Talk:Monty Hall problem/Archive4

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A simple way to explain, immediately, why this craziness works

~Switching is ultimately analogous to saying 'I think that door X does NOT have the car'. If I'm right (2/3 chance), then I win it.~ In the beginning, I choose which of the three I think is the car-- here I have a 1/3 chance of winning. Imagine that at this point, the three doors are separated into those which I did not choose and the one I chose. We already know that there is a 2/3 chance that the car is in the section of two doors that I did not choose. By revealing a goat in the section I did not choose, my decision is between saying that one of the three is a car, and saying that one of the two I did not choose is a car-- I place my bet on the section of two by choosing the only remaining door in it. If EITHER was going to be the car, I would be choosing it by switching. only if NEITHER was the car would sticking result in me winning.

it's simple: to win the prize, assuming a strategy of always switching when asked if you want to, then you must make a wrong choice first up. therefore to win (if you will switch), you must choose a the booby prize first. the probability of this is 2/3. easy!!

Could this also explain the solution?

The player has three choices. One is the correct door. Thus, 2/3rds of the time, they will make an incorrect choice. Therefore, they should switch 2/3rds of the time, regardless of which door was opened by the host. It seems like one door being opened is a red herring. The player is still dealing with the intial choice (between three doors) and not two doors, even though the problem seems this way. Basically, I'm just asking if the fact that one door is opened has no bearing on the solution at all. So, what do y'all think? —Preceding unsigned comment added by Lord Kuat (talk • contribs)

Well, you have to know one door NOT to switch to, otherwise you switch from one 1/3 chance of winning to another 1/3 chance of winning. You don't have to actually know the door the host proves is a goat, I suppose; you could choose to switch to a random other door, and if it was the one the host would have pointed out, he'd have to say "sorry, not that one". But that comes to the same thing really. You could say, "host, please switch me to another door, but pick one with a goat not to choose for me", not knowing which one the host would have revealed. But then you are still using the host's information that he knows one door which has a goat behind it. In the choices offered, the information about which door certainly has a goat is the thing that makes switching valuable. --Mike Van Emmerik 22:06, 24 April 2006 (UTC)
It is much better (at least for many readers, as I doubt I am alone) to describe this in terms of the probability of being wrong. -- Bilbo1507 17:15, 28 April 2006 (UTC)
Not sure if anyone's still working on this page, but it seems to me that the intro makes quite a big deal of the fact that the host 'knows' what's behind the doors. This makes the thing sound like a trick: it's not structurally necessary for the host to know anything, is it? The fact is, the door that gets opened after you've made your first selection IS a goat; that's the point. I think it confuses the issue to talk about the host's knowledge etc - the problem just wouldn't exist if the door didn't reveal a goat, no? Or am i wrong? I thought I was right when I started typing but now I'm confused again... :-) Ajcounter 16:11, 25 August 2006 (UTC)
Sorry, in reply to myself I now see why the host thing is being laid on so thick, it's to prevent people misunderstanding the rules of the game. Ajcounter 16:16, 25 August 2006 (UTC)
The host thing is "laid on so thick" because it is critically important to the problem. It is structurally necessary for the host to know what is behind the doors. The point is, if the host knows what is behind the doors the only random event is the player's initial selection. If the host doesn't know what's behind the doors, opens one, and (fortuitously) shows a goat, the players chances are 50/50 between the original door and the other unopened door. This is a different problem because if the host knows what is behind the other doors there is no chance the game ends with the host accidentally opening a door showing the car (which will happen about 1 time in 3 if the host opens one of the remaining doors randomly, without knowing what's behind them). If the host doesn't know, there's a 1/3 chance the player picked the car, a 1/3 chance the game ends (with the host showing the car), and a 1/3 chance the other unopened door has the car. Most people's intuition does not seem to distinguish these two cases (host knows vs. host doesn't know), which leads to their intuition leading them to the wrong answer. There's more discussion of this point in the thread currently at the bottom of this page. -- Rick Block (talk) 19:48, 25 August 2006 (UTC)
OK, get it now. Thanks a lot! Ajcounter 13:33, 26 August 2006 (UTC)
Hmm, so it was good that I was so dense that night thus giving Rick so many opportunities to explain it all in such fine detail. Yeah, damn right!  :-)) --hydnjo talk 22:54, 26 August 2006 (UTC)

More "catchy"?... sexy etc. ?

When I first heard about the Monty Hall Problem, I was very drawn into it by its simplicity ... this article, while certainly fantastic, has an introduction so full of disclaimers and strings that the problem is convoluted and no longer interesting, and catchy, IMHO.... is it possible that we can rework the article to state the problem simply at the beginning and then put all the assumptions and stuff a bit later ..... it just seems to me that over time people have inserted little disclaimer words to the point where the problem itself is no longer fascinating. -Abscissa 05:36, 4 March 2006 (UTC)

The problem is that it tends to go in a cycle: someone comes along, thinks the description is too verbose and complicated, and pares it down. Someone else comes along, thinks that it fails to account for this possibility or that one, and expands it again. -- Antaeus Feldspar 22:42, 4 March 2006 (UTC)
What I am suggesting is a very simple statement of the problem at the beginning which, IMHO, the article currently lacks. -- THEN a detailed description of the constraints, etc. ? ... - Abscissa 04:44, 6 March 2006 (UTC)
That's actually what I'm talking about: at times when we have had a very simple statement of the problem, people have asserted that it needs changing because it doesn't specify one constraint or another. Which constraints are you thinking should be moved or removed to make the initial description simple? -- Antaeus Feldspar 17:54, 6 March 2006 (UTC)

Game theory redux

I know there was some conflict about this some time ago, I thought I would send some feelers out. Would anyone mind if I took this article out of Category:Game theory and put it in Category:Decision theory? Thanks! --best, kevin [kzollman][talk] 05:12, 8 March 2006 (UTC)

It is my opinion that it belongs in neither category. The problem is specifically one of probability. One could just as easily say it belongs in the category, "Engilsh Language Problems" or "Game Show Related Trivia" etc. etc. It is a probability problem, plain and simple, not a game. - Abscissa 17:34, 8 March 2006 (UTC)
I agree that its not a game theory problem, but can you explain why you think its not a problem in decision theory? --best, kevin [kzollman][talk] 20:01, 8 March 2006 (UTC)
This problem is designed to illustrate a principle in probability, not one in decision theory. It's not a "real" problem or game. Decision theory/game theory have a branch of "theory" because their problems are related to real-life examples. The same way I would say the three cards problem is not related to game theory or decision theory. I am glad you agree that it is not part of game theory... the MHP *definately* has nothing to do with game theory. I 100% support removing it from the Game Theory category. If you think it should be moved to Category:Decision theory I will abstain from any objection... but as it stands I strongly strongly support anyone who wants to remove all references to game theory from the article. In fact, the article, after re-reading it, is really a huge, huge mess IMHO. -Abscissa 05:02, 9 March 2006 (UTC)
And in only eight months since its Mainpage-FA exposure! hydnjo talk 14:31, 9 March 2006 (UTC)
Don't know if that is sarcasam or not -- but this article in its current state is nothing remotely like the way it was when it was a FA. - Abscissa 14:58, 9 March 2006 (UTC)
My point exactly. The hundreds of edits since July 23 2005 have not made this article better and I suspect that it would not withstand FA scrutiny in its current state. It may be time to bring it through the process once again, it is a good subject. hydnjo talk 04:45, 10 March 2006 (UTC)
Not sure, but perhaps this is related to Second law of thermodynamics. -- Rick Block (talk) 04:55, 10 March 2006 (UTC)
Time to vote for a complete rewrite? I'd do it, but someone would revert it I'm sure... -Abscissa 04:51, 10 March 2006 (UTC)
Rick, I think that the reason for this (besides the Second law) is because as the non-believers raise their arguments, someone goes in an makes a patch to address that particular nit or pick and so the article no longer flows in a coherent way, death by a thousand cuts or something. And Absicissa has a point about a rewrite which is why bringing to FAC status may be the way to go. Just think Rick, you get to do it all over again.  :-) I'd like to hear from Antaeus about this, he has been putting a great deal of effort into this "project". hydnjo talk 05:12, 10 March 2006 (UTC)

Article Rewrite

There is some discussion above about rewriting the article, from scratch, incorporating most of what is currently in the article but editing it heavily... (see above) but there are some strange things about the article right now that someone needs to look at. I propose the new article look something like (or at least incorporate these elements in some order, if someone can rework it to be slightly better):

  • A brief statement of the problem (and why it might be considered significant) (FOUR SENTENCES MAX!)
  • The origin of the problem, the history of the problem. But NOT variants of the problem. -- I would start with Marilyn. And probably finish with her.
  • A statement of the problem with the express contraints.
  • The solution to the problem, with subsections on the various ways of understanding the problem. Perhaps also, starting with the simplest and ending with the most complex. Like Bayes's Theorem? WTF? Is someone who does not understand the solution at this point seriously going to understand that?
  • Similar problems with:
  • Further history of the problem (e.g. Gardner, &c.)
  • Links to similar problems (three card problem, boy girl problem, &c.)

In summary, I think this article could be 50% of the size its now and 200% of the quality. Are there some people who would be willing to work on this with me? Or others who think it is a bad idea and that the article is best in its current form? - Abscissa 06:27, 10 March 2006 (UTC)

I have no argument against the need but I am somewhat troubled by the suggested process. The proper way of dealing with a seriously deteriorated FA is to list it at WP:FARC (see some of the candidates there) with the reasons that it no longer meets WP:WIAFA. Only after a consensus is achieved should the article be rewritten. After that, if we have the desire to do so, the article can be submitted for WP:PR and/or WP:FAC. --hydnjo talk 12:34, 10 March 2006 (UTC)
Sure, sounds good to me. - Abscissa 12:52, 10 March 2006 (UTC)
Abscissa, I've asked Rick Block and Antaeus Feldspar to comment on this proposal. The reason being that they have been principals in bringing this article to FA status and trying to keep it together. Submitting an article for WP:FARC is a pretty serious step in that it may not be promoted again even after a rewrite. Perhaps they might feel that we should just try, by rewriting, to push this article back to FA status. Let's let them at least weigh in on this. hydnjo talk 04:51, 11 March 2006 (UTC)
Sure, OK. I'm going to turn into a pacifist here. I'll just point out that this article so totally does not meet the criteria for being a FA anymore. But it's not the fault of Rick Block and Antaeus Feldspar. Also, I notice that the article says you should switch, because the odds change to 2/3 if you switch. Clearly someone needs to correct this... if there are two doors, the odds would be 50:50! Just kidding. Though I can't promise I won't nominate the article... -Abscissa 05:26, 11 March 2006 (UTC)
I don't think the article has deteriorated to the point that WP:FARC is necessary (although I certainly wouldn't try to prevent anyone with strong feelings about this from doing so). It has distinctly grown since its WP:FAC days and could stand some pruning, but I'm not sure a complete rewrite is necessary. If you'd rather start from scratch than prune down perhaps you might try a draft in a private area and invite comments. I think I agree with Antaeus that this article will always be somewhat of a compromise (it takes a fair amount of effort to keep it factually accurate). I hope this doesn't sound discouraging - continuous improvement is what Wikipedia is all about. -- Rick Block (talk) 05:46, 11 March 2006 (UTC)
As points of comparison here's the article version when it was promoted to FA and this is immediately after one day of Mainpage exposure. I'm not suggesting that the article be reset but in the spirit of Rick's feelings, come up with a proposal that is an updated and improved version that retains enough of the the original structure so as not to be considered a "new" article. This could be done by either starting with the original and editing up or as Rick suggested pruning back the current version. hydnjo talk 12:54, 11 March 2006 (UTC)
Sorry, I never meant to suggest a "new" article really, just moving sections around, a lot of cutting and pasting. But some sentences need to be rewritten. -Abscissa 17:27, 11 March 2006 (UTC)
Abscissa, what about Rick's idea. Start a user sub-page for the entire article and do whatever editing you want without interference except for the folks here who want to contribute. That way, we can all agree that the edits fall within the limits of editing and do a replacement all at once. If you do that, please post your user sub-page name here for us. --hydnjo talk 18:59, 11 March 2006 (UTC)
I think Rick's idea of working on a new version in a subpage is probably the best idea. (I'll probably write more when my current grogginess lifts; I'm writing right now to confirm my interest.) -- Antaeus Feldspar 22:22, 11 March 2006 (UTC)
As to the question "Like Bayes's Theorem? WTF? Is someone who does not understand the solution at this point seriously going to understand that?" I have to say "yes." This is the *only* mathematical statement of the problem's solution, and therefore it is the only unequivocally convincing one to a mathemetician. It is, in fact, precisely the form of solution that I came to the page looking for. —Preceding unsigned comment added by 64.132.59.83 (talk • contribs)
64.132.59.83. I don't think that there was any strong feeling about removing Bayes' theorem and thanks for your input. --hydnjo talk 20:44, 31 March 2006 (UTC)

Rewrite issues: reasonable vs. unreasonable interpretations

I think that if we are going to tighten up the description of the problem, we need to separate the constraints needed to make the problem work into those which are "non-obvious" (those where, if the problem is stated carelessly, a reasonable person might assume something contrary to those constraints) and "obvious" constraints (those that a reasonable person would concede are already identified in the problem statement.

To give examples, based on the current problem statement:

  • A non-obvious constraint would be that when it is said "the game host opens another door revealing a goat", it means "Monty opens a door because it reveals a goat", not "Monty opens a door that happens to reveal a goat". The latter is a reasonable interpretation, and it actually changes the answer: If, after the player chooses a door, Monty chooses randomly between the other two doors to determine which one he will open, then two times out of six, he will choose the door that has the car. The statement "the game host opens another door revealing a goat" then means that we are only dealing with those four out of six cases where Monty did not reveal the car, and out of those four cases, exactly as many win by staying as by switching.
  • An obvious constraint would be that that the host "always makes the offer to switch". The problem statement already says "The host then offers the player an option to switch to the other closed door"; we're told that this is what happens. Any speculation that "well, maybe Monty only offers the player the chance to switch if he knows the car is behind the player's door" is just that -- speculation for which there is no support at all in the problem statement.

