

- Add (2) block diagrams, showing: 1) Sample Space before 1st choice and 2) Sample Space before 2nd choice

- Explicitly explain how, even though it visually looks like a 1:2 chance at the 2nd choice diagram, it's the information acquired when the door is opened (no car there), which allows the player in the game to calculate that "switch" is better.

3) That's it - I want to make the article more accessible to visual thinkers - and I think this will help.

Tweedledee2011 (talk) 06:39, 4 January 2014 (UTC)

PS: Here's how I would explain it:

In this game, the probability of the car being somewhere behind the set of the three doors is 1 (a probability of 1 means it's a certainty). When the first choice is made, that choice is 1 in 3 because there are three doors and one car. When the host opens a door, the open door has no car; therefore, there is no possibility that it has the car. As a result, all of the possibilities for the car are now held behind the remaining two doors. And, even though at the 2nd choice, it visually looks like (and is) 1 car and 2 doors, we have more information than that available to us. We know that the original choice was 1:3 (and stays at 1:3, because the car does not move), the total must equal 1 and the removed door equals zero. Therefore, the other door must be 2:3, to make 3:3, which equals 1. So it's better to switch.

Tweedledee2011 (talk) 06:48, 4 January 2014 (UTC)

What you are now talking about is the reason that many people get the wrong answer. There are not many reliable sources that have actually scientifically investigated this topic. One that I can think of is Krauss and Wang. They did experiments on a group of people and did provide some answers although there was no single clear conclusion as to why most people got the problem wrong. One thing that is clear is that everybody has their own pet way of intuitively understanding why the answer is 2/3 and their own view on why most people get the wrong answer.

What you now seem to be saying is that many people assume that probability is always evenly distributed between the available alternatives. That is undoubtedly true and it is confirmed by reliable sources and clearly stated in the article already, we have '...nearly all people still think each of the two unopened doors has an equal probability and conclude switching does not matter (Mueser and Granberg, 1999). This "equal probability" assumption is a deeply rooted intuition (Falk 1992:202). People strongly tend to think probability is evenly distributed across as many unknowns as are present, whether it is or not (Fox and Levav, 2004:637)'. What do you want to add to this? Martin Hogbin (talk) 12:04, 4 January 2014 (UTC)

Martin - what I want to add is (2) very simple block diagrams, or possibly one; along with a very simple walk-through of the reasoning. All fact-based, no esoteric explanations. Tweedledee2011 (talk) 15:18, 4 January 2014 (UTC)

How is what you're proposing different from the image already in the article (shown to the right of the "Carlton" section)? And, by the way, your text above (We know that the original choice was 1:3 (and stays at 1:3, because the car does not move), the total must equal 1 and the removed door equals zero. Therefore, the other door must be 2:3, to make 3:3, which equals 1.) is actually incorrect. The original choice was 1:3, and (as it turns out) stays 1:3 but not because the car does not move. If you want to discuss this further, please start a thread on the /Arguments page. -- Rick Block (talk) 17:03, 4 January 2014 (UTC)

Are you kidding? The fact that the car does not move, is the ONLY thing keeping the odds of that door locked in at 1:3. If you move the car, you change the sample space, and if you do that, you change the odds of that door. And BTW, we are not disagreeing on the math. Rather, we are only disagreeing on some potential changes to the narrative of the article. No one is trying to take away the "correctness" of the article from you.Tweedledee2011 (talk) 21:17, 4 January 2014 (UTC)

Rick is right. All sorts of things can change the odds of the car being behind a given door. The host could give you a clue verbally or by his actions, for example. He could even tell you it was not behind a specific door. It does not need the car to move.

Completely wrong. Once the player chooses a door, and the host opens a door (leaving 2 unopened), nothing except moving the car changes the odds of where the car is actually located. Verbal clues would only add information which would aid in locating the car, but that's not the same thing as where it car is. The location of the car is fixed from the start of the game and the odds of it being in a location do not change. Only the odds of finding it do. More information, makes it easier to determine where the car is, but that does not move the car. At all times during this game, there are two sets of information: Set #1, which is where it car actually is - that's an unknown value until the correct door is opened. And set number #2, which is what we know either directly (by observation - such as the open door) or by inference (through probability calculations) Tweedledee2011 (talk) 23:35, 4 January 2014 (UTC)

There are no odds on where the car is actually located; it actually is behind one door, for certain. The most useful paradigm of probability for the Monty Hall problem is the Bayesian one, in which probability is, by definition, a state of knowledge.

One well know example of this fact is this one. I take a card uniformly at random from a pack of cards and ask several people what is the probability is that I have chosen the king of hearts. I tell Alice that the card I have chosen is red, Bob that it is a heart, Catherine that it is a king, and Dave nothing at all. From the point of view of each of these people, what is the probability that the card that I have chosen is the king of hearts? What are your answers?

I should add, before someone else says it, that this discussion should not be on this page. I suggest the arguments page. Martin Hogbin (talk) 09:17, 10 January 2014 (UTC)

The point is that you have you own pet way of explaining why the odds are not 50:50; a way that works for you. I have my own thoughts also, as do Rick and many others. WP, on the other hand, works on reliable sources. If you can find a source that gives your explanation and get a reasonable consensus of editors then it can be added here. Personally I do not think it adds anything to the article. Martin Hogbin (talk) 23:18, 4 January 2014 (UTC)

My words used to describe things are no more or less "pet" than anyone else's editing here. A substantial number of the words in this article are not verbatim quotes from reliable sources. And to the degree they are not, then they too fit your definition of "pet". Tweedledee2011 (talk) 23:35, 4 January 2014 (UTC)

I've started a thread on the Arguments page, see Talk:Monty_Hall_problem/Arguments#Sample_space. -- Rick Block (talk) 01:57, 5 January 2014 (UTC)

External link

Tweedledee, the article already shows the external link to University of California San Diego, Monty Knows Version and Monty Does Not Know Version, An Explanation of the Game.

The red parts of the outer ring showing where the contestant should switch to win the prize, and the blue parts showing where the contestant should stay to win the prize.

The host intentionally shows a goat and never the car, and the aberrant variant
The host does not know where the car is, wasting half of the potential winning cases when he accidentally reveals the car and by that discards such plain winning event (black parts). That link is already in the article, just have a look. Gerhardvalentin (talk) 13:32, 4 January 2014 (UTC)

Gerhard - I do not deny that those illustrations exist. Rather, I am only saying that, unlike what you might think, the meaning of them is not easily apprehended by novices. Hence, my hopes of adding some extremely basic, but clear additions to the article Tweedledee2011 (talk) 15:30, 4 January 2014 (UTC)

But it is not particularly clear. I did not mean to insult you by calling you solutions and explanations 'pet'. Everybody has their own preferred solutions and explanations including their own opinion of why people get the answer wrong. The article is very substantially based on reliable sources. It is not made up of verbatim quotes, and neither should it be, but I think you will find that all significant assertions and diagrams are backed up by reliable sources. Your proposed solution is not backed by a source.

You may think your solution very clear but the fact that several editors have found it very hard to understand what point you are making must give you a clue that it is not as clear to other people as you think. That problem is not unique to you. Most of us long term editors have tried to think of ways to make the correct solution obvious on first sight but we have all failed to some degree. That is the beauty of this problem in is extraordinarily counter-intuitive. Martin Hogbin (talk) 15:40, 5 January 2014 (UTC)

Martin -First, without my diagrams being agreed upon, I have no solution to offer. Second, if you stick with this dialog until you understand me, you'll find that it's you who are being counterintuitive. But you don't see that because you are well versed in probability, so you can't see where you are shifting context and are not actually following the premise as explained in the problem. As plain as I can tell you, the answer of 'switch does yield 2:3 wins' is correct, but it cannot be explained from the player's perspective in the way you are articulating it. From the perspective of the player - which is the perspective that the question requires we adopt; "Suppose you're on a game show, and you're given the choice of three doors..." the additional information you are relying on to make your explanations, would not be available to you. Come over to the arguments page, and if you stay long enough, perhaps you'll be able to move past your own counter-intuitiveness here. Tweedledee2011 (talk) 01:46, 6 January 2014 (UTC)