I think that one of the problems we have seen with this article, and one of the reasons it tends to degrade over time, is that nearly every reader comes along and thinks everything should be spelled out more. I think, if we're to keep the article quality high over the long term, we need to thresh out what does need to be spelled out and what doesn't, and resist the temptation to let "well, someone might think Monty really loves one of the goats better than the other!" creep in. -- Antaeus Feldspar 15:40, 12 March 2006 (UTC)

How about if we move the Parade letter description of the problem to the lead? Since it's a quote it should clearly not be changed. Then, the paragraph in the lead that discusses the solution can introduce most of the critical constraints (like the existing second paragraph). Then, later, we explain the entire set of constraints no matter how obvious (like the initial placement of the car is random). This leaves the second paragraph of the lead open to debate as to which constraints should be included, but I think the usual quibbles are covered by the top three, i.e. "knows where the car is, always opens a goat door, always offers the switch". Thoughts? -- Rick Block (talk) 17:00, 12 March 2006 (UTC)
That's an excellent idea -- that's probably the most prominent description of the problem, and if we'll be covering anyways whether the description was missing vital constraints that led to the widespread disagreement, we can tackle them right there. -- Antaeus Feldspar 17:35, 12 March 2006 (UTC)
"The contestant chooses one of three doors, behind one of which is a car. After selecting a door, the game show host reveals a goat in order to show the contestant what he would have happened if he had chosen that door, and to build excitement. With two doors left, the game show host offers the contestant the choice to switch doors to build excitement on the show. Should he switch?" ........ In any event I agree that the the complete restraints should not be stated at the very first part of the article. - Abscissa 19:39, 12 March 2006 (UTC)
P.S. I [think] I disagree with Antaeus Feldspar in that I think the first part of the article SHOULD be about getting people "disagreed" and interested in the problem. Then the article can go into all the things Marilyn forgot to state... all the people who wrote into Marilyn complaing were trying to save face. - Abscissa 19:44, 12 March 2006 (UTC)
No Abscissa, this isn't the game show, the very idea of getting folks "disagreed" is what the whole game show was about. Here is where we try (sometimes without success) to enlighten, not mislead or "gotcha". The more clearly we expose the faultiness and show truthiness (I couldn't resist) the better. We're not about the game show staging because we have no sponsors wanting us to keep the readers here during the commercials. This article should be relentless at every opportunity to keep the reader on-track, as relentless as were those who intended to misdirect and confuse. --hydnjo talk 20:38, 12 March 2006 (UTC)
Yes, I think Hydnjo puts it well. If we were writing feature articles for a newspaper, then I think the strategy of using a teasing "hook" to get people first would be a good one. But for Wikipedia, I think the style that's more appropriate is indeed to "relentlessly" present our information. I personally try to edit articles with the thought "The reader could stop after any sentence; if they were to stop after this one, would I have told them the most important things to know about the subject in that time? -- Antaeus Feldspar 23:47, 12 March 2006 (UTC)
I've moved the Parade quote to the lead. I changed the quote to match the images which present different door numbers (I'm considering redoing the images to match the original quote - requires relocating the door hinge to the right - sigh). -- Rick Block (talk) 04:28, 13 March 2006 (UTC)
Because the Parade description really does leave some major holes, I've separated the Mueser and Granberg constraints out from the rest of the "The problem" section and moved it higher, to just under the TOC. What remained in "The problem" was about the history of the problem, so I renamed it to indicate that. I'm strongly of the opinion, especially looking at it now, that the history section should be moved to much later in the article, after the "aids to understanding" section, but I wanted to give others a chance to see what I'd done and discuss. -- Antaeus Feldspar 15:54, 16 March 2006 (UTC)
Agree with Antaeus' logic, the more up front with the constraints the better. The first bullet point from Mueser and Granberg needs a little touchup. Is it a quote or a paraphrase? hydnjo talk 02:49, 17 March 2006 (UTC)
There's some duplication between the "history of the problem" section and the later "origins" section; they both mention Gardner's Three Prisoners Problem. What would people say to cutting down or out the digression in "origins" about three-card monte, combining "history of the problem" and "origins", and putting the history in chronological order? Perhaps also moving the "Straight Dope" appearance in the history section, too? -- Antaeus Feldspar 18:24, 17 March 2006 (UTC)
Antaeus Feldspar, you have been so in touch with this article over the months that we can't wait for you to fix the things that you think need fixing. Have at it... please. Read no sarcasm in my comment, I'd really like to see you take the lead in fixing the problems with this article. If I could I would. Thanks,  :-) hydnjo talk 22:08, 17 March 2006 (UTC)

Hey guys, have you considered starting from the version of this article that reached Featured Article status? (Rather than starting from scratch.) --P3d0 21:50, 3 April 2006 (UTC)

Antaeus Feldspar said: "An obvious constraint would be that that the host "always makes the offer to switch". The problem statement already says "The host then offers the player an option to switch to the other closed door"; we're told that this is what happens. Any speculation that "well, maybe Monty only offers the player the chance to switch if he knows the car is behind the player's door" is just that -- speculation for which there is no support at all in the problem statement."

The fact that it happens in this particular case is irrelevant. Calling the idea that someone might offer a deal only in certain situations "speculation" is bizarre. I'll give you a deal: I just had Excel give me a random integer between 0 and 9. I'll bet you $100 that it was 3. I'm even willing to give you even odds. Will take this deal? Or will enage in the "speculation" that perhaps I would only offer this deal if I already know the random number is indeed a 3? Hmmm?Flarity 01:06, 13 July 2006 (UTC)

page name, problem or paradox

The page was recently moved to Monty Hall Paradox. I just moved it back. Based on a quick google search, "problem" is the predominant name by which it's known (68,000 to 650 hits). -- Rick Block (talk) 20:22, 26 March 2006 (UTC)

I support a redirect at Monty Hall Paradox and I support the article text mentioning it as a paradox but I agree, it is far better known as the "Monty Hall problem", not the "Monty Hall paradox". -- Antaeus Feldspar 04:13, 27 March 2006 (UTC)
I have strong feelings about this, it isn't a paradox, and it's only a problem because of the hype, drama, and misdirection surrounding its presentation. I'm drawing a line in the sand on this one, it's no paradox. hydnjo talk 04:26, 27 March 2006 (UTC)
And DavidNS1128 please tell us why you think tat this a paradox. --hydnjo talk 04:32, 27 March 2006 (UTC)
Actually, we've been through this before. There are actually two meanings of paradox. The better-known one is of course the inherently self-contradicting situation, but an also-correct meaning (more frequently used in mathematics) is the situation where there is a perfectly correct answer that does not contradict itself, but that answer does contradict the common intuition about what should be correct. -- Antaeus Feldspar 15:21, 27 March 2006 (UTC)

The question is flawed and must be answered at a meta-level !

That's correct, the display of one of the two doors in the 2/3 set of doors has no effect on the odds, or the choice to use the switch betting execution sequence. Everyone is wrong, the correct answer is that the original question - even with the additional "precision" constraints - is flawed.

[ forgive me if I keep repeating myself ]

The reality is that you are choosing between the set of one doors or a set of two doors, and this accounts for the 1/3 and 2/3 probabilities of getting it right. It is a stated given that we 100% know that one of the doors in the set of two doors is a goat. Therefore revealing it does not add any information useful for determining what to do next. What it does "reveal" is that the unshown door in the set of two doors is now the last chance for that set of two doors to hit its 2/3 probability.

The one-at-a-time exposure of the set of two cards is the insideously confusing part that causes people to run through various cases or to run a switch betting sequence, as if the revealing of the door is related to the choice of whether to switch

The switch strategy is not just a strategy, but a changing of the game such that instead of having the original choice of a set of one doors with its 1/3 probability, you decide you want to have the advantage of choosing the set of two doors for their 2/3 probability. It may not look like it when you only have one door choice other than your first door choice because one door has been revealed, but that's how it works. You are not deciding to switch based on the content revealed or whether the content was revealed on a particular door in the set of two doors. You are deciding to switch unconditionally to pick the remaining door in the 2/3 probability set to gain its 2/3 probability set advantage.

The original question implies a falsehood by prominently asking the switch question in the same breath as the irrelevant exposure of a door in the set of two doors not initially chosen. As if revealing both of the doors in the set of two doors by exposing them one at a time affects the decision to switch. The question actually being asked is if the person wants to really choose two doors or stick with their single door choice. To not point that out in a response causes said response to be obfuscated. You can put your hand in your pants and do a little dance, or you can declare that the reason to switch is to get the two chances to win and that this has nothing to do with the exposed door.

Yet everyone seems to think the exposure of one door does matter, else why would they give the forward-looking switch betting sequence as an explanation of why to switch, instead of pointing out you are getting two picks out of three, something people understand instinctively? Is it difficult to explain why switching unconditionally is the same as getting two picks out of three?

I guess so, when the question is flawed by seemingly implying that the revealed door is the basis for stopping to ask if they want to switch.

It is, as the person I am responding to said, a red herring.

Revealing both of the doors in the set of two doors simultaneously makes it obvious that it does not change the equation because both will end up exposed anyway. What, does someone want to claim if the doors are not revealed exactly simultaneously that at some time component of discrepacy that the odds fairy will come in and start cursing? Revealing both of the doors in the set of two doors simultaneously also makes it clearer that you are getting two picks to achieve the 2/3 probability.

If you get two picks out of three then you have 2/3 probabilities, plain and simple.

Everyone's answer that involves a betting sequence is obfuscating their answer, it's really just a matter of getting the bet changed to the 2/3 probability set of doors to get the equivalent of two picks. This also applies to the simulation program examples, it is bogus to run such a sequence as proof that the exposed door has something to do with the probabilities. You are either choosing to go with the set of one doors or the set of two doors.

Your first choice was known to be at a 1/3 probability, so switching gets you the 2/3 probability, it's got nothing to do with the exposed door but with the probability set it is in.

The switch betting sequence is subordinate to the probability set memberships, the switch betting sequence does not affect those probabilities from being the ruling factor.

The switch betting sequence is a waste of breath in anyone's attempt to explain what is happening.

If you get two picks out of three then you have 2/3 probabilities, plain and simple. You can get two picks out of three by switching every time from your first choice, which was locked in on a 1/3 probability. It doesn't matter if one of those two picks was revealed to be a goat, it is 100% guaranteed that one of them would be a goat.

It's an irrelevant flaw in the original question to mention the goat detail, like a trick question. If one quickly explains that simply being asked if they would switch - and switching - was the equivelant of getting two picks out of three, then most of the current Wekipedia page on this can be deleted as overblown. There's no need to huff and puff about the various case scenarios when the instinctive two picks out of three phrase can be used.

From the Wikipedia entry:

  1. The most common objection to the solution is the idea that, for
  2. various reasons, the past can be ignored when assessing the probability.
  3. Although ignoring the past works fine for some games, like coin
  4. flipping, it doesn't work for all games. The most notable counterexample
  5. is card counting in some card games, which allows players to use
  6. information on past events to their advantage.

Probabilities past never affect the current game. The card counting example is bogus because the information revealed by played cards determines a new "deck" with different probabilities of getting certain cards. It is not an example of the past probabilities affecting current probabilities.

The original question is flawed by mentioning an exposed door as if it were a consideration in deciding whether to switch.

What prize do we win for finally getting this right? ;-)

Oh, I guess I also have to give the one short answer that will wipe out all the other answers. It is:

Switching from your initial choice, which is locked into a 1/3 possibility of getting the car, is the same as getting two picks out of three, regardless that one of the other two picks has been exposed at the time you are asked whether to switch, because both of the other two picks will be exposed anyway. If both of the other two picks are exposed simultaneously then it it more obvious to most people that you are getting two picks. Switching unconditionally changes your bet from the 1/3 probability to the 2/3 probability that is instinctly understood by saying that switching gets you two picks out of three. The original question (even with the additional constraints spelled out) is flawed because naming an irrelevant door exposure in the same breath as asking whether to switch is obfuscating the question.

by Double Think 10:31, 3 April 2006 (UTC)
Under different constraints, the initial probability does change. For example, if the host picks randomly and happens to open a "goat door". In this case, the two remaining doors have equal chance. Yes, the problem setup is intended to mislead, and (with sufficient constraints) the choice is the contents of the one door vs. the contents of both the other doors, but the problem itself is not flawed. -- Rick Block (talk) 02:16, 3 April 2006 (UTC)
Okay, but...

If the problem-question is not considered flawed, yet you acknowledge the problem-question reduces down to a choice of one door versus a choice of two doors, then all the answers that go through the different cases or go through a switch betting sequence are flawed because they are not the simpler and clearer reply that unconditionally switching is the same as getting two picks out of three. It's as if the problem-question is a troll for overly complicated answers. Why should this Wikipedia page be stuffed with unnecessarily belabored answers? ---SGP

I'm not sure exactly what change you're suggesting to the article. Are you objecting to the probability diagrams that follow the Venn diagrams? (BTW - you can "sign" your comments by entering "~~~~"). -- Rick Block (talk) 03:46, 3 April 2006 (UTC)
Yes, the page is too complicated, I was still confused despite all the "Aids to understanding"

Yes, once a primary and clear explanation is in place to state what is purposefully misleading in the problem-question and that the simple choice of unconditionally switching gives you both of the other doors and so doubles your odds, then a number of explanation examples can go. Because it is infinitely clearer to say getting two picks out of three is the best strategy compared to sticking with the original pick which is stuck at 1/3 probability.

Then at least the following can go, or be moved to a "belabored proofs" page: the probability diagrams that follow the Venn diagrams, the 'Increasing the number of doors', 'Combining doors', "Bayes' theorem", 'Opposing player', 'Simulation', and 'Card game experiment'.

The Venn diagram text can be changed to document the 'two picks out of three' more emphatically, and once someone gets that, then you don't need all sorts of arcane additional explanations. KISS.

If not that, then I'd like to make my own explanation for youz to add in the Aids to Understanding section, but as merely a one-line link to a separate page, because it would be embarrassing to be next to all that other mess.

Double Think 09:50, 3 April 2006 (UTC)

The issue is that different people "get it" based on different explanations. The one you seem to prefer (opening the door is simply a distraction) is simply not understood by many people. The usual problem with this explanation is that some people see two closed doors and internally morph the problem into a new problem consisting of a random choice between two doors. Getting such people to understand the choice is between 1 door or 2 doors (one of which is now open) is the entire reason the article is as long as it is (it started significantly smaller, and people keep adding explanations that match how they ended up understanding it). There truly seems to be no universally understood explanation (the first one listed under "The Solution" seems pretty compelling to me - 3 possibilities, you win by switching with 2 and by staying with 1). I'm not at all saying I don't think the article can be improved - but simply whacking all the explanations but one is probably too drastic. -- Rick Block (talk) 23:14, 3 April 2006 (UTC)

Forget everything else - this - is BY FAR the simplest explanation. Why people don't get it is totally beyond me.

How can you win? - You can win in one of two ways, a) by picking the car and not switching, or b) by picking a goat (either goat) and then switching. You have two chances out of three of picking a goat. So you're better off switching, because if you switch you are going to win two out of three times (i.e. every time you pick a goat first) e.g.

  1. Choose a door; you're lucky it's a goat! but you had a 2/3 chance of choosing a goat so the odds were on your side.
  2. Monty reveals the other goat. Only one goat remaining.
  3. You switch; it has to be the car as the other goat is now gone.
  4. You win!
  5. If you had chosen not to switch you would have lost.
  6. By not switching you're stuck with that 2/3 chance of getting a goat. Switching meant that you turned it into a 2/3 chance of winning the car.
  7. Easy peasy. Jooler 23:42, 3 April 2006 (UTC)
  • Hey Jooler, thanks again for your insightful "Easy peasy" explanation. I think your earlier post is somewhere in archive heaven so thanks for posting it again. Please feel free to post your analysis from time to time to be sure that remains in view. hydnjo talk 00:02, 4 April 2006 (UTC)

The Neverending Story...

... or so it seems. This does present an everlasting challenge, whenever things become dull in your whatever world you can always be assured that coming back to here you will be challenged anew. There will never be an end to this so, c'mon back whenever and we'll be waiting to provide an argument to your most recent foil. Best wishes and good fun to and for all, hydnjo talk 03:08, 4 April 2006 (UTC)

Simplified explaination

Long mathematical formulas and diagrams and such are interesting, but I think in the end they cloud the issue. It took me a year to realize the answer is "switch", partly because I too tried to get very detailed in the explaination. Then I realized it boils down to this:

After a door with the booby prize is opened, there are two doors left and you are given the choice of whether to switch or not. If you switch, you are switching "right" and "wrong" - if you have a winning door, you WILL switch to a wrong door, and vice versa. As the odds of you initially choosing a wrong door are 2/3, then switching will, in the end, give you a 2/3 chance of having the winning door.

Yessss! but, just as you have seen the light there are others that have not. Your "simplified" explanation is simplified because that is the perspective that was the clearest for you. Others, for whatever reason, will view things differently and respond to another approach. The article provides several rational approaches to understanding the (what is to you) simple and obvious. The "long and mathematical formulas and diagrams" are more than interesting, they provide insight to others who do not see it quite your way. Please be patient with all who are having difficulty, if it were that obvious to everyone then this article wouldn't exist. It exists only because the understanding is counter-intuitive to many. hydnjo talk 17:57, 12 April 2006 (UTC)

My comments were not intended to insult anyone, and I apologize if they come across that way. I was just trying to suggest a simpler explaination than any I saw, because I feel that at times, an explaination can be overly complex. Verbally, I've had good luck with that explaination, and was just offering a simple variation of the explanation.

Venn diagrams?