Maybe you're just wrong.Flat Out let's discuss it 02:35, 6 January 2014 (UTC)

Not likely. Why? Because I'm not trying to be "right", Rather, I am trying to substantiate a valid starting point for a logical discussion. Perhaps if you were to participate in the conversation, rather than just make snide comments, you would know that. Tweedledee2011 (talk) 07:01, 6 January 2014 (UTC)

With respect, you have used up two pages arguing your point and still don't have any support for your position. Flat Out let's discuss it 09:57, 6 January 2014 (UTC)

And you have inserted 2 lines of pointless argumentation - which does not advance any dialog at all. Tweedledee2011 (talk) 14:43, 6 January 2014 (UTC)

Tweedledee, it is clear from your comments above that you are not clear on the meaning of the word 'probability'. I suggest that you discuss this subject on the arguments page before going any further. Martin Hogbin (talk) 09:38, 10 January 2014 (UTC)

Martin - please keep your comments directed on the article, not your opinion of me.Tweedledee2011 (talk) 00:42, 11 January 2014 (UTC)

Too much vos Savant

Her contribution was to make the problem known to the public. Nothing else. I feel the article misleads the reader into believing that she solved the problem. Dryfee (talk) 17:44, 2 April 2014 (UTC)

Making the problem known to the public and the huge outcry of readers who claimed her answer was wrong was what made this problem particularly notable. The article does give some history and explains that the problem was vary similar to some earlier ones, which had been solved. Martin Hogbin (talk) 16:00, 3 April 2014 (UTC)

Not about the article but need some help

I have moved this section to the arguments page. That is still not really the right place but better than here. Martin Hogbin (talk) 18:08, 1 May 2014 (UTC)

92% of the general public

The below sentence, from the media furor section is badly written and confusing, I can't find the relevant source on it to fix it which is why I didn't boldly] replace it.

She received thousands of letters from her readers; 92% of the general public, 65% of universities, and many with PhDs, were against her answer.

I think a solution, although more wordy, would be

She received thousands of letters from her readers of which 92% of those from the general public, 65% of those from university graduates, and many from PhDs, disagreed with her answer.

thoughts? SPACKlick (talk) 11:32, 12 May 2014 (UTC)

I've also just noticed this sentence from the same Section has a similar issue.

Nearly 100% of those who carried out vos Savant's shell simulation changed their minds. About 56% of the general public and 71% of academics accepted the answer.

although without sourcing I can't work out if those percentages are overall, overall after a sub-population did the shells test or of those who did the shells test which would contradict the nearly 100% in the preceding included sentence.SPACKlick (talk) 11:36, 12 May 2014 (UTC)

people keep removing my N-door animation under the N-door section

The text from N-doors:
D. L. Ferguson (1975 in a letter to Selvin cited in (Selvin 1975b)) suggests an N-door generalization of the original problem in which the host opens p losing doors and then offers the player the opportunity to switch; in this variant switching wins with probability (N−1)/[N(N−p−1)]. If the host opens even a single door, the player is better off switching, but, if the host opens only one door, the advantage approaches zero as N grows large (Granberg 1996:188). At the other extreme, if the host opens all but one losing door the advantage increases as N grows large (the probability of winning by switching approaches 1 as N grows very large).

@Ningauble (Undid revision by C0NPAQ – Application of the illustrated concept to the Monty Hall problem is unclear. Please discuss on the talk page before reinserting this animation.)

I don't know what to say. It exactly is what the text says and it even incorporates a real-world application. It even underlines the Monty Hall problem red in the last step of the underlying algorithm. How can it be any more obvious? Are you trolling me? Please just stop. C0NPAQ (talk) 18:09, 19 February 2014 (UTC)

Oppose addition of figure. The figure is confusing, and it is not a clear analog of the MHP. The contestant must be ignorant of history for the placement to appear random. The car is always behind door 1945. Replaying the game with the same question will always place the car behind door 1945. The hints apply a total order to the doors. The figure does not help. Glrx (talk) 00:25, 20 February 2014 (UTC)

I kindly insult your intelligence. C0NPAQ (talk) 00:40, 20 February 2014 (UTC)

Oppose. Really sorry C0NPAQ, but your off-topic own original research isn't any improvement at all for this article about a simple but world-famous "paradox". Gerhardvalentin (talk) 22:18, 20 February 2014 (UTC)

Hm, let's see what is going on here ...

I made a gif illustrating the N-door version of the monty-hall problem for 5 doors particularly, with all p loosing doors shown as mentioned in the N-door section of the article where I added it on the side.
It gets removed for unknown reasons
I improve the gif by adding goats and stuff to make it dead obvious to anyone how that relates to the topic
Someone says the gif is confusing because the illustration doesn't show all the random variations of elimination in the probability space (hahahaha, with 5 doors imagine) and says other stuff that makes no sense to me if you think about it for a second (which just sounds like trolling with irrelevant pretenses to me)
Another guy comes along and coins a simple illustration gif that exactly shows what's mentioned in the subsection as off topic' and original research, or in other words, tries to use wikipedia policies as a cheap excuse to make an argument.

I don't know if you just have a hard time explaining yourselves or if you really mean it. Anyway, the usual wikipedia weirdness. C0NPAQ (talk) 23:10, 20 February 2014 (UTC)

C0NPAQ, It's really quite simple. The addition of the "when did WWII end?" element isn't well explained or graphically clear. The fact that you need to modify the clues for each line to keep the problem the same as monty hall is also a little confusing (you have to watch the animation several times to see why particular boxes are removed). There is also no explanation of what the odds at the top or the odds next to each example refer to. It adds many complex elements to the problem, doesn't relate to any published version of the problem or any published discussion of it so it's not appropriate for inclusion. original research is a relevant policy but even if it wasn't OR, it would still be a confusing graphical representation of the problem. SPACKlick (talk) 11:25, 12 May 2014 (UTC)

YES, I know. I am so sorry that I made an illustration *illustrate* something I should have rather made it pointless so that you need to read the article instead to understand what it is about. If you just look at a short animation 3x in a row and instantly know as much as the thousand words in the explanation text from the article that's just *WRONG* it is really simple relevant policy. 11:41, 12 May 2014 (UTC)11:41, 12 May 2014 (UTC)11:41, 12 May 2014 (UTC)11:41, 12 May 2014 (UTC)11:41, 12 May 2014 (UTC)37.201.225.98 (talk)

You miss our point, unless you are already reasonably familiar with the MHP your illustration is very hard to understand in terms of MHP. You didn't make the illustration illuminating. However a simplified version of it could have some merit.

James May - Recent Discussion

In the Recent Discussion section under James May there is this line

In the end, he was doused 40 times, while his colleague Sim, who had to pick the remaining beer can, was doused 60 times. The resulting percentage was roughly what they expected, but the explanation James May gave was not correct.^{[citation needed]}

I've found a transcript of the show [Here] and the quote is

Let's say I choose this can, there's a 33% chance that that is the safe one, and there's a 66% chance that the safe one is in those two. But this is where it all becomes a bit corrupted because Viet Tom, the game show host, comes in to remove a can but he has to remove a dangerous one, let's say it's that one. Now, of course, that one could be dangerous as well but there's now a 50/50 chance that that is the safe one. Somehow, 17% of the odds disappear with that can because Tom has to take a dangerous one. So I should swap from my original 33% chance to my new 50% chance. And the amazing thing is, if you extrapolate the ratio 50 to 33, you arrive at 60 to 39.6, or only 0.4 adrift from the result we got. That's less than half a tin of beer out from what our theory would predict. So if you are a proper statistician, please do not write to us on manlab@bbc.co.uk because our brains hurt very badly already.