The diagrams shown under Monty_Hall_problem#Venn_diagrams are not Venn diagrams at all. I'd rename the section, but I'm not sure what those diagrams are called. Any ideas? --P3d0 19:28, 12 April 2006 (UTC)

Euler diagrams? Or just "diagrams", since it has more diagrams than just the "boxes within boxes" Euler-style diagrams. – Wisq 21:22, 12 April 2006 (UTC)
I don't understand. Do Venn diagrams require that an overlap exists between the sets? The diagrams are shown to illustrate and to help the reader visualize two mutually exclusive sets of doors. hydnjo talk 19:11, 13 April 2006 (UTC)
Ok, maybe I don't understand. [here] is the article version I was talking about. Which of those diagrams is a Venn diagram? --P3d0 17:35, 28 April 2006 (UTC)

MS Excel Experiment proves this true KeyStroke 17:48, 17 April 2006 (UTC)

This has to be an April Fools joke gone wild. Here is an experiment providing 10,000 random observations implemented in Microsoft Excel:

In cell A1 put formula: =TRUNC(RAND()*3)+1 In cell B1 put formula: =TRUNC(RAND()*3)+1 In cell C1 put formula: =TRUNC(RAND()*2)+1 In cell D1 put formula: =IF(AND(B1=3,C1=2),1,IF(AND(B1=3,C1=1),2,IF(AND(B1=2,C1=1),1,IF(AND(B1=2,C1=2),3,IF(AND(B1=1,C1=1),2,IF(AND(B1=1,C1=2),3,0)))))) In cell E1 put formula: =IF(AND(B1=3,D1=2),1,IF(AND(B1=3,D1=1),2,IF(AND(B1=2,D1=1),3,IF(AND(B1=2,D1=3),1,IF(AND(B1=1,D1=2),3,IF(AND(B1=1,D1=3),2,0)))))) In cell F1 put formula: =IF(A1=E1,"Y","") In cell G1 put formula: =IF(A1=B1,"Y","")

A1 represents the door which has the car B1 represents the players first choice C1 represents Montey's choice between the two donky doors D1 represents the door that Monety opens to reveal the donkey E1 represents your second choice door if you choose to change F1 indicates if your second choice is correct G1 indicates if your first choice is correct

propogate these formulas for 10,000 lines (or so)

In cell F10001 put formula: =COUNTIF(F1:F10000,"Y") In cell G10001 put formula: =COUNTIF(G1:G10000,"Y")

In my spreadsheet I get 3,384 observations of success if you change (for 33.8% rate) vs. 3,283 observations of success if you keep your original choice (for 32.8% rate) Clearly indicating that your second choice only has about a 50% chance of being right.

KeyStroke 21:07, 14 April 2006 (UTC)

Your formula for D1 is incorrect (and the fact that stay vs. switch combined only add up to .666 might have given you a clue something was wrong). For example, the very first clause of the IF says if you pick door 3 and Monty's random choice value is 2, then Monty opens door 1. Door 1 might have the car. I suspect you might be able to fix this by adding Monty's fixed choices before the clauses you have. In particular, if you pick door 3 and the car is door 1, Monty (unconditionally) picks door 2. I'm not sure the exact Excel syntax, but something like =IF(A1<>B1,6-(A1+B1), ...everything else you have ) should fix it. This clause says, if you didn't pick the car then Monty's choice is 6 minus the sum of the door you picked and the door where the car is (which is an algebraic way to come up with the door Monty must open rather than use 6 if-then-else's). If you fix it and run it again, please post your revised results! -- Rick Block (talk) 22:45, 14 April 2006 (UTC)
The formulas are not incorrect... your assumptions are. First it is an invalid assumption that the total of wins from changing your choice, plus the total wins of not changing your choice should add up to 100%. Since they both approx. 33.3% this tells us that changing your original choice does not help, nor hurt, your chances of winning. Secondly, the 'scenario' modeled by the MS Excel model is not one of Monty dramatically opening the door not chosen but one of simply finding out if the latest choice you made was correct. While it is true that my formulas can likely be improved by taking an algebraic approach as opposed to a complex set of nested if-then-else statements I still hold that the result is valid. Conclusion: change or no change (of your original choice) does not (in the long run) help nor does it hurt your chances of winning. I suspect that the other proofs provided do not take into consideration that Monty should make a random choice between which of the two donkeys he shows (but I cannot prove that at this point). There are three random elements: your first choice, the choice Monty makes of which donkey to show, and your choice to change your first choice to the remaining door not chosen nor shown. The sum of all those random elements leads to the conclusion that you always have a 33.3% chance of being right, regardless if you change your choice or not. KeyStroke 13:11, 16 April 2006 (UTC)
(interject to keep close to the above comment) Monty always exposes a goat, leaving only two doors. One of them must be the car and they are your first choice and the door you can switch to. One of stay or switch MUST win in each case, so the sum of the Y's in columns F and G must be the same as the total number of trials. If you look at your spreadsheet, you'll see some (about 1/3) of the lines don't have a Y in column F or G. In these cases, according to your formulas, Monty is opening the door showing the car! The problem statement says this never happens. In each of these cases Monty isn't making a random choice but must open the only door with a goat (since in these cases Monty is revealing the car, your initial choice was a goat), and switching would have won the car. The fix is as I suggest above (I actually got on a Windows machine and tried it - and have fixed the syntax). The point is Monty's choice is only random if the initial choice is the car, which your formula for D1 overlooks. -- Rick Block (talk) 17:29, 16 April 2006 (UTC)
Sorry, KeyStroke. You're simply wrong on this. I didn't trace out where you made your mistake, but it's there somewhere. Will this seem arrogant to you? I have no doubt it will, but at the same time, once you see why switching wins 2/3rds of the time, it's so simple you can't doubt it anymore. And it doesn't have anything to do with which donkey or goat Monty chooses to show.
Arrogant no. Ignorant? Yes. Only a fool always assumes himself to be correct. Often people see something and yet fail to understand what they have seen.
Maybe this demonstration will help you -- you might even be able to do it in Excel, and if you do, let us know -- we might be able to get an image for the article from it. Draw a pie chart. Right now that pie chart is undivided -- it's representing 100% of the possible outcomes of the experiment. We'll now divide it into thirds; color one third red, and the other two thirds green. The red section represents the one-third chance that the player will pick the car on the first try; the green sections represent the player picking goat 1 or goat 2.
Now, if the player initially picks the car, then and only then does Monty have a choice of which goat to show. If you think this might matter, then you can divide the red section into two red sections: one representing Monty showing goat 1, one representing Monty showing goat 2. You can divide it in any proportions you like: even chances for both goats, goat 1 three times as frequent as goat 2, goat 2 five times as frequent as goat 1 -- anything you like. However, since Monty can only choose between two goats when the player has picked the car, the total of the two red sections will always be the same: 33.3% of the complete chart. And when the player picked the car initially, staying is the only move that will win. That is why it doesn't matter which goat Monty picks; it never changes the total percentage of red on the chart, only subdivides it differently.
Now let's look at the green sections of the chart. When the section is green, it means that the player picked a goat initially. When this happens, Monty has no choices. He must open the door revealing the other goat. With that goat eliminated, the player has only one door he can switch to -- the door containing the car. Switching always wins when the player picks a goat initially.
This is why I say it's so simple. When you understand the problem constraints correctly, you see that if the player picks the car initially, switching loses -- no exceptions. And if the player picks a goat initially, switching wins -- no exceptions. Since the player picks a goat initially 66.6% of the time, that's how much of the time switching wins. -- Antaeus Feldspar 15:50, 16 April 2006 (UTC)
You do point out where I may have gone wrong. I may have had Monty chose to show the car instead of always showing the goat. I will work on revising the MS Excel formulas when I get extra time today. KeyStroke 14:58, 17 April 2006 (UTC)
Let me save you the trouble of worrying about it and dealing with these fools much longer. The wiki definition does ignore the fact that
The error in my spreadsheet is now corrected. I did 'program' it to allow Monty to reveal the car, as opposed to always revealing a goat. The correction is to the formula in cell D1 which should be: =IF(A1=B1,IF(AND(B1=3,C1=2),2,IF(AND(B1=3,C1=1),1,IF(AND(B1=2,C1=1),1,IF(AND(B1=2,C1=2),3,IF(AND(B1=1,C1=1),2,IF(AND(B1=1,C1=2),3,0)))))),IF(AND(A1=3,B1=2),1,IF(AND(A1=3,B1=1),2,IF(AND(A1=2,B1=3),1,IF(AND(A1=2,B1=1),3,IF(AND(A1=1,B1=2),3,IF(AND(A1=1,B1=3),2,0)))))))
With this correction the outcome is now 66.4% success rate when switching and 33.6% success rate when keeping your first choice. I stand corrected. The premise is valid. KeyStroke 17:48, 17 April 2006 (UTC)
  • Thanks KeyStroke for getting back to us with your revised conclusion. Others may have just dropped it and not done what you have so admirably done. You may want to consider watchlisting this article so as to help out with the perpetual arguments against all logic and proof.  :-) hydnjo talk 22:56, 17 April 2006 (UTC)
I have a lot of respect for someone who comes back and admites they were wrong like that. thanks - jesse_d_s

To those who troll this article and repeatedly revert to these primitive explanations

The problem is not with your understanding of math, but your dealings with general logic. A 2/3 chance of choosing a "Not car" is not in every way equivalent to a 1/3 chance of choosing each goat. Goat 1s, Goat 2s and Non cars are not equivalent. The Goats can be said to be non cars, but all non cars are not goats, the goats are a subset of noncars etc. The wiki definition deals with Goats 1 and 2, not non cars (as does the real problem). Whether or not you deal with goats 1 and 2 or non cars determines whether or not you learn something new when Monty reveals a goat. In the case of the wiki explanation you gain new information because it deals with goats, but the sample space ignores the new information. It is utter nonsense to say you have a 1/3 chance of choosing goat 1 at a time when you know you did not choose goat 1 because monty has revealed it. A sample space that reflects the point in time when the decision is made should take this into account, and the sample space given in the first wiki explanation does not.

You can claim that treating the 2 goats as "non cars" is equivalent to dealing with the goats in every way necessary to determining which choice is better (a metaphor of sorts) but you need to make sure and identify what makes the two situations equivalent for this purpose or people will use it when it doesn't work. Even places where it makes it clear it is going the "2/3 non car" route, it doesn't identify why it works in this case and wouldn't in others. And then theres just the question why deal with a metaphor rather than solve the actual problem anyways? —Preceding unsigned comment added by 69.180.7.137 (talk • contribs)

As rigorously described in the article, when Monty reveals a goat you have gained no new information. Monty always reveals a goat and so that action can't possibly be informative or of any use since it a precondition. The precondition is that Monty will open a door revealing a goat. hydnjo talk 23:07, 17 April 2006 (UTC)
The 'new information' is information of detail, specificly which of the two doors you did not choose will Monty reveal. The information of "Monty always reveals a goat" is meta-information. It is not the revelation of the goat that is significant, it is which door that is. If your first choice is a goat then he must reveal the other goat making the door not-chosen-and-not-revealed to be the car. So, the new information you have is that the door Monty reveals is not the one with the car. KeyStroke 13:52, 18 April 2006 (UTC)

Here is the easiest way I can think of to demonstrate the problem with your reasoning. Make 2 seperate VENN diagrams. One with 5 Events and one with 3. On the one with 5, draw 3 events which represent your choice between goat 1, goat 2, and the car. Then draw two events of Monty revealing each goat overlapping the events that you chose the car and the opposing goat. You should have one third chance that you chose each goat and monty reveals the other, and then 1/6 chance that you chose the car and monty reveals the other. Now given that monty reveals one of the two goats you cross out the one sixth chance monty reveals the other goat, and the 1/3 chance you chose the revealed goat. You will be left with 1/3 that you chose the other goat and 1/6 you chose the car, which rebalances to 2/3 and 1/3. Notice the signifigance of the information you recieved when the goat was revealed and how it only by chance happens to end up at the same place you started.

Now, with the other ven diagram label 1 event "Not car chosen", 1 event "Car chosen" and one event which encompasses all "Monty reveals that you did not choose one of the not cars". Now although it happens to bring you to the same answer, you might want to ask yourself which is the ACTUAL problem? You might also ask yourself what special conditions allow the 3 event diagram to reach the same conclusion and how those conditions might be absent in similar problems. Then ask yourself why you would use this second explanation anymore than you would say multiplication is just like addition, but only when dealing with 2s that is.

Prior Information on the hosts bias

I took the liberty of removing the statement that switching would be beneficial even in the original unconstrained problem of the Parade magazine, because it is not true. If we want to include the prior information of the bias of the host we have the following alternatives:

  1. The host is paid and only opens a door showing a goat if you selected the other goat, and asks you to switch, in which case switching always is beneficial. If you chose the car in the first place he simply presents you with your prize.
  2. The host is hostile and only presents a goat and asks you to choose if you chose the car. If you chose a goat he simply presents to you that you have not won. Therefore switching is never beneficial.

Assuming a noninformative prior, and marginalising over it, we can see that switching does not give any benefit unless more information is presented on the bias of the host. --Janus Coriolanus 09:55, 19 April 2006 (UTC)

COMMMENT BY ANOTHER USER (Not Janus): Well, Monty Hall is a real person. We could study his life and speculate how he acts under different conditions. This would result in a better prior. The article is quite nice, by the way.

It occurs to me that this fundamental bit about Monty always switching is unknowable by the person playing the game. We see that if Monty chooses to open a door randomly, then when he happens to open a goat door, we do not actually have an advantage to switching. Now after Monty opens the door, he can swear up and down he had to open the door but we really have no way of knowing the probability of Monty's honesty. Could people who "don't get it" be intuitively screening out the fundamental assumption about Monty always switching because this fundamental assumption is unverifiable? 64.42.233.61 22:59, 17 November 2006 (UTC)

This is dealt with in the unambiguous formulation because Monty explains what he is going to do before the first choice is made. Therefore, it doesn't really matter if the contestant didn't know whether Monty always opened a door, as long as he knows, in advance, that Monty is going to open a door this time. Davkal 23:30, 17 November 2006 (UTC)

Two Players Problem

What would happen if you were to have three players each choose a door, one of them is revealed to have a goat and that player is sent home. We are now left with two players each having chosen a door. From the perspective of either player, one would say there is still only a 1/3 probability that they chose the correct door, and a 2/3 probability that the car is behind the other door - which is clearly impossible. I believe this is reconciled by the fact that in this situation there was no guarantee originally for each person that they wouldn't be kicked out. So, initially, there is a 1/3 probability of choosing the car, a 1/3 probability of choosing a goat, and a 1/3 probability of both choosing a goat AND being sent home. Once they discover they haven't been sent home, they now have a 1/2 probability of having chosen the car and 1/2 the other goat. This would only be the case, of course, when you have three contestants to begin with. The 1/3 - 2/3 split would still be in effect with only one contestant.