I think the final comment is a tacit admission that their explanation is wrong. I also think it's trivial maths that their explanation is wrong. I've put this here incase people are unhappy with the citation. SPACKlick (talk) 13:58, 12 May 2014 (UTC)

I believe the citation request was for a reference saying the explanation James May gave was not correct. I know it's wrong. You may know it's wrong. But if Wikipedia is going to say it's wrong we need a reliable source that says this. Without such a source, the right thing to do is to simply delete the comment. -- Rick Block (talk) 15:23, 12 May 2014 (UTC)

Well in answer to that I'd offer the following. 1) James May's comment about statisticians not commenting is a tacit admission of fault. 2) Given the large number of sources proving the correct answer is 1:2 (2/3) not 2:3 (3/5) we have a lot of sources that James May's explanation is wrong. SPACKlick (talk) 09:22, 13 May 2014 (UTC)

We could include their comment about not wanting statisticians to write to them. We could include a comment that says nearly all reliable sources say the correct answer is 1:2. But we can't say their answer is wrong unless we have a source that says this (and a source saying some other answer is right is not the same). -- Rick Block (talk) 15:17, 13 May 2014 (UTC)

I disagree with your interpretation of wikipedia policy here. If James May said that World War II ended in 1954 we wouldn't need a source specifically saying "James May was wrong about the end date of WWII when he said it was 1954" it is a well sourced fact that the war ended in 1945 and that fact is inherently exclusive of all other years being correct. We have a source saying what answer James gave, all we would need was a source explicitly saying that answer was wrong, whether or not it attributes that answer to James May or not, and I believe we have plenty of sources which discuss their solution being the unique solution or the solution to the exclusion of other solutions. SPACKlick (talk) 08:08, 14 May 2014 (UTC)

The explanation is clearly bogus; but I do not know whether is customary on the program to horse around in a nonsensical manner, in the spirit of good-natured whimsy, or whether it was a lapse of judgment in this particular instance. In either event Rick is right: it is not Wikipedia's place to comment on their tomfoolery.

It might be acceptable to state what the explanation was, and let our readers draw their own conclusions; but I am not sure this silly program needs to be mentioned in the article at all. The request for statisticians to leave them alone seems to indicate they do not wish to be taken seriously. Where is the educational value in even mentioning it? ~ Ningauble (talk) 18:54, 14 May 2014 (UTC)

I've restructured the paragraph a bit, so it doesn't look like we're referencing the statement "the explanation James May gave was not correct" to himself. And if we're going to keep this statement, I still think it needs a source. -- Rick Block (talk) 01:30, 17 May 2014 (UTC)

Conditional probability

I notice that the former formulation of the solution by calculating the conditional probability has disappeared, and as a consequence, someone reintroduces it. How come it disappeared?Nijdam (talk) 10:56, 4 June 2014 (UTC)

It was deleted with this edit by Gill110951 (talk · contribs). There was some (not much) discussion about this here. -- Rick Block (talk) 13:48, 4 June 2014 (UTC)

I think it is reasonable in the section titled "Bayes' theorem" to include an expression using the notation associated with that theorem in addition to or, better, integrated with the verbal explanation. However, the recently added paragraph is not very clear, nor entirely correct. ~ Ningauble (talk) 15:24, 4 June 2014 (UTC)

I read the recently added paragraph, I am pretty sure it is correct -- just not immediately obvious that it is helpful or relevant. It also suffers from a lack of logical clarity because he switches from talking about the probability that the first choice is correct as P(A) to talking about P(A|B) where B is not "a goat has been revealed": but it is not even clear that these events are from the same event space, which would invalidate the comparison.

But even if we overlook this, the paragraph is still unclear because we really need to see not just P(A|B) but also P(~A|B) which would be the chance the car is behind the other door and if this is larger, we should clearly switch. But getting from P(A|B) to P(~A|B) is non-trivial if done right: you can't just say P(~A|B) = 1 - P(A|B), you have to show it is so.