Hmm, this is a real puzzler. I don't see why each player wouldn't think they're better off switching. Since they both have all the same information, this is absurd. --P3d0 21:06, 21 April 2006 (UTC)
One condition of the original Monty Hall problem is that Monty cannot choose to reveal your door. If you pick a goat (2/3), Monty has no option about what goat to pick, and hence 2/3rds of the time, his choice reveals the answer. But if you show up late, as mentioned below, you have 1/2 of picking the correct door (of the two remaining).
Here, Monty always has a choice, so his choice doesn't tell you anything about your own door. Hence unlike the original problem, your choice has no effect on Monty's choice. Monty could choose first (but not show it or tell you), and then your three players could pick their doors. One instantly gets the boot, and the other two have the inherent 1/2 that comes with showing up after Monty chooses.
I'm currently adapting my Monty Hall pseudo-quantum-entanglement program to see if I can model this. It should be able to model all possible outcomes (universes) and give me an idea of the probably of each and how they interact. But the most fundamental difference so far is that Monty always has two doors to choose from; Monty's choice is only "entangled" with the players' choices after he makes it. – Wisq 21:25, 26 April 2006 (UTC)
Each player has a 1/3 chance of picking the car (and won't be eliminated) and a 2/3 chance of picking a goat with a 50-50 chance of being eliminated. So, each player has a 1/3 chance of picking the car, a 1/3 chance of remaining in the game with a goat, and a 1/3 chance of being eliminated. Assuming you're not eliminated, you have an equal chance of having picked the car or a goat. The other remaining player is in the same situation, so there's no benefit to switching. The difference between this version and the version described in the article is in this version there's no guarantee that you're not eliminated after you make your initial pick. -- Rick Block (talk) 21:31, 26 April 2006 (UTC)
No, the chance of being eliminated when you pick a goat is 3/4, not 1/2. Something that is glossed over here is that P2 can't choose P1's door, so P2 only has 2 choices. Despite this, P2 has the same 1/3 chance of initially picking the car. In the case when P1 picks a goat, P2 only has a choice of two doors, one with a goat and one with a car. P2 independently chooses the car (1/2) or goat (1/2). For the 1/2 of the time when P2 does choose the car, there is a 100% chance that P1 is eliminated. Only for the 1/2 of the time when P2 chooses the other goat does the 1/2 chance of either player randomly being eliminated apply. So when P1 chooses a goat, P1 is eliminated with probability 1/2*1 + 1/2*1/2 = 3/4. Sanity check: every game, exactly one player is eliminated. P1 is eliminated 2/3*3/4 of the time, or 1/2 of the time. Let's check P2. First case: P1 has picked the car (1/3); P2 is always eliminated. Case 2: P1 picks a goat (2/3); P2 chooses the car (1/2) and is not eliminated, or the goat (1/2) and has a 50% chance of being eliminated. So that's 1/3*1+2/3*1/2*1/2 = 3/6 = 1/2. So each player has the same 50% chance of being eliminated, as you would expect. Note that with your numbers, P1 is eliminated 1/3 of the time. Since neither player knows where the car is, going first has no advantage, so presumably by your numbers P2 is also eliminated 1/3 of the time. But then, 1/3 of the time no player is eliminated, which doesn't agree with the rules of the game.
So now P1 (and by symmetry P2) has a 1/3 chance of picking the car, 1/2 chance of being eliminated, and only 2/3*1/4 = 1/6 chance of having picked the goat and not being eliminated. "Being" the remaining player effectively doubles your original probabilities; you have to scale up the probabilities so that they still sum to unity. The one probability that doesn't change with the elimination is whether you initially picked the car. So that's now 2/3 after scaling. Another perspective (ignoring the scaling): for the surviving player, your original 1/3 chance of picking the car is untouched, but 3/4 of the chance that you picked the goat is gone. The 1/3 chance that you initially picked the car is twice what is is left of the chance that you originally picked the goat. The other 1/2 of probability is that you are eliminated, so that doesn't affect your decision of whether to switch or not. --Mike Van Emmerik 23:56, 6 May 2006 (UTC)

But now consider the 3 prisoners problem in the same light: Prisoner A goes and asks the guards to tell him which of B and C are being released, and is told that C is. Prisoner B then goes and asks which of A and C are being released and again is told that C is. Now, from each of their perspectives, they each believe the other has a 2/3 chance of being executed. The next time they see each-other - since they hate each-other - they each decide to gloat to the other. How is this problem reconciled?

There is no need to reconcile this one. The two prisoners have different information, so their judgement of the probabilities differ. Just like if I flip a coin, and show it to you but not to me. Now you know there's a 100% chance it's heads, while I believe there's a 50% chance of heads and tails. The three-prisoner situation is more subtle, but the point is that the two people can validly have different judgements of the probabilities. --P3d0 20:25, 21 April 2006 (UTC)
Think of it like this. When there is one player, the host reveals information to him, but with two players, nothing new is being revealed. Imagine 1000 doors and 1000 players, where one gets sent home. The new information is "split" between the new players. In your example, the chances that they picked the car are actually 1 in 2, not 1 in 3. - Abscissa 06:12, 23 April 2006 (UTC)
Sorry, I don't buy it. That argument is way too abstract to be reliable. --P3d0 13:12, 24 April 2006 (UTC)
Like in the Boy or Girl Problem, everything depends upon which point you ask for the probability (what information you are given). If Monty shows you a goat THEN you choose a door, the odds are 1 in 2.
Imagine YOU are the only player, but two other people will keep what are behind the other doors. This is essentially what you are asking -- in this case, swithing is still the best thing (all we have added is that two other people will keep the remaining prizes.) If you add another player who can make decisions, then the game doesn't really work because Monty must reveal a goat. The odds are only ever 2 in 3 for switching if Monty reveals information to you. He cannot do this to two people, in advance, while also with a guarantee that neither will be eiliminated by revealing a goat. You may say, that there is another player, so we can just use him instead, but we must have chosen him in advance to assess which outcome is probable... otherwise we are just playing the first version of the game I mentioned, where two "dummies" exist and just get to keep the remaining prizes. (E.g. if I ask you, what is the chance that red will come on the roulette wheel three times in a row the next three times I spin it (assuming no green), the chances are 1/2 * 1/2 * 1/2 or 1/8. But if red comes twice in a row, and I ask you, what are the odds that red will come the next time, the odds are 1/2, not 1/8.) - Abscissa 13:44, 24 April 2006 (UTC)
Also see the Three card problem. Even if you are a Monty Hall Veteran, this will give you some trouble at first -- it's all about prior knowledge. - Abscissa 17:16, 24 April 2006 (UTC)

Everybody has one

Opinions on this subject are like well... (fill in your own example), everyone seems to have one. hydnjo talk 22:44, 23 April 2006 (UTC)

(copied from hydnjo's talk:)

= Monty Hall Problem? ==
Are you sure that the odds are 1 in 3? I am pretty confident that it's actually 1 in 2 because there are two doors, so each door must be a 50% chance. Consequently I have changed the entire article to reflect this.
Just kidding... :-) The early bird gets the worm, but the second mouse gets the cheese! - Abscissa 00:36, 24 April 2006 (UTC)
(copied from Abscissa's talk:)
==Monty Hall, Monty Hall, Monty Hall,Monty Hall, Monty Hall, Monty Hall, continued...==
Actually, the odds of losing by not swapping has more to do with the number of doors at the outset than anything else. Starting with only three doors seems to be the most likely way to confound the greatest number of people, and that is the point after all. Imagine if you will that the "show" were to be lengthened so as to accommodate say one hundred doors with Monty dutifully revealing ninety-eight goats each night - wow! wouldn't that be a thriller, no viewers, no sponsors and no show. So, lets reduce the number of doors to the minimum - three. That way we can maximize the confusion (along with our profit). Clever. Look at the folks here at WP that are unwilling to "see the light", some marketeer got wealthy over that one!
It's still fun however, especially if we get a convert now and again. ;-) hydnjo talk 01:22, 24 April 2006 (UTC)
  1. Acquire two goats and a car and some large doors, with frames.
  2. Reveal one goat.
  3.  ????
  4. Profit!

- Abscissa 12:27, 24 April 2006 (UTC)

POV

This article is blatantly biased against goats. --Bonalaw 09:08, 24 April 2006 (UTC)

Well it certainly gets my goat. hydnjo talk 21:04, 24 April 2006 (UTC)
Now, now, guys. Stops making asses of yourselves. Icey 19:11, 29 May 2006 (UTC)

solution wording

user:Iamtheari and I started a discussion about the wording of the solution section, copied here. -- Rick Block (talk) 19:18, 28 April 2006 (UTC)

Hi - Can we talk about this sentence from Monty Hall problem?
As long as the player always switches his choice, there are only three possible scenarios, each with equal probability (1/3)
The scenarios that are listed are the only possibilities and have no dependency on what the player decides to do (switch or not). Since there aren't any other possibilities the qualifying clause (As long as the player always switches his choice) is unnecessary (could perhaps be Whatever the player decides to do, ...). I see in the summary of your change re-adding this clause that you seem to think it makes an assumption explicit. There is no such assumption. Is the issue the sentence at the end of each scenario that says whether you win or lose by switching? I'd be fine with deleting those as well since the text following the bullet list explains the win/lose in the scenarios. I'll watch your talk page for bit, feel free to reply here. -- Rick Block (talk) 12:49, 28 April 2006 (UTC)
Howdy! Thanks for the message. I think that the way it is worded now is the best. Without the prefatory language that I added, some people (who happen to be both intelligent and mathematically-trained) found the three possibilities confusing. The language "Switching wins the car." does not make it explicit enough, and the three possibilities that are listed do not explicitly handle the player not switching his choice. Therefore, the three possibilities listed do assume that the player will switch.
I originally even bothered to read this article when a friend brought it up yesterday afternoon and another friend was confused by it. The confused friend said that the changes I made cleared things up perfectly for him. Therefore, I think that it was a valuable change.
Valuable or not, I am all for reducing unnecessary redundancy within a single paragraph (because of the nature of the problem itself, the redundant explanations are "necessary" to a sufficient degree to keep them in). However, I do not feel that it is unnecessarily redundant to have both the prefatory language that I added and the language at the end of each bullet point.
It may be possible to word it better, but I think that it is best to keep it around. On a side note, thank you for polishing up the other paragraph that I added. I did it in a hurry and it is easier to read now. Ari 14:00, 28 April 2006 (UTC)

I think the current wording is misleading - it seems to be saying other scenarios might be possible if the player does not choose to switch. The scenarios are meant to be an exhaustive enumeration of all alternatives. Perhaps this could be made more clear with wording like:

Depending on the player's initial choice of door there are only three possible scenarios, each with equal probability (1/3)

and, in addition, removing the Switching wins the car statements from the bullets and clarifying the wording following the bullet list. The idea is there are three things that might happen. If you switch then you win the car in two out of the three and if you don't switch you win the car in only one out of the three. -- Rick Block (talk) 19:18, 28 April 2006 (UTC)

Clarifying when switching wins the car and when it doesn't helps explain that 2/3rds of the time it helps to switch. There is no ambiguity in reiterating when the player wins. There are, in fact, 6 possible outcomes (this was my confusion -- I'm the guy that Iamtheari mentioned). The possibilities are:

  • The player picks the door hiding goat number 1. The game host shows the other goat. Player switches and wins.
  • The player picks the door hiding goat number 1. The game host shows the other goat. Player does not switch and loses.
  • The player picks the door hiding goat number 2. The game host shows the other goat. Player switches and wins.
  • The player picks the door hiding goat number 2. The game host shows the other goat. Player does not switch and loses.
  • The player picks the door hiding the car. The game host shows either of the two goats. Player switches and loses.
  • The player picks the door hiding the car. The game host shows either of the two goats. Player does not switch and wins.

Removing the sentence I added at the end of each simply obfuscates the explanation. This is a very difficult problem to grasp intuitively; this is what makes it neat, but this means it requires very explicit explanation in terms that are obviously correct. Redundancy is not always a bad thing. Do not assume the reader will just connect the parts of it out on their own.

-- Bilbo1507 23:11, 28 April 2006 (UTC)

The way I view it, at the point the player decides whether to switch or not there are only 3 possible states (reflecting the player's initial choice). Switching or not is then a binary choice affecting the chances of winning. Combining the choice of switching with each of the possible states makes it look like there are 6 independent states, but in reality there are only 3 which become 2 sets of 3 based on the (single) switch/stay choice. Would it be clearer to present the bullet list as the three possible states at the point the player is asked to decide whether to switch, with an intro sentence like:

There are three possible scenarios, each with equal probability (1/3), leading to the point at which the player is asked whether to switch

The idea is to make it clear the player has not chosen yet, but is on the brink of doing so. Then the ramifications of each choice (stay or switch) can be explained. -- Rick Block (talk) 01:14, 29 April 2006 (UTC)

I think that you are on the right track, but the wording still isn't quite the best. How about:
Before the player is offered the opportunity to switch, there are three possible states that the game can be in, each with an equal probability of 1/3.
This makes it explicit and unambiguous that the three possibilities are before the player has any chance to switch. The wording "leading to the point" is ambiguous in this context. Ari 01:54, 29 April 2006 (UTC)

I don't understand your resistance to explicitly iterating the 6 possibilities. There really are 6 possibilities at the end of the game, and what we really care about is how the choice impacts the end of the game. If you don't follow through to the end of the game for each possibility, it obfuscates the problem. If you do, there's no harm done; both explanations are correct, it's just that one is clearer. However, maybe it would be more palettable to iterate them this way (and adjust the surrounding wording accordingly):

The player wins 2/3rds of the time if he always switches:

  • The player picks the door hiding goat number 1. The game host shows the other goat. Player switches and wins.
  • The player picks the door hiding goat number 2. The game host shows the other goat. Player switches and wins.
  • The player picks the door hiding the car. The game host shows either of the two goats. Player switches and loses.

The player loses 2/3rds of the time if he never switches:

  • The player picks the door hiding goat number 1. The game host shows the other goat. Player does not switch and loses.
  • The player picks the door hiding goat number 2. The game host shows the other goat. Player does not switch and loses.
  • The player picks the door hiding the car. The game host shows either of the two goats. Player does not switch and wins.

I still really really really think that the original iteration is too obfuscated. The first time I read this, I was absolutely convinced that the whole thing was misanalyzed. It is seldom worse to have a couple extra sentences than to lose the reader entirely. It helps a lot to explain it in terms of the 2/3rds chance of getting it wrong the first time. What is the downside to a little extra?

-- Bilbo1507 03:12, 30 April 2006 (UTC)

PS, I think I like splitting it into two groups of three better -- and I wouldn't have thought of that without this discussion.  :) -- Bilbo1507 03:17, 30 April 2006 (UTC)
Apparently too hastily, but I updated the wording before the above comments were posted (based on Ari's comments above) . The iteration is now meant to be clearly at the point the player is deciding whether to switch or not (after the initial choice, and after the host has opened whatever door is being opened, but before the switch/stay choice is made). Most of the article is focused on this specific point. I am very interested in making this section absolutely clear. I'd also strongly prefer to avoid repetition and to have this section be quite succinct. IMO, the critical insight is that at the decision point the player is, with equal probability, in one of the 3 states now described without knowing which and the switch/stay analysis . Switching has one outcome (2/3 win), staying has another (1/3 win). Whether we enumerate the states once and then say what happens with both choices or enumerate the states twice combining the results in with the states doesn't seem (to me) to make much difference with regard to clarity (I actually think shorter is less subject to misunderstanding, but I understand you might have a different opinion). I've made another slight change so the switch/stay analysis is more explicit. Do you find this version clear? -- Rick Block (talk) 16:12, 30 April 2006 (UTC)

Ya, I think it's at least as clear as I could make it now. I do like your most recent improvements.

Honestly, I've been pondering it so much now it's hard to say now. It'd be interesting to see at what point while reading a new reader finds an intuitive understanding. For me, this was when I read the part about a 2/3rds chance of picking a goat at first, so I think adding that was a big improvement, as it didn't exist in the version I initially read. Then, on re-reading, Increasing the number of doors helped me understand it more intuitively, but the bullet list confused me (which is why I wanted to add "switching wins" or "switching loses"). I think your latest modifications address this better than my modifications. Thanks. End of thread for me.  :) -- Bilbo1507 03:03, 1 May 2006 (UTC)

Thanks for pushing this. I think it is better now. -- Rick Block (talk) 03:52, 1 May 2006 (UTC)
I like it, too. Good work (and much better than my initial edit, to be sure). :) Ari 04:13, 1 May 2006 (UTC)

Randomly deciding whether to switch

I've taken a large chunk out of the Venn diagram section. The sequence of diagrams seemed a bit long-winded, so I've replaced them with one that attempts to cover everything more concisely. Also, a lot of this deals with deciding whether to switch by flipping a coin; I don't really see what this discussion adds to the topic.

Image:Monty3.jpg
Note that switching at random is quite distinct from just keeping the original choice. Having arrived at A, B, C, or D, if the contestant then blindly flips a coin to choose whether to switch or stand pat, then there is a 50% chance of ending up with the car. However, the choice doesn't have to be made at random. The coin flip gives a 50% chance of being the option with 1/3 likelihood, and 50% of being the option with 2/3 likelihood, so they balance.
There are three cases. The following diagram extends the tree in the event that the contestant elects to flip a coin.
Image:Monty4.jpg
The result is one of twelve outcomes, with the probabilities indicated. The sum of outcomes with the car is 1/2, and the sum of outcomes with the goat is also 1/2.
The following diagram shows what happens if the contestant never switches the original pick:
Image:Monty5.jpg
In this case, the outcomes arriving at a goat add to 2/3, while the car is 1/3. Contrast that with the third case here in which the switch is always chosen:
Image:Monty6.jpg
This time the car is a 2/3 possibility, and the goat only 1/3.