That said, this is still a mildly interesting result because it shows that after Monty opens the door, the chance of winning with the original choice has not changed: it is still 1/3. 76.212.2.138 (talk) 04:47, 17 June 2014 (UTC)

I'm not sure that your claim about P(A|B) = 1 - P(~A|B) being non trivial is true. A & ~A are exclusive and exhaustive (the car has to be somewhere either in first choice or not in first choice and not both) so P(P(A|B)+P(~A|B) <>1)=0 surely? It seems trivial to me. SPACKlick (talk) 09:26, 18 June 2014 (UTC)

Shorter Lede

A suggestion for reducing the lede I will Be Bold and make this change but i leave the section here for discussion.SPACKlick (talk) 09:39, 18 June 2014 (UTC)

The Monty Hall problem is a brain teaser, in the form of a probability puzzle (Gruber, Krauss and others), loosely based on the American television game show Let's Make a Deal and named after its original host, Monty Hall. The problem was originally posed in a letter by Steve Selvin to the American Statistician in 1975 (Selvin 1975a), (Selvin 1975b). It became famous as a question from a reader's letter quoted in Marilyn vos Savant's "Ask Marilyn" column in Parade magazine in 1990 (vos Savant 1990a):

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, 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 switch to the other door. (vos Savant 1990a). Contestants who switch have a 2/3 chance of winning the car, while contestants who stick to their choice have only a 1/3 chance.

Many readers of vos Savant's column refused to believe switching is beneficial despite her explanation. After the problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine, most of them claiming vos Savant was wrong (Tierney 1991). Even when given explanations, simulations, and formal mathematical proofs, many people still do not accept that switching is the best strategy (vos Savant 1991a). Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until he was shown a computer simulation confirming the predicted result (Vazsonyi 1999).

The problem is a paradox of the veridical type, because the correct result (you should switch doors) is so counterintuitive it can seem absurd, but is nevertheless demonstrably true. The Monty Hall problem is mathematically closely related to the earlier Three Prisoners problem and to the much older Bertrand's box paradox.

Shorter "The Paradox" Section

Below is a suggestion for a reduced "The Paradox" section. Everything up to and including vos Savants response should remain the same, and then the wordiness can be reduced with the below. Again I'm boldly deleting the excess but leaving this here for discussion. SPACKlick (talk) 09:51, 18 June 2014 (UTC)

The Correct Answer is based on the premise that the host knows which door hides the car and intentionally reveals a goat. Players initially have a 2/3 chance of picking a goat. Those who swap always get the opposite of their original choice, so those who swap have 2/3 chance of winning the car (Carlton 2005). If the player selected the door hiding the car (1/3), then both remaining doors hide goats and the host may choose either door at random, and switching doors loses. On the other hand, if the player initially selected a door that hides a goat (a 2-in-3 chance), then the host's choice is no longer random, as he is forced to show the second goat only, and switching doors wins.

Shorter Vos Savant and the media furor Section

Continuing my quest to shorten this bloated page, the next section. I couldn't find much to trim in the second controversy section although I think the wordiness can be reduced by someone with more effort to devote to it. SPACKlick (talk) 10:11, 18 June 2014 (UTC)

Vos Savant wrote in her first column on the Monty Hall problem that the player should switch(vos Savant 1990a) She received thousands of letters from her readers; The vast majority of which, including many from readers with PhDs, disagreed with her answer. During 1990-1991 three more of her columns in Parade were devoted to the paradox (vos Savant 1990–1991), and the discussion was replayed in other venues (e.g., in Cecil Adams popular "The Straight Dope" newspaper column, Adams, 1990), and reported in major newspapers such as the New York Times (Tierney, 1991).

In an attempt to clarify her answer she proposed a shell game (Gardner 1982) to illustrate: "You look away, and I put a pea under one of three shells. Then I ask you to put your finger on a shell. The odds that your choice contains a pea are 1/3, agreed? Then I simply lift up an empty shell from the remaining other two. As I can (and will) do this regardless of what you've chosen, we've learned nothing to allow us to revise the odds on the shell under your finger." She also proposed a similar simulation with three playing cards.

Despite further elaboration, many readers continued to disagree with her, but some changed their minds and agreed. Nearly 100% of those who carried out vos Savant's shell simulation changed their minds. About 56% of the general public and 71% of academics accepted the answer.

Vos Savant commented that though some confusion was caused by some readers not realizing that they were supposed to assume that the host must always reveal a goat, almost all of her numerous correspondents had correctly understood the problem assumptions, and were still initially convinced that vos Savant's answer ("switch") was wrong.

===The little green woman===

To help explain the thought process leading to the equal probability conjecture, vos Savant asked readers to consider the case where a little green woman emerges from a UFO at the point when the player has to decide whether or not to switch. The host asks the little green woman to point to one of the two unopened doors. Vos Savant noted the chance of randomly choosing the door with the prize is 1/2, because she does not know which door the player had initially chosen.

Page too long

At this point this page looks more like an attempt to convince people that can't accept facts than an encyclopedic article. I think the page should be shortened to a succinct description of the problem, the origin of the problem and the response it got and then a few proofs. 2A02:1810:908E:1400:6021:B780:75F4:6A83 (talk) 12:32, 29 May 2014 (UTC)

Agreed and I did a scan to see what could be removed. The article is currently 24 pages long with images (size 11 font), and 22 without. The lede is over a page, so that should be trimmed.

My suggestion for a shorter lede is a new section below, and I feel it could probably afford to be reduced further. SPACKlick (talk) 09:38, 18 June 2014 (UTC)

Ok the recent work has trimmed it to 21 with images, 19 without. Still to do.

Improve clairty of the proportions of letters in the Media Furor section
Remove further unecessary duplication in simple solutions (Economist maybe?)
Remove duplication in the conditional probability section (extended vos savant maybe?) (Direct calculation gets 2 sections... could be combined to reduce wordiness)
Neaten up strategic dominance (possibly add 12 strategy table)
Reduce excess repetition in other host behaviours
Tidy history slightly.

Any help on this would be appreciated SPACKlick (talk) 11:56, 24 June 2014 (UTC)

Reduced Sections, Solutions

Continuing to reduce the wordiness of the article, I turn to the various solutions sections.

Vos Savant. I removed the last sentence because it was superfluous

Carlton Combined the two paragraphs, the picture descriptions were trimmed. Realigned the pictures

Adams and Devlin Removed variations of the problem for later sections (to reduce repetition), reduced paraphrasing of included quotes.

Vos Savant - increasing the number of doors. Removed excessive description of a problem we've already had half an article on

The Economist It's too big, and it just repeats everyhting from before but I've left it alone

Simulations Trimmed repetition in descriptions.

SPACKlick (talk) 11:09, 18 June 2014 (UTC)

I think some of these sections can be combine, for instance, Carlton Adams and Devlin are very similar in content. We really just need to present the simple solution, explain how conditional solutions differ but come to the same result and move on to variations and criticisms. Having 4 tables, 6 images and 3 lines of equations feels excessive. SPACKlick (talk) 11:49, 18 June 2014 (UTC)

Removed sectioning of simple solutions because it added little but made the page look clunky. I think we can remove the entirety of the economist section as it adds no new information over and above the previously presented solutions and that diagram is HUGE. SPACKlick (talk) 11:23, 19 June 2014 (UTC)

I'm also not sure of the benefit of the extended vos Savant table in the article...

As it has been a week with no defenders, I'm removing the Economist and the Extended Vos Savant as they are essentially repetitions of other solutions which serve only to bloat the article. SPACKlick (talk) 11:33, 26 June 2014 (UTC)

Devlin 2005: failed verification

The third paragraph under Bayes' theorem in the current article includes the following statement:

"he later (Devlin 2005) retracted his 'combined doors' solution"

This statement completely misrepresents what the cited source actually says. Devlin 2005 does not retract anything, it defends his 'combined doors' solution.