~ Booyabazooka 21:38, 1 May 2006 (UTC)

I'm pretty sure this discussion was added in response to an online criticism of Wikipedia's article (which I can't seem to find). In general I like the change, if you're looking for feedback. -- Rick Block (talk) 23:34, 1 May 2006 (UTC)

moved from section on the solution

A user has added:

In a matrix its much easier to see:


              Door
        Case  A        B      C
         1.   goat     goat   car
         2.   goat     car    goat
         3.   car      goat   goat

to the section on the solution (following the bullet list of alternatives). This is problematic, since in the rest of the article the doors are numbered (not lettered) and the cases are not exhaustive. It's not obvious to me how to clearly present this in a table. Anyone have any better ideas? -- Rick Block (talk) 17:40, 14 May 2006 (UTC)

I don't understand what value this adds at all. All it seems to illustrate is the premise that a car is randomly placed behind one of the three doors, and that's not the point of confusion in this problem. ~ Booyabazooka 21:31, 14 May 2006 (UTC)

I could see adding a little table before the At the point the player is asked sentence enumerating the six possible arrangements of two goats and a car behind three doors, and changing the solution section to read as follows:

The answer to the problem is yes; the chance of winning the car is doubled when the player switches to another door rather than sticking with the original choice.
Door 1 Door 2 Door 3
Goat 1 Goat 2 Car
Goat 1 Car Goat 2
Goat 2 Goat 1 Car
Goat 2 Car Goat 1
Car Goat 1 Goat 2
Car Goat 2 Goat 1
There are six possible arrangements of two goats and a car behind three doors.
Whatever door the player picks results in a 1/3 chance of picking Goat 1, a 1/3 chance of picking Goat 2, and a 1/3 chance of picking the car. At the point the player is asked whether to switch there are three possible situations corresponding to the player's initial choice, each with equal probability (2/6):
  • In the 2 cases in which the player originally picked the door hiding goat number 1, the game host has shown the other goat.
  • In the 2 cases in which the player originally picked the door hiding goat number 2, the game host has shown the other goat.
  • In the 2 cases in which the player originally picked the door hiding the car, the game host has shown either of the two goats.
If the player chooses to switch, the car is won in four out of the six cases. A player choosing to stay with the initial choice wins in only two cases. Since in four out of six equally likely cases switching wins, the odds of winning by switching are 2/3. In other words, a player who has a policy of always switching will win the car on average two times out of the three.

I think this is fundamentally the same point user:Bilbo1507 was getting at, above. -- Rick Block (talk) 22:32, 14 May 2006 (UTC)

Thought I'd test this out

I found this problem really interesting, so decided to scribble a little PHP script to simulate the game. My first run of 50,000 games without switching shows the chance of winning to be 33.367%. My first run of 50,000 games with switching shows the chance of winning to be 66.62%. So it is indeed true, switching does increase your chances of winning from 1/3 to 2/3. I know we all know it's true already, but I just like to waste my time!

If anyone would like to see, here is the script in action: [1] and here is the code: [2]. Icey 19:30, 29 May 2006 (UTC)

the unambiguous statement

A game-show host has placed a car behind one of three doors and a goat behind each of the other two. You have no means of determining which door conceals the car. The rules are as follows: you must provisionally choose one door; the game-show host must then reveal a goat by opening one of the other two doors; you can now stick with your original choice or switch to the other remaining door. You win whatever is behind the final door you choose. Do your odds of getting the car increase by switching?

I think this is a much shorter and clearer version of the unambiguous statement of the problem.Davkal 15:52, 19 June 2006 (UTC)


Intelligent people will prefer the unambiguous statement (either this one or the current one) because without all of this information the problem is underconstrained. As it stands now, you must read past the first spoiler before finally getting to the clearly stated (unambiguous) version of the problem. What a bummer!

From article page

this was added to the article page, I moved it here Rx StrangeLove 20:03, 23 June 2006 (UTC)

Here's why the "paradox" is an erroneous approach to a mathematical solution. The falacy is that the goats are not treated as unique. The following example shows the flaw in the 2/3 win conclusion. Let's say behind the 3 doors are 3 socks (Red, Blue, and Green) and my objective is to find the red sock. There are 4 distinct scenarios: My Pick: Red Red Blue Green Reveal: Blue Green Green Blue

================================================

Remaining: Green Blue Red Red

So, if I switch to the remaining door, I only have a 50 / 50 odds of success and the whole Monty Hall "paradox" is not; unless of course, we don't give credence to the goats being different. QED (JW) —Preceding unsigned comment added by 216.138.96.149 (talk • contribs)

It has nothing to do with socks verses goats; it's about the knowledge the host has when he reveals a door. Your example is skewed - you pick the red sock 50% of the time. Try six examples (still neatly divisible by 2, in case the rest of the world is wrong and you are right). Like this:
You:         R R G G B B   1/3
Host:        G B B B G G
If stick:    R R G G B B   1/3
If switch:   B G R R R R   2/3
--Mike Van Emmerik 20:55, 23 June 2006 (UTC)

Error in reasoning

I suspect that responding to this situation will be a little like spiting into the wind. I will do so anyway. Those who reason that there is a 2/3 chance of success by changing have made the following error in reasoning. Probabilities are mathematical tools for estimating events with incompete information. They are not properties of events or objects. Events or objects either exist or not. There is a 100% chance of success in the first step since no matter which door is chosen, the player will advance to the same second step. In the second step, one door is removed leaving two doors (one success and one failure). Without any additional information, this leaves a random selection of two or 50/50 chance of success. —Preceding unsigned comment added by StevenQ (talk • contribs)

Umm, I think that your statement: "Without any additional information," doesn't properly characterize the MH problem. Going back to the "second step", you have left out the part that door that is "removed" (opened by the host) is known by the host to reveal a goat. That, is quite a bit of "additional information". --hydnjo talk 18:34, 24 June 2006 (UTC)
I'll assume you've read the article and remain unconvinced (BTW - you're in very good company, there were hundreds of math professors who responded to the Parade magazine article about this problem who claimed the 2/3 answer is incorrect). The key to understanding the solution is the realization that the fact that there are two choices does not mean the choices have equal probability. For example, if I deal 1 card to you from a 52-card deck and keep the rest and then while I'm looking at the 51 I've kept I turn over 50 that are not the ace of spades, you have one card and I now have one card. The chances that we each have the ace are not 50/50. If youdisagree with this, please try it (maybe 10 or 20 times) and let me know how it turns out. The sameprinciple applies here. You have a 1 in 52 chance of getting the ace (1 in 3 chance of initially picking the car) and I have a 51 in 52 chance (the car being behind one of the other two doors is a 2/3 chance). Turning over 50 cards (opening a door) doesn't change your initial probability even though only 2 unseen choices remain. If I originally had the ace (if the car is behind one of the 2 unchosen doors), by turning all but one over (opening the other door) I've simply shown you which of my 51 cards the ace must be assuming I had it to begin with (which door has the car if it's behind either). If I don't turn the cards over, but honestly tell you which of my cards is the ace if I have it we end up in the same position. Similarly opening a door is equivalent to the host honestly telling you "if the car is behind either of these doors, it's behind this one". The fact that you can see the cards or that the door is open doesn't change the probabilities. -- Rick Block (talk) 19:06, 24 June 2006 (UTC)
I'll admit that I was wrong. I looked at it from another angle as a result of the card illustration. The problem boils down to 1/52 chance of success (51/52 chance of failure) in first choice. The second step lumps all remaining choices into one. Given the oportunity to choose between the first card chosen and all the other cards, there is a 51/52 chance of success in swithcing. 69.15.83.121 22:13, 24 June 2006 (UTC)StevenQ
Is there something in the article we could change to make this more clear? In particular, the card analogy is already described in the "Card game experiment" section. -- Rick Block (talk) 22:21, 24 June 2006 (UTC)
I think that the casual reader's eyes have already glazed-over before they get to the card game experiment in the article. It would be good to have a stand-alone explanation somehow, just as you've done so successfully in this section. --hydnjo talk 22:48, 24 June 2006 (UTC)
Possibly you could include the statement that excluding one door or one card is equivalent to combining (stapling) the two remaining together so that choosing one is equivelant to choosing both thus combining the probabilities.--69.15.83.121 23:03, 24 June 2006 (UTC)StevenQ--69.15.83.121 23:06, 24 June 2006 (UTC)
Please don't read this as "it's perfect just the way it is" (it clearly isn't since people are still confused by the existing explanations), but this is basically the intent of the "Venn diagram" section (with pictures for the more visual folks). Any suggestions for clarifications in this section? -- Rick Block (talk) 23:11, 24 June 2006 (UTC)


I hate to do this, but I am returning to my original conclusion. A Probability is an estimate, not a property of an event. It is based on the best information which is 1/3 to start with. Once one of the doors has been removed, the problem has changed and the best information has changed. The original estimate is now obsolete and cannot become part of the second step. The probability can be revised to 1/2 since there are two doors to choose from instead of three. (One could also estimate the original probability at 1/2 to start with since the remaining two doors could be combined into one choice in the first step).

StevenQ, the correctness of this article's conclusion has been confirmed numerous times by experiment. It has empirically proven beyond any reasonable doubt. 1/2 is most certainly the wrong answer. We may be able to help you understand why this is, but it is futile to argue that 1/2 is the right answer unless you claim that the countless experiments are wrong. --P3d0 03:07, 27 June 2006 (UTC)

---Returning to the 52 card illustration, the player originally has the choice of one card out of a deck of 52. An estimate of 1/52 chance of success can be estimated. The ace of spades is a success and any other is a failure. After one is selected, the dealer removes one of the failure cards and gives the player a choice of choosing any one of the remaining 51 cards (The original selection or one from the deck). The best information has changed, there are now 51 cards to choose from. Any one of the 51 cards can be the ace of spades. The player now estimates a 1/51 chance of success. The process can be repeated and the player chooses from 50 cards with a 1/50 chance of success. Each step can be repeated until there are two cards left, the orginal choice and one in the deck. The estimate can now be revised to 1/2.--StevenQ 03:20, 25 June 2006 (UTC)--User:StevenQ 02:36, 25 June 2006 (UTC)

I can only presume you didn't actually try the experiment I suggested. Rather than argue about abstract notions that are exceedingly difficult to explain, can you please try the experiment I suggested (shouldn't take more than 5 or 10 minutes to do 20 iterations). I'd be willing to bet an arbitrarily large sum of money that in 20 iterations (with 52 cards) you don't end up with the ace more than 5 times. A user wrote an Excel simulation (see above) which you can run as well. There is no doubt at all that the answer presented in the article is correct. I fully understand that it is not intuitive, however please try to see the unintuitive (correct) answer. -- Rick Block (talk) 03:35, 25 June 2006 (UTC)
BTW - I just tried this 20 times with a deck of 52 cards. I got the ace only once (actually getting it even once shocked the hell out of me!). As you do this, I think you'll find that it is easier to look at the one card dealt to the "player" rather than sorting through the 51 cards left with the dealer (finding 50 cards that are not the ace to discard) to determine the outcome. -- Rick Block (talk) 04:04, 25 June 2006 (UTC)
... and if you'd like formula for the actual odds of a random card left in the dealer's hand being the ace after each discard, it's (51/52)*(1/n) where n is the number of cards left. With no cards discarded (n=51) this simplifies down to 1/52 (which I suspect is what you expect). With only one card left this simplifies down to 51/52 (which I'm gathering you really don't expect). Your probability of having the ace remains constant at 1/52 (which is 1 - (formula*n) regardless of the value of n). -- Rick Block (talk) 05:15, 25 June 2006 (UTC)
...and about Rick's bet with StevenQ, the odds in favor of Rick winning his arbitrarily large sum of money is about 204:1 (I think). Nice odds Rick! --hydnjo talk 11:41, 25 June 2006 (UTC)

I have never been more confused in MY ENTIRE LIFE

Seriously. My head is hurting with a degree and manner i never thought possible. Once one door has gone, it becomes a 50-50 and you are neither better or worse off by switching. The point of me writing this is to illustrate more than anything else, that this whole article, and it's discussion is to me (and I am supsecting a large amount of other people) very headache inducing. I am about to go and lie down. No analogies of cards or socks seem to be helping my condition. I must lie down as I seriously suspect my brain may be haemorraging at this very moment. I am not a maths professor. But I have been a programmer for large amounts of my life and have never had problems understanding logic. My point is simply this: any logic you are presenting (if there is any, and this whole thing isnt just some evil joke by a bored maths professor), is not being presented in a logical manner. I certainly cannot see any logic in this. Whether that is the problem or the way it is presented, I will leave in the hands you people who seem to insist in the truth of this so called logic. In conclusion, I am going to lie down in the KNOWLEDGE that I am right. Though I leave the door open for you to try and prove to me otherwise, I suggest taking a course in english literature or common sense beforehand. -Pete —Preceding unsigned comment added by 88.104.105.244 (talk • contribs)

First of all, take comfort in knowing that no matter how hard you think about it, it won't physically damage your brain. If anything it's good for you. Second of all, I too was thoroughly convinced it should be 50-50 at first, and I had to think about it for about a week in many different ways before I understood. This is the way I understand it now: The door Monty chooses is not random, because he never chooses the one with the car. Therefore, Monty's choice provides information to you. Not definite information, because you still don't know for sure which of the two remaining doors has the car, but probabilistic information. There are two possibilities, but they are not equivalent and not equally likely, because of this probabilistic information available to you. —Keenan Pepper 01:53, 27 June 2006 (UTC)
In an entirely logical manner, your first choice has a 1/3 chance if being the car (right?). What, exactly, happens after that and changes this probability? The point is (as Keenan suggests), the subsequent event (opening the other door) simply tells you which of the other two doors must have the car. The initial odds don't change. There's no difference between opening the door and a completely honest host (who knows what's behind the other two doors) not opening the door but telling you, "if the car is behind either of these doors, it's behind this one". Your initial choice is still a 1/3 chance. There's clearly a 2/3 chance the car is behind one of the other two doors. If some omniscient intelligence can tell you which one of these two it must be, isn't it rather obvious that this door must be the 2/3 chance? Capisce? -- Rick Block (talk) 03:50, 27 June 2006 (UTC)
Actually, Capite? --hydnjo talk 04:00, 27 June 2006 (UTC)
See Capisce. —Keenan Pepper 04:13, 27 June 2006 (UTC)
OK, I get it now. My family used capice? informally but used capite? during formal reprimands. Sort of like capice? = get it? vs capite? = do you agree and understand? Subtle but different. --hydnjo talk 04:34, 27 June 2006 (UTC)
Actually, let's make this stronger. You go to a lotto vendor. You buy a $1 ticket with either random numbers or numbers you picked. An honest, omniscient oracle (if you know one of these, PLEASE let me know) tells you, "if you didn't pick the right numbers, the winning numbers will be <insert winning numbers here>, would you like to switch to my numbers?" Well, you say, my numbers and your numbers are equally likely, it's a 50/50 chance, I might as well stay with my numbers. Let me suggest that if you're ever faced with this scenario, take the oracle's numbers and I'll gladly pay you $10,000 for the $1 ticket. -- Rick Block (talk) 04:45, 27 June 2006 (UTC)
The Many Doors problem illustrates the "problem" with a greater number of doors than the traditional three doors MH problem (which is the least intuitive presentation and why it is the most popular example). --hydnjo talk 05:41, 27 June 2006 (UTC)

I think the simplest way to explain it is this. Firstly, the odds don't change. The door you intially pick has a one in three chance of being right and this never changes. What the game show host effectively offers is the choice between sticking with the door you originally picked OR CHOOSING BOTH OF THE OTHER TWO DOORS (2 chances in 3 of getting the car)- but since there will always be at least one goat behind these two doors his showing you this (the goat) is the empty move on which the trickiness of the puzzle turns. Forget the fact the game show host shows you the goat - would you like your first choice of door or the other two, would you like a 1 in 3 chance or a 2 in 3 chance. The puzzle is created by making the 2 in 3 chance (the 2-door choice) look less than it is- the empty move of revealing the goat.Davkal 07:03, 27 June 2006 (UTC)


As regards the article, I think the solution/answer is muddied by all the stuff about what information the game-show imparts by showing you the goat. This is the sleight of hand/misdirection that makes the puzzle difficult and we would do well in the article to not focus so intently on the very thing that confuses. What is effectively on offer here is a choice of one door (the original choice) versus both of the other doors - a choice that, combined with an honest/unambiguous host, nobody would have a problem making. The puzzle works as a puzzle because the two-door choice is masked by the game-show host's shenanigans (here's a goat do you want to switch). To recap: would you like your original door or the other two - obviously the other two... (then much later and in small print so that the point has had time to sink in, wait for it, wait for it, ...even though one of the other two must have a goat behind it, e.g. this one here, opens door to reveal goat.Davkal 07:23, 27 June 2006 (UTC)


For anyone who's confused, read the article for the Three Prisoners Problem. In terms of the probabilities involved it's exactly the same problem, but framed in a way that makes the fallacy much more obvious. --Bonalaw 07:37, 27 June 2006 (UTC)

Think about this puzzle: same show, same prizes, different format. 1000 doors, 999 goats, 1 car.