He notes that "I also received emails from readers who looked back at that earlier piece and were absolutely convinced that the solution I had given (in that earlier column) was wrong." Then, after reviewing some history of the problem, he states the following, in a paragraph unto itself:

"There are several ways to explain what is going on here. Here is what I think is the simplest account."

This is immediately followed by a statement of his 'combined doors' solution, after which he unequivocally stands behind his 'combined doors' solution, and even ridicules those who do not understand it, with the following remark:

"Now, if you have never seen this problem before, or have still not managed to 'see the light', there is really little point in you reading on. (Besides, you probably can't resist spending your time instead emailing me to tell me my reasoning is fundamentally flawed.)"

He then emphasizes that his solution relies on the fact that Monty deliberately shows a goat, showing that it would be a mistake to use the "combined doors" logic if the door were opened by someone who does not know where the car is. He criticizes this error by saying "Moreover, you, the contestant, know that Monty plays this way [always showing a goat]. This is crucial to your reasoning, although you probably never realized that fact."

He then ends the column by giving a solution using Bayes' theorem (recall "there are several ways to explain what is going on here" above), which he prefaces with the following witty remark:

"Confused? As sometimes arises in mathematics, when you find yourself in a confusing situation, it may be easier to find the relevant mathematical formula and simply plug in the appropriate values without worrying what it all means."

These are not the words of someone who is retracting the "combined doors" solution in favor of using Bayes' theorem, as stated in the Wikipedia article. These are the words of someone who fully stands behind the "combined doors" solution and who actually belittles plugging numbers into Bayes' theorem as a substitute for understanding why the answer is true.

I have also tagged the opening sentence of the paragraph with {{failed verification}} because Gill (2011) does not use an argument similar to the preceding paragraph as a patch for the "combined doors" solution, and does not even cite Devlin. None of the opinions about Devlin 2003 expressed in the paragraph are attributed to sources that actually say these things.

Now therefore, I propose to delete the entire paragraph from the article because it is full of unsupported and outright false statements (and move the citation of Gill (2011) to the following paragraph, which does discuss that paper). If anyone believes my reading of Devlin (2005) or Gill (2011) is incorrect, please quote specific passages in those sources expressly stating what I have said those sources do not state. ~ Ningauble (talk) 23:22, 22 June 2014 (UTC)

@Ningauble: I fully supprt you. Nijdam (talk) 08:04, 23 June 2014 (UTC)

The Devlin link seems not to work for me but from the quotes you have given it would seem that you are correct and that the claim that Devlin retracted his 'combined doors' solution is wrong and should be removed from the article as you suggest. Martin Hogbin (talk) 12:09, 23 June 2014 (UTC)

Thanks guys. I will wait a little longer for responses before acting. I don't know who originally added this content.

(I had replaced dead links because maa.org moved the articles to an archive. The links I obtained from Google search use https, and my browser complains that the site does not have a valid certificate but displays the page anyway. Other browsers might refuse to load it because it is a security risk. With a little experimentation I found that the site answers to http as well, without frightening browsers, so I have updated the reference links accordingly. Let me know if you are still unable to read it. ) ~ Ningauble (talk) 13:14, 23 June 2014 (UTC)

Done. The paragraph has been removed without objection. ~ Ningauble (talk) 13:05, 26 June 2014 (UTC)

Pigeons and bird-brained people

I do not understand this edit claiming the pigeon study {{failed verification}}. The cited source studies humans in Experiment 3, finding they do not rapidly learn to always switch. ~ Ningauble (talk) 12:43, 26 June 2014 (UTC)

The simple explanation is I am bird brained and must have been skimming by the time I got to experiment 3. I completely missed that there were human participants... boy do I feel sheepish. SPACKlick (talk) 13:36, 26 June 2014 (UTC)

Yes, having now actually read the whole paper whilst awake I can see this is the conclusion they come to. I think some of their experimental method could be questioned but not without WP:OR. Have removed tag. SPACKlick (talk) 13:48, 26 June 2014 (UTC)