Host: Would you like to pick one door or 999 doors.

Contestant: Obviously 999 doors.

Host: Why?

Contestant: Because 999 doors gives me a 999 in 1000 (99.9%) chance of getting the car.

Host: Excellent. You are aware that at least 998 of the 999 doors you pick must have goats behind them.

Contestant: Obviously, there's only one car after all.

Host: So if I wanted to I could open 998 of the 999 doors you chose and show you goats.

Contestant: Of course you could

Host: and you still want to go for the 999 door option?

Contestant: Yes.

Host: Clever boy, here's your car keys. Hope you've got room for 998 goats.

SonofDavkal 07:39, 27 June 2006 (UTC)

Python Simulator

I wrote a python script simulation of the monty hall problem for anyone still skeptical. It uses the rules of the monty hall problem and determines how many wins for a given number of games. I have it currently set to 100k games (takes about 30 - 60 seconds on a 1 ghz OS X laptop). Check it out, just save the below code to a file and name it monty.py. In console mode navigate to the directory and type python monty.py to run the program. Iterations can be changed by changing the number on the variable "rounds".

I thought about adding this to the article but I am not sure how to do it or if it'd even still be appropriate. Anyone can feel free to add the below code to the article.

#!/usr/local/bin/python


import sys,string,random
#print 'Enter the number of games you want to play:'
#rounds = raw_input()
rounds = 100000
print 'Now processing',rounds,'games'
i = 1
wins = 0


 while rounds != i :
   # print 'The while loop is starting';
    #set the prizes behind the doors, 1=Donkey, 2=Donkey, 3=Car
    doors = [0,0,0]
    doors[0] = random.randint(1,3)
    #print 'Door 1 contains',doors[0]
    door_setter = random.randint(1,3)
    
    while door_setter == doors[0] :
        door_setter = random.randint(1,3)
   
    doors[1] = door_setter
    #print 'Door 2 contains',doors[1]
    door_setter = random.randint(1,3)
    
    while door_setter == doors[0] or door_setter == doors[1] :
        door_setter = random.randint(1,3)
    doors[2] = door_setter
    #print 'Door 3 contains',doors[2]
    #pick a door
    door_selection = random.randint(0,2)
    #print 'The door selected is',door_selection
    #show a door with a donkey
    donkey_door = 0
    
    while donkey_door == door_selection or doors[donkey_door] == 3 :
        donkey_door = random.randint(0,2)
    #print 'the donkey door is',donkey_door    
    #switch doors 
    new_door = 0
    while new_door == door_selection or new_door == donkey_door :
        new_door = random.randint(0,2)
    #print 'the new door is',new_door    
    #see if there is a win
    
    if doors[new_door] == 3 :
        wins += 1
        #print 'Win'
    i += 1
    #print i
   #print rounds
    #print 'The while loop is finishing'
#win_perc = wins / rounds
print 'Out of',rounds,'you won',wins,'of them!'

--Tastywithmilk 03:07, 2 July 2006 (UTC)

The code doesn't belong in the article, but there is an underdeveloped Wikibook for this sort of content, Algorithm implementation. I've transwikied your code there, and added a link to it in the External links section. ~ Booya Bazooka 00:02, 3 July 2006 (UTC)

The whole thing is a fake, since the parameters change

Once Monty opens a door, that door is out of play. Therefore, we are no longer dealing with the odds for 3 doors. The whole premise is a sham.

There are 2 doors left, and the player has chosen one of them. He has a 50:50 chance of winning the car, no matter which door he chooses (i.e. stay with the original pick, or take the other of the 2 doors). Since the third door has been opened, the 3-door problem no longer exists, and we are only dealing with a 2-door problem.

I understand the theory behind the 3-door solution, and it would make sense if there were in fact 3 unopened doors (i.e. 3 choices). However, once the door is opened, the 3-door problem has evaporated, and therefore you cannot apply the 3-door problem to it.

This is why most people do not like statistics, because the answer for a single event can be different depending on how you ask the question. Illustration: a coin flip. Flip a coin, and you have a 50% chance for either heads or tails. However, suppose that the flipper has beaten the astronomical odds and has already flipped 20 heads in a row. The odds of the next flip are different depending on if you ask question #1 "What are the odds that the coin will flip heads vs. tails?" or question #2 "What are the odds that he will flip 21 heads in a row?" The first is 50%, the second is a very, very small number - but it is the same event. It seems that we have even odds of getting a very improbable event.

In reality, we are asking no such thing. The odds for 21 heads in a row only are calculable prior to the first flip. Once there are 20 heads in a row, the odds that there will be 21 heads in a row is 50%, because the coin only has two sides.

So, back to Monty's doors. Prior to any doors opening, there is a 1/3 chance that a given door has the car, and 2/3 chance that a given door has a goat. Once Monty opens a door and shows a goat, that door is no longer calculable - we KNOW that there is a goat there. So, we re-calculate our odds based upon the remaining unknowns. There is 1 chance that the remaining doors are car-goat and 1 chance that the remaining doors are goat-car. These are the ONLY choices - Monty has removed the third choice by opening the door and revealing the goat, just like we changed the parameters of the coin flip question by having already started with 20 heads in a row.

The only reason that Monty's Problem is a problem at all is because someone has fooled us all into accepting the wrong question about it (that once Door #3 is opened, that it still counts as an unknown).

--68.111.248.203 10:34, 10 July 2006 (UTC)

If your view were correct, then we would expect the switching strategy to succeed half the time; however, this is not what happens. Numerous experiments have confirmed that the switching strategy wins 2/3 of the time. This is empirical fact. --P3d0 13:01, 10 July 2006 (UTC)

The point made above would be correct if Monty didn't know what was behind the doors and opened one at random. That is, in the case where one "goat door" was opened at random the odds of the car being behind either of the others would indeed become 50-50. But, and this is the important point, if Monty opened a door at random then one in three times he would reveal a car and the game could not continue as described. In the other 2 out of three times where the game could continue the odds would be 50-50 but a 50-50 chance in 2 out of 3 cases still equals a 1 in 3 chance overall. The whole puzzle only works because Monty knows what's behind the doors and therefore never shows a car - his choice of door is therefore not random and doesn't affect the odds during the game the way a random choice would in those cases where a goat was randomly picked by Monty.Davkal 13:36, 11 July 2006 (UTC)


No, regardless of whether Monty knows the odds would still be 50-50. The fact that the door was not randomly chosen means nothing at all, either way one door is being taken out of the equation. Therefore the odds would change. Also in opposition of the article, after one door is opened the remaining not chosen door's odds do not raise to 2/3 because another door was opened. First of all the odds would change but imagining the odds were still out of 3 then whats to say that instead of the unchosen door's odds raising to 2/3, the chosen door's odds don't go to 2/3. This entire solution is flawed. 66.30.160.186 01:42, 9 December 2006 (UTC)

I think you're not understanding that the actual probability, as confirmed by numerous experiments, is truly 2/3. That part is not in question. --P3d0 02:04, 9 December 2006 (UTC)
Perhaps you could explain why, in actual fact, you win 2/3 of the time by switching if the odds are 50/50 and the solution flawed?Davkal 01:53, 9 December 2006 (UTC)
160.186, let me ask you this question. Suppose I show you two closed jars -- one red, one blue -- and tell you that there is a ball hidden inside one of the two jars. Without further knowledge, you may assume the chances of the ball being hidden in either jar are 50%/50%. However, suppose I open the lid of the red jar and let you see the completely empty inside. Wouldn't you now conclude that the chances of the ball being in the red jar are 0%, and that the chances of the ball being in the blue jar are 100%?
Okay, so now suppose we do it again, except this time, I let you know that there's only a 2/3 chance that I even put a ball in one of the two jars. Without further knowledge, you can assume that the 2/3 chance of there being a ball somewhere between the two jars is split evenly -- so there's a 1/3 chance it's in the red jar, and a 1/3 chance it's in the blue jar (and of course a 1/3 chance that it isn't in either jar.) Now I do the same thing, I open the lid of the red jar and show you it's not in there. Just as in the previous scenario, the chances that the ball was in one of the two jars now becomes the chance that the ball was in the blue jar (since we know it's not in the red jar.)
Okay, so now suppose that I add a third jar -- a green jar. This time the change is so trivial it barely counts as a change: there's only a 2/3 chance that I put the ball in the red or the blue jar, and the 1/3 of the time that I don't put it in one of those two jars, I put it in the green jar instead. Now I open up the red jar and show you that the ball isn't in there. Does the fact that I showed you where I put the ball if I don't put it in the two jars you knew about change the result from the previous scenario? I think it's pretty obvious it doesn't; if it isn't going in either the red or the blue jar, it doesn't matter if it gets tossed in a green jar or in a fire or on a trash heap. So, just as before, when we see that the red jar is empty, we know there's a 2/3 chance it's in the blue jar instead.
Now, realize that this is the Monty Hall problem. Whichever door the player picked initially is the green jar, and whichever door Monty opens is the red jar; the remaining jar is the blue jar. The chance that the ball/car is in the green jar/player's door is 1/3 -- and it always stays 1/3. The chance that the ball/car is in the red or blue jar/the two doors the player didn't pick is 2/3, and that doesn't change; what does change is that once the red jar is revealed not to contain the ball/car, the chance that it did goes to the blue jar instead. -- Antaeus Feldspar 05:05, 9 December 2006 (UTC)
Do you agree the card version is an analogous situation - take 13 cards including the ace of spades, shuffle, deal one to yourself and 12 to Monty. Monty looks at his hand and shows you 11 that are not the ace of spades, keeping only one. Now you have one and he has one. Is this also 50-50? If you say yes, please actually try this maybe 10 or 20 times playing the part of Monty. What we're saying is Monty will end up with the ace 12 times out of 13. This game (with the cards) is decided as soon as the cards are dealt. The actual version, with the car, is decided as soon as the player selects the door. -- Rick Block (talk) 03:12, 9 December 2006 (UTC)

"A Super Quick Explanation"

An anon recently added this:

== A Super Quick Explanation ==
To win the car, by switching when given the choice (which will always be given), you must originally choose a door with a goat - and the probability of choosing a goat door first is 2/3. To win the car by not switching (when given the choice - which will always be given) you must choose the door with the car first. The probability of initially choosing the car door 1/3.

I don't think this improves the article overall, especially because of the place it was added, but mostly because it's redundant with already existing sections. —Keenan Pepper 14:30, 10 July 2006 (UTC)

I agree. It simply repeats the very first explanation, doesn't it? -- Rick Block (talk) 14:33, 10 July 2006 (UTC)

I think the whole section on explnation and aids to understanding should be shortened. The article as it stands is a bit of a muddle. I think the quick explanation offered above is probably the shortest (and best) I have seen. If it is already in the article I think it should follow immediately from the solution in the "problem and solution" section almost exactly as written above. I think there are way too many aids to understanding which merely confuse matters and virtually none of them actually do the job (see the above discussion for proof of this). The Venn diagrams and Bayes Theorum could then be included in another section - "formal proofs/solutions" or something of the sort. I think it might also be useful to be clearer about the point I made above about how the puzzle actually works. That is, the puzzle is generated by the confusion caused by the host opening the door - once it is seen that this is, in effect, a false move which merely masks the the actual 2-door option of switching by making 66-33 look like 50-50 this would also help to clarify the solution. As noted, at present the article has become too unwieldy, jumps about all over the place and is confusing to read. I would try to change it but don't want to start editing out vast chunks of others' work. Davkal 13:11, 11 July 2006 (UTC)

I also think we should archive this page and put a proviso after the experiment claim above (which should be kept) saying that a) anyone who wishes to dispute the article should provide some explanation of how countless experiments validate the "switch" answer rather than merely suggesting a different reading of the probabilities. And b) anyone who doesn't believe the experiments show this should try 100 times on a simulator and see what score they get. Davkal 13:47, 11 July 2006 (UTC)

I like the super-quick explanation. And, I believe it leads to the following criticism of the article. Suppose you know the host is honest, and will always give you the opportunity to switch. If you decide in advance that you will always switch, then the odds of winning are 2/3. That is, if you play the game 99 times, you will probably win about 66 cars. The way the problem is stated in the article, however, you make the decison whether to switch after the first door has been opened. I am not sure this is the same thing.M dorothy 04:08, 14 July 2006 (UTC)

It doesn't matter when you decide to switch. If you do switch, you have a 2/3 chance of winning the car. If you don't switch, you have a 1/3 chance. -- Rick Block (talk) 04:58, 14 July 2006 (UTC)


Thank you for the above comments on my 'super quick solution' and I apologise if I was out of order posting it on the main article page. I am a physicist and I love to share this problem. It is, however, difficult or impossible to convince the average person of the truth of what you are saying using basic probability calcs (eg Bayes' Theorem) or the many doors variation. I have found, invariably, that when you explain as I did ("...to win you must choose a wrong door first, of which there are 2 out of 3...") the layman's eyes immediately light up. Try it for yourself. That's why I wrote it.. as I think it's the best layman's guide to understanding.

I agree with the above because I did try it and it did work. I therefore think that the super quick explanation should be used (almost exacly as stated above) immediately as the first explnation of the problem. I then think all the other "aids to underatsnding" should be removed except for the formal stuff (Bayes theorum etc.) which should be put it a section of their own called "Formal proofs" or something. As I have said above, at present the article is too unwieldy and I didn't want to simply remove peoples work without discussion - but, since nobody seems to want to offer a reason for keeping all the confusing stuff, I will amend the article in light of these comments by the end of the week. Davkal 10:14, 18 July 2006 (UTC)

The text in the section "The Solution" was painstakingly worked out (mostly here, on the talk page). Doesn't the second paragraph of this section say the same thing as the proposed "super quick explanation" (not quite in the same words)? Are you suggesting saying something like this in the lead (section before the contents)? Before ripping and hacking, please be somewhat more specific about your proposed changed. -- Rick Block (talk) 13:32, 18 July 2006 (UTC)

Bad

This is the most long winded and confusing explanation to this problem I have ever heard in my life. It is much easier explained, as above, that you have 3/2 chances of picking a goat (MORE chance of picking a goat) and when the other goat is taken out of the equation (and you are left with your door, that has MORE chance of having a goat and the other door which may have the car) then the other door suddenly has MORE chance of holding the car. JayKeaton 23:18, 14 July 2006 (UTC)

More is needed for an unabiguous statement of the problem

I've been mulling this problem over a bit, and I'm a little bothered by a detail concerning the host protocol for when the contestant initially chooses correctly. In this case, the host has a choice of two doors to open. The unspoken assumption is that he will randomly choose one to open. The article makes that assumption in the "Decision Tree" section of the article, and again in the "Bayes' Theorem" section.

I am concerned about the possibility of host bias in this case. What if Monty likes door number three, and will choose it whenever possible? Or consider the situation in which the Monty simply scans the doors from left to right and opens the first door with a goat that the constestant did not pick. Or consider the situation where he chooses the door with the highest number with probability 3/4.

I am not here to argue the general premise that "you win with probability 2/3 by switching." That statement, which is the general premise of the article, is correct. Because it doesn't specify any specific doors, the bias, if any, would be "averaged out" in the conditional probability equations which govern the behavior. In other words, if x, y, and z are the doors, and the contestant chose x and then switched, there are four outcomes in the sample space:

CxHy: The car is behind x, host opened y, contestant switched to z and lost

CxHz: The car is behind x, host opened z, contestant switched to y and lost

CyHz: The car is behind y, host opened z, contestant switched to y and won

CzHy: The car is behind z, host opened y, contestant switched to z and won


We know only that P(CzHy) = 1/3, P(CyHx) = 1/3, and P(CxHy) + P(CxHz) = 1/3

Conditional probability laws tell us that the probability of winning by switching in general is given by dividing the winning outcomes over the possible outcomes: (P(CyHz)+ P(CzHy)) / (P(CzHy) + P(CyHz) + P(CxHy) + P(CxHz)) = (1/3 + 1/3) / (1/3 + 1/3 + 1/3) = 2/3.

If the question statement were, "You begin by pointing to a door. I show you that one of the other doors contains a goat," then we would not need this discussion. However, that is not what the "unambigous" statement of the problem says. The problem clearly says, "You begin by pointing to door number 1. The host shows you that door number 3 has a goat." Because the question uses specific numbers, we can put x=1, y=3 in equations above:

The probabilty of winning by switching in the above defined instance happens if the car is behind door 2. We must also consider the reduced sample space; i.e. P(C2H3) given that the host opened door number three. It becomes P(C2H3) / (P(C2H3) + P(C1H3))

The answer to this question is no longer 2/3 if the host exhibits any kind of spoken or unspoken bias when he has the choice. Bias is defined in the equations above by allowing P(CxHy) to be different from P(CxHz). If there were no bias, P(C1H2) = P(C1H3), and since they must each be 1/6, we get 2/3 again for the specific instance.

But if bias is allowed, and P(C1H3) is not necessarily equal to P(C1H2) then, the picture is different. If Monty scans from 1 to 3 and picks the first possible door, then the odds are 100% that the contestant will win by switching to door number two. On the other hand if Monty picks the door with the highest number with probability 3/4, then P(C1H2) = 1/3 * 1/4; and P(C1H3) = 1/3 * 3/4. Then the odds of winning by switching to door number two are P(C2H3) / (P(C2H3) + P(C1H3)) = (1/3) / (1/3 + 1/3 * 3/4) = 4/7, less than 2/3.

In order to be less ambiguous, the description attributed to Mueser and Granberg must be further refined to state, for example, "If I have a choice of doors to open, I will make my choice completely at random." Otherwise, it allows for intended or unintended host bias, which does in fact affect the outcome as you can see for yourself by verifying the problems I have defined above.

Thank you for your attention.

R. Niles, San Jose, CA. —Preceding unsigned comment added by Rniles (talk • contribs) 00:07, 17 July 2006

The description attributed to Mueser and Granberg is a direct quote from their article about this problem. Changing the quote really isn't an option. I agree there is still some ambiguity, although since the "thoroughly honest host" hasn't divulged any bias I think we may be safe in assuming the choice of door to open is random (if neither remaining door has the car). Keeping the initial description and this unambiguous description as quotes eliminates a tremendous number of complaints about this article. Unless you feel very strongly about this, I think I'd prefer to keep the descriptions as is. We could perhaps add a footnote explaining this issue. -- Rick Block (talk) 04:23, 17 July 2006 (UTC)
I agree that direct quotes need to be preserved. However, the point of having an "unambiguous" description is to eliminate the need to "assume." In this sense, I think that Mueser and Granberg simply overlooked the issue of host bias, and if they had become aware of it, they certainly would have excluded the possibility of host bias it in their formulation of the problem. I am a bit of a purist, and if the goal is to state the problem unambiguously, and an ambiguity, such as host bias, which affects the result of the problem, can be shown to exist, then it must be accounted for. I don't mind a footnote. I'd sure be happier with a separate section entitled "Host Bias" (which could be brief and discuss a simple example such as my above case where the host chooses the higher numbered door with probability 3/4, and by the way, I just corrected some errors in my original example). I'd be happiest of all if Mueser and/or Granberg decided to recognize the ambiguity and update their statement. I'm not sure if they are even interested in it after all of these years. Personally, I only read this article because I happened to come across Marilyn vos Savant's new website while researching something else, and she has resurrected the problem by publishing the initial article there and the contentious debate that ensued. Insofar as that debate was prompted by ambiguities in the problem statement and the tacit assumptions necessary to proceed, it is important to point out any and all of these assumptions if and when we become aware of them. -- R. Niles —Preceding unsigned comment added by Rniles (talk • contribs) 01:50, 17 July 2006
An update. I've just read Mueser and Granberg's paper. They were certainly aware of the possibility of this host bias, since it is mentioned in their paper. They write, "The solution normally provided assumes ... that there is an equal chance that the automobile is behind each door, that if the automobile is behind doors 2 or 3, the host chooses the door without the prize, and if the prize is behind door 1, the host chooses doors 2 and 3 with equal probability. In this case, the probability of winning by sticking is 1/3, and the probability of winning by switching is 2/3...Morgan et al. (1991) were the first to provide a formal analysis showing that the answer normally given by those stating the puzzle relied on unstated assumptions. They pointed out that if the host favored either door 2 or door 3 when the prize was behind neither ... the probabilities of winning depended on the host's particular protocol. (Emphasis mine.) Given the attention to this detail in their paper, it is likely that they intended that their "thoroughly honest host" example imply a random choice when the prize is behind neither door. However, I don't believe that assumption is clear. A host can be honest, in that he doesn't want to cheat you, yet still exhibit an unconscious bias, if for example he likes the number three better than the number two. It's not easy for a human being, honest or not, to make a perfectly random choice on his own. I would suggest that the section which quotes M&G, entitled "The problem, with all constraints explicit" be enhanced to clearly indicate that we are operating under the assumption that "if the prize is behind door 1, the host chooses doors 2 and 3 with equal probability" That should eliminate any uncertainty, and to do so should be straightforward since it's quoting a small amount of additional material from the same paper already being quoted. --Rniles 09:01, 17 July 2006 (UTC)
I've updated the intro to the description. Look OK? -- Rick Block (talk) 00:59, 18 July 2006 (UTC)
That's great Rick. It gets the point across, but do you think it needs a comma after the word "door" ? Rniles 18:27, 18 July 2006 (UTC)
I added a comma - BTW, you could have as well. I don't "own" this article anymore than anyone else does. -- Rick Block (talk) 21:16, 18 July 2006 (UTC)
Sorry, I just thought that out of respect, I just shouldn't show up out of nowhere and start editing featured articles on Wiki. Rniles 22:03, 18 July 2006 (UTC)
No need to apologize. You did exactly the right thing. Bring it up on the talk page, discuss for a while, ... I was specifically referring to the comma - typos and grammar corrections are welcome pretty much anywhere. Bigger changes are generally OK as well, but you're right to be more cautious with a featured article. Even this is not really a big deal, since any change not viewed as an improvement can be reverted very easily. It's all about collaborative editing. The basic model is much like the Extreme Programming philosophy of "everyone owns everything". At Wikipedia's scale this tends to break down a little bit (everyone with an internet connection and a browser owns all the Wikipedia articles??), but this is the underlying philosophy. -- Rick Block (talk) 22:50, 18 July 2006 (UTC)
Rniles, you might want to take a look at Wikipedia:Be bold. —Keenan Pepper 23:24, 18 July 2006 (UTC)

changes to the article

The problem with discussing the specific changes I think should be made is that there are so many. What I would like to do is rewrite the whole thing, retaining about 70% of what's already there - adding very little more, but significantly changing the order and adding some new headings. Given the length of the article I don't want to print it all out here on the talk page for discussion but given that the changes I have in mind would all compliment one another (I think) it is hard to discuss them one by one or overall without looking at the entire article. I should say that I think the article as it stands is very good, but it bears the hallmarks of something that has grown unwieldy through too many seperate changes being made - ie. (partial) repetition, tangential points placed centrally. In other words, to a newcomer (like myself) it seems a bit of a jumble and I'd like (with everyone's cooperation) to try to tidy it up a bit - but tidying up only looks good once one has finished. Davkal 15:50, 18 July 2006 (UTC)

How about drafting something in a separate subpage, perhaps Monty Hall problem/revised? -- Rick Block (talk) 16:46, 18 July 2006 (UTC)

What I have in mind should be there tomorrow night - I hope that the points I am trying to make will be clearer then. Davkal 21:56, 18 July 2006 (UTC)

I have put a first draft in the above page. I still think a few things could be changed but will leave it as is for the moment. Davkal 09:54, 19 July 2006 (UTC)

So, how would you like feedback? Others edit the same draft? -- Rick Block (talk) 13:09, 20 July 2006 (UTC)

I don't really like the idea of two drafts going at once. I think the changes to my draft article were simply the removal of repetition and extra explanations and the reordering of the text slightly. I think the current article could simply be replaced with the draft one and then edits made here. As noted, it is still, in essence, the same article.Davkal 10:44, 21 July 2006 (UTC)

The main change seem to be replacing these two paragraphs:
At the point the player is asked whether to switch there are three possible situations corresponding to the player's initial choice, each with equal probability (1/3):
  • The player originally picked the door hiding goat number 1. The game host has shown the other goat.
  • The player originally picked the door hiding goat number 2. The game host has shown the other goat.
  • The player originally picked the door hiding the car. The game host has shown either of the two goats.
If the player chooses to switch, the car is won in the first two cases. A player choosing to stay with the initial choice wins in only the third case. Since in two out of three equally likely cases switching wins, the odds of winning by switching are 2/3. In other words, a player who has a policy of always switching will win the car on average two times out of the three.
Another way of thinking about this explanation is by observing that the player begins the game with a 2/3 chance of picking a door with a goat behind it. In this case the host must show the other goat, so switching turns the initial 2/3 chance of picking a goat into an equivalent chance of winning the car. Switching loses only if the player picks the car first (1/3 chance). The total expected outcome for the player who always switches is therefore a 2/3 chance of winning the car.
with this one:
The reason for this is that to win the car by sticking with your original choice you must choose the door with the car first, and the probability of initially choosing the car is one in three. Whereas, to win the car by switching you must originally choose a door with a goat first, and the probability of choosing a goat door first is two in three.
I think rather than delete the explicit enumeration of cases, something like the new paragraph could be inserted before this enumeration. There's a style issue as well (addressing the reader should generally be avoided). Perhaps the following:

The Solution

The answer to the problem is yes; the chance of winning the car is doubled when the player switches to another door rather sticking with the original choice. The reason for this is that to win the car by sticking with the original choice the player must choose the door with the car first, and the probability of initially choosing the car is one in three. Whereas, to win the car by switching the player must originally choose a door with a goat first, and the probability of choosing a goat door first is two in three.

At the point the player is asked whether to switch there are three possible situations corresponding to the player's initial choice, each with equal probability (1/3):

  • The player originally picked the door hiding goat number 1. The game host has shown the other goat.
  • The player originally picked the door hiding goat number 2. The game host has shown the other goat.
  • The player originally picked the door hiding the car. The game host has shown either of the two goats.

If the player chooses to switch, the car is won in the first two cases. A player choosing to stay with the initial choice wins in only the third case. Since in two out of three equally likely cases switching wins, the odds of winning by switching are 2/3. In other words, a player who has a policy of always switching will win the car on average two times out of the three.

There have been some other changes to the article since this version was forked as well, so rather than replace I'd suggest making any changes incrementally. -- Rick Block (talk) 15:03, 21 July 2006 (UTC)

Alright let's start with that change. What about moving the formal stuff into the new section - I think it's better to separate this as I feel that the explanation just goes on and on at present.Davkal 16:53, 21 July 2006 (UTC)

By the formal stuff do you mean the bullet list of cases? I think it's important to include this in the solution section since it really is the basis for the solution (in some sense, it's just a slightly more detailed version of the explanation you want added). Would you be OK with the text above (with the enumerated cases) being the entire text of the "Solution" section? I like the idea of keeping this section as short as possible (and, btw, the rearrangement you've suggested). -- Rick Block (talk) 18:30, 21 July 2006 (UTC)


If there was no other point to my drafting the article in the alternative section than a mere delaying tactic I will simply edit it into the main article tomorrow and take it from there. It appears from your comments that you have not even read it. Davkal 00:16, 22 July 2006 (UTC)

Excuse me? I have read it. I quoted part of it above. If you're not referring to the enumeration as part of the formal stuff (which, by the tone of your comment just above I gather you're not) then I think we're pretty much on the same page here. Your rearrangement, which I said I liked, includes introducing the new heading "formal explanations". I was just trying to clarify how much you were including in "the formal stuff". I think drafting the new version was quite useful. Thank you for taking the time to do this. I think now we're just talking about how best to implement these changes (and copying over the existing article is always an option - I'm recommending against that so that the changes are easier to follow and because other changes have been made - not as a delaying tactic). -- Rick Block (talk) 01:25, 22 July 2006 (UTC)
I've added the new solution description. Your version deletes the "Aids to understanding" heading (leaving the content under "Solution") and the entire "Increasing the number of doors" section (right?), and then contnues with the "Formal explanations" section. I do know a lot of people find the "lots of doors" explanation convincing. It's not a "formal explanation", and I don't think belongs under "solution". Rather than deleting it, do you have a suggestion for a different place it could go? -- Rick Block (talk) 02:52, 22 July 2006 (UTC)

Original Assumption

I think this line could fit in somewhere in the problem description: "Let's assume you want a car more than a goat". It is, obviously, already understood by all reading the problem, but nevertheless clarifies the problem's goal. Wwwhhh 02:39, 20 July 2006 (UTC)

No, it distracts from the article. We can all come up with dozens of assumptions, like "let's assume the goat is not a magic wish-granting goat". DJ Clayworth 16:56, 21 July 2006 (UTC)

That's taking it a bit far, I think the line is actually quite integral to the problem, especially since the article adresses if it to your "advantage" to switch doors, hence there must be some mention of which, out of a car and a goat, would be to your advantage to choose. Wwwhhh 05:41, 23 July 2006 (UTC)

I agree that this assumption is far more important than the "magical goat". However, in the unambiguous formulation of the problem the question is "Do the player's odds of getting the car increase by switching to door number 2" and so if considered in those terms then it does not really need to be said. Davkal 09:44, 23 July 2006 (UTC)

Haha, this is why I'm not a math student

Holy hell, am I ever confused. I mean, the statement The reason for this is that to win the car by sticking with the original choice the player must choose the door with the car first, and the probability of initially choosing the car is one in three. Whereas, to win the car by switching the player must originally choose a door with a goat first, and the probability of choosing a goat door first is two in three. makes sense...yet it completely contradicts...everything! My head is going to explode. Anyways, fun article, kudos :) Sherurcij (Speaker for the Dead) 18:10, 28 July 2006 (UTC)

Isn't it a hoax?

See page 54 and further at http://www.skeptics.com.au/journal/2003/1.pdf) User:faragon 16:39, 31 July 2006 (UTC)

Did you read the document you linked? Our article corresponds to appendix B from the document you linked, which does indeed conclude that you win 2/3 of the time by changing (which is the same conclusion our article reaches). --P3d0 15:10, 31 July 2006 (UTC)

I think the points being made in the cited article are just going over the same hidden assumptions that are clarified in the unambiguous formulation and explained briefly in the explanation section.Davkal 16:02, 31 July 2006 (UTC)

Excuse me, you were right. I apologize for the inconvenience, was my fault, due to precipitation. User:faragon 22:46, 31 July 2006 (UTC)

A new way to wrap your mind around this (for all of the naysayers)

In high school, one of my math teachers introduced this problem to the class. It stumped me at first, but then I thought of it in these terms: Imagine if you were presented with one million doors, and you had to select one. Then, 999,998 of them were taken away, leaving only two doors, one being the winner. Honestly, think of the chances that you just so happened to pick the winning door. They're next to nothing. Think of the situation in much larger terms like these and it will suddenly make a lot more sense to you. Bp28 06:19, 2 August 2006 (UTC)

Thanks, but this isn't new. It is the very first "Aid to Understanding" given in the article, and it has been in the article for ages. --P3d0 17:03, 11 September 2006 (UTC)

Table

no switch
Door 1 Door 2 Door 3 Result
car goat goat win
goat car goat lose
goat goat car lose
switch
Door 1 Door 2 Door 3 Result
car goat goat lose
goat car goat win
goat goat car win


door chosen

Removed this table, because it's a bit unclear, and, frankly, I think we already have enough explanations and visual aids. ~ Booya Bazooka 05:07, 9 August 2006 (UTC)

Aids to understanding

Removed this text:

The solution to the problem relies on the fact that the host can and must reveal a goat behind one of the two doors not initially chosen by the player. Indeed, if it is only by luck that the host reveals a goat behind one of the two doors not initially chosen by the player then the probability of winning after switching to the remaining door and the probability of winning after staying with the initially chosen door are both 50%! This is counterintuitive because at first glance it appears that the decision the player has to make should not depend on whether the host knew the goat was behind the door that was opened.

I think we need to focus on improving the explanations we have; adding a plethora of similar explanation paragraphs is not going to improve the article. ~ Booya Bazooka 21:41, 19 August 2006 (UTC)

Agreed. Also, the removed text is wrong. --hydnjo talk 03:24, 20 August 2006 (UTC)
I agree that it's not particularly useful, but wrong? I think it's saying if the host doesn't know what's behind the doors and opens one (and happens to reveal a goat), then the result is a 50/50 chance. This seems right to me. -- Rick Block (talk) 03:37, 20 August 2006 (UTC)
The set of two doors that the host gets is twice as likely to contain the prize as does the set of one door that the player gets. If the host's prompting headphone fails at the critical moment that he was to receive his cue, well then he just wings it (the show must go on). If luckily (for him) he reveals a goat well then nothing has changed the 2:1 odds in favor of the host's set of two doors containing the prize. --hydnjo talk 04:00, 20 August 2006 (UTC)
If the host's choice is random, without the foreknowledge of what's behind what door, he has a 1/3 chance of revealing the car. The possibilities are the player picked the car (1/3 chance) and then the host opens one of the goat doors, or the player picked a goat (2/3 chance) and then the host with a 50/50 chance shows the other goat or the car. The player has a 1/3 chance of having picked the car, a 1/3 chance of having picked a goat and the host having shown the car, and a 1/3 chance of having picked a goat and the host having shown a goat. If you assume the car has not been shown, there's a new "universe of possibilities" consisting of only 2/3 of the total number of outcomes. Among these, the player has the car 1/2 the time. The host's knowledge of what's behind the doors is critically important to the problem. -- Rick Block (talk) 04:21, 20 August 2006 (UTC)
Alright then, the host's cueing headphone becomes very noisy at the critical moment and as the show must go on he opens the door that he thinks that he was prompted to open without really being absolutely sure but luckily (for the audio engineer) he reveals a goat. Now, whether or not he chose randomly is open to speculation. He listened for the cue and only thinks he's opening a goat door (with crossed fingers). His set of two doors is still a better bet. No? --hydnjo talk 04:35, 20 August 2006 (UTC)
Random or not, that is the question - but if you want to introduce "degrees" of randomness, if he heard with a 0% certainty, then the odds are 50-50. If he heard with 20% certainty the odds are slightly better to switch. If he heard with 80% certainty the odds are much better to switch. If he heard with 100% certainty the odds are the familiar 1/3 vs. 2/3. Bayes theorem can be used to compute the exact probabilities in all of these cases - an exercise left for the reader :). -- Rick Block (talk) 04:53, 20 August 2006 (UTC)
Oops! You're right (of course), I see where I went astray. Thanks Rick for jostling me out of my midnight haziness. :-( --hydnjo talk 05:11, 20 August 2006 (UTC)

I just looked at this for the first time. As many people have already done, I confirmed this in a Monte Carlo experiment on an Excel spreadsheet. Pretty cool! It is certainly not intuitive at first, but it starts to make sense the more you work at it. Oh well, I know I'm sounding redundant here, but what a great article and discussion of a fun math problem. Thanks to Wikipedia and its contributors. 11:40pm EDT 23 Aug 2006.

Long-run game

On 8/24/06, Martin Gradwell <martin.gradwell@gmail.com> (presumed User:88.108.234.187) wrote:

Hi,

I'm a relatively light user of Wikipedia, so I don't always know the best way of going about things, so, for instance, I'd discuss this on the talk page associated with the article, except it's a long page and I don't know where to add stuff to it. Should new stuff on a talk page always be added at the bottom? I don't know, and if I put stuff in the wrong place the odds are that you won't see it. If email is inappropriate then I apologise for the intrusion, but I don't see any other way.

Anyway, when you removed my paragraph, you noted that "this is repeated above, and the long-run game theory stuff is irrelevant. The host's behavior is either specified, or entirely unknown".

I don't understand this. What part of my paragraph repeated anything that was above it? I have examined the article careully without finding anything, either above or below, bearing any resemblance or relation to anything that I had written. Your new table does encapsulate a small part of what I wrote, but it was not present at the time when I wrote.

My paragraph might be irrelevant to the fully constrained problem as commonly stated, but then so is two-thirds of your table. My paragraph was within a section titled "The General Case", and was entirely relevant to that section. Your table lists three specific cases, one of which merely duplicates the specific case which is talked about ad nauseam throughout the rest of the article.

Your other two cases are entirely unrealistic. For instance: "The host, to maximize winnings, only offers the option to switch when the player has chosen incorrectly." -"Switching is always successful.": In this case, why doesn't the host, to save time, simply offer a car to anybody who approaches him? Why bother with doors, and goats, and a game show format? OK, you could say that we aren't discussing a real gameshow here, just trying to frame a mathematical problem in an interesting way . But your variation on it isn't even a mathematical problem! What could be problematical about a gameshow host who will always give away a car, no matter what? What could be interesting about it, apart from the address you apply to to get on that show? And since there is no such address, this being a fictitious show and all, the whole thing is even more boring than the usual statement of the "problem".

And "the long-run game theory stuff is irrelevant", you say. And indeed it is, in the fully constrained version of the "problem". But the fully constrained version is one of the most uninteresting "problems" there is, at least from a mathematical standpoint. It is only interesting from a sociological standpoint. The fact that so many people get it wrong, even Ph.D's, tells us something about the nature of intelligence, or about our educational system. The fact that they are impervious to repeated detailed explanations, and will become quite angry when it is suggested that the rest of the world is right and they are wrong, tells you something about personality. But none of this has anything whatsoever to do with maths.

Anyway, I didn't talk about long-run game theory. I taked about the host optimising some quantity, e.g. maximising ratings or minimising the prize budget. He could try to do this over the course of a single game. He would have to, if his options were as boring as the ones you suggested, since the show would certainly be cancelled after one airing.

To have a real show, the host would have to use psychology. For instance, he could invite a whole crowd of contestants each of whom is familiar with, and thoroughly accepting of, the conventional "vos Savant" and Wikipedia wisdom. Each of them gets to play, in turn, while those who have not yet played are kept in a soundproof booth to keep them from learning how their colleagues have done. Every single one of them gets a goat, wtihout exception, even though the car was available, and some of them actually chose it initially. Every single one of them changed his choice if/when given the opportunity to do so. And the host has produced a really entertainig show, where lots of people have received a well-deserved come-uppance, for the cost of a few goats. Of course this show would NOT be part of any long-run game. Repeating it, with the same parameters, would defeat the object. But a long run of the game show COULD be interesting, provided that the host doesn't stick rigidly to one strategy which makes it optimal for the guests to also stick rigidly to one strategy.

OK, that's more than I intended to write, but I do get carried away. Should I post this on the talk page, or what? Please feel free to discuss this either via email or on the talk page, and I will follow your lead. If you use the talk page, please just drop me a very brief email to let me know that the talk page has been updated.

Thanks in advance,

Martin Gradwell

I based that table on the paragraph before the one you wrote; that's the paragraph that contained a lot of the repetition. The first half of your paragraph seems to enumerate the same host-behaviors that the previous paragraph had already described.
"To have a real show, the host would have to use psychology." Yes, but this article isn't about Let's Make a Deal. It's about a specific logical conundrum. The examples of extremes in host-behavior serve only to demonstrate why the constraints must be made clear. If the host's constraints are lifted, then there is no way to answer this question with any degree of mathematical precision, so we'd be far off-topic in discussing it. ~ Booya Bazooka 19:36, 24 August 2006 (UTC)
The solution to additions such as this is really quite simple. Is this "general case analysis" described in a reference somewhere? If so, then go ahead and add it (with a reference). If it is not, then it's original research and should not be in this article. We're not developing an original analysis of the problem here. We're providing an encyclopedia article about a notable problem that many folks have heard about and about which there are many, many published papers. This distinction is lost on many Wikipedia contributors. The point is not what YOU think about this problem. The point is what you might find about this problem that's already been published elsewhere (please see WP:NOR for more details about this). -- Rick Block (talk) 03:20, 25 August 2006 (UTC)

The second paragraph of the "general case" section previously mentioned possible alternative host behaviours, without suggesting any possible motivation for any of them. Why, for instance, might the host "only offer the option to switch when the player has chosen incorrectly"? The new table at least attempts to remedy this deficiency by adding things like "to minimize the show's budget for prizes" or "to maximize winnings", but the table is still no more than an incomplete enumeration of options available to the host in the general case. And "maximising winnings" by giving away a car whenever the opportunity arises hardly makes for either an interesting gameshow or an interesting mathematical problem.

The host can minimize the show's budget for prizes by only offering the option to switch when the player has chosen the door containing the car. But he can achieve the same goal, against a contestant who plays optimally, by offering the option to switch on up to half of the occasions where the contestant initially chooses a goat. Allowing a switch on exactly half of the occasions where the contestant chooses a goat would also have the merit of being scrupulously fair, not favouring the proponents of any one strategy over another. In that case, you have initially a one in three chance of winning a car regardless of whether you are a switcher or a sticker.

Rick, I could be wrong, but judging from what I read in the discussion page I suspect that the vast majority of this article is "original research", in that it contains the contributors' own insights on how the problem cna be most simply explained. E.g. the python simulator was written by the person who provided it. It was not written elsewhere and just cited here. So why should the section on the "general case" be singled out? The important thing is that "Encyclopedic content must be verifiable." Verifiability would seem to imply a need for citations, and that is true in most disciplines, but it is not true in mathematics. You can verify for yourself that 2+2=4 without having to reference prior work.

The section on the "general case" could simply be removed, but then the article would clearly be incomplete, and could potentially mislead some unfortunate readers into winning a flock of goats when they could have won a fleet of cars. Martin Gradwell 09:00, 25 August 2006 (UTC)

The general case section section should probably be renamed... "General" is sort of a misnomer. The purpose of this section is to demonstrate why some have criticized vos Savant for not explaining the problem specifically enough (a verifiable enough claim, if people truly have put forth such criticism). And then it provides a not-too-detailed explanation for why the unspecified problem doesn't have the same answer as the "real" problem, by giving a few examples of alternate behavior. There's no point in expanding this explanation, since it's already clear enough. Also, note that the Python simulator is not part of the article; it's an external link to Wikibooks, where not all the same rules apply. ~ Booya Bazooka 22:29, 25 August 2006 (UTC)

I can't see anything in the "general case" section that is not dealt with earlier in the article. Davkal 10:39, 25 August 2006 (UTC)

"Monty Hell" problem??

I edited the Monty Hall problem to include the following (see below) - I called it ""Monty Hall 2".

However, this was then removed by someone who said that Monty Hall 2 is the same as the "Monty Hell Problem".

Having looked at this, it is not clear that the two are indeed the same.

Not sure where (if anywhere) I should put my contribution. (It has a reference - in the Financial Times this week - which shd be pretty impeccable?!

(I also made other changes to the standard Monty Hall problem, with the intention of making this clearer. However, these edits have also been reverted, possibly unintentionally)

My initial additions follow (these have since been removed, along with other editorial changes)

A further problem which is also attributed to Monty Hall is the following "Monty Hall 2 Problem":

Suppose that there are two doors. You know that each door conceals a sum of money, say x and y. You know that y = 2x.
 However, you do not know which door has x, and which has y.
 Which door do you pick?

The paradox is that whichever door you pick, it seems profitable to swap, as this gives you

  • a 50% chance of doubling your amount (if you selected the x-door),
  • a 50% chance of halving your amount (if you selected the y-door)

i.e. whichever door you start with, switching has a positive expected gain.

Thus, whichever door you start with, the other one is 'better' - a seeming paradox.

For a discussion of this, see Financial Times, c.23 August 2006 (Full reference to follow.)

John, The Monty Hall 2 problem as you state it is clearly nothing like the Monty Hell problem, as can be checked by anyone willing to follow a couple of links and read a paragraph or two. That illustrates one of the problems of Wikipedia, that it aspires to be authoritative yet cannot be correct even on something so simple and easily checked. However, Monty Hall 2 is clearly a distinct problem, unrelated to the Monty Hall problem except for the similarity of name and the presence of prizes hidden by doors. Because of that, it belongs, if anywhere, in an article of its own. If it was left here then there would be various statements (e.g. the one at the top of this talk page, "The conclusions of this article have been confirmed by experiment") which would be inapplicable to it, and these would become misleading.
There is a problem with the proposed contents of the new article: It is quite simply not true that "whichever door you start with, the other one is 'better'". If this is the sum total of the analysis by the Financial Times, then the Times article is wrong, and misleading. Since it should not be the function of an encyclopedia article to mislead its readers, any article here should give a fuller analysis of the problem. But, in the absence of a fuller analysis elsewhere, this would constitute "original research". This illustrates the other problem with Wikipedia: That, where no correct analysis exists, we are apparently forced to either slavishly relay error, or simply to ignore the problematic topic entirely, even when a simple correction is possible. The original idea of Wikipedia as I understand it, that if contentious issues were argued through sufficiently then a concensus would eventually arise, was sound. But now, with the "no original research" rule, contentious issues cannot even be raised unless they have already been resolved elsewhere, and the role of the wikipedian is simply to transcribe without adding value. Martin Gradwell 10:34, 27 August 2006 (UTC)
The concept being described as The Monty Hall 2 problem has a WP article titled Two envelopes problem and is referenced in this article's "see also" section. hydnjo talk 12:18, 27 August 2006 (UTC)

An additional misconseption that the original problem could give

Am I the only one that thought, upon reading the original problem, that the host would always open a door that was different from the one you picked? I thought that switching would just make the host randomly open one of the other doors. —The preceding unsigned comment was added by 71.125.180.245 (talk • contribs) .

Which description did you find confusing (and, BTW, the host does always open a door that's different from the one you originally picked)? -- Rick Block (talk) 03:05, 1 September 2006 (UTC)

The original problem from the collumm confused me. It led me to think the host would open a different door even if you switched.

IE.

I pick a door. Host opens a different door, shows me goat, and asks if I would like to switch. I decide to switch. Host opens another door and asks me if I want to switch agian.

Like that. —The preceding unsigned comment was added by 71.255.210.15 (talk • contribs) .

Did you think there was an unlimited number of doors? Or that the things behind them could change places?? Otherwise you'd quickly run out of doors. —Keenan Pepper 14:06, 5 September 2006 (UTC)

I thought that the host would always open a door that was NOT the one you choose, and the problem was figuring out how he chose which door to open. This is easily understood to be incorrect once you read the more in-depth description of the problem.

Where have I gone wrong?

I apologize for asking a question that I'm sure has been answered elsewhere in the lengthy talk page, but I don't have the patience to read the entire discussion; my question is this (I should say that I haven't run the experiment and I don't doubt the material truth of the counterintuitive solution, I only want to know where my reasoning has gone wrong): I understand that when the contestant first chooses a door, the odds of it containing the prize are 1/3; however, once one door that doesn't contain the prize has been opened (say, door C out of doors A, B, and C), the contestant is left with two doors to choose from, making the probably of choosing the correct one 1/2. So let's say the contestant flips a coin and determines to pick door A and not door B, since it was 50/50 anyway, so he had no reason to prefer one over the other. So if door A was the same door he chose when his odds were 1/3 (that is, before the host opened one of the losing doors), then it seems he has no reason to change his choice after all. . . .I don't know, maybe this is not a real objection or disagreement at all, but I'm pretty sure I'm doing something wrong. 152.23.84.168 04:11, 11 September 2006 (UTC)

Yes, your statement that however, once one door that doesn't contain the prize has been opened (say, door C out of doors A, B, and C), the contestant is left with two doors to choose from, making the probability of choosing the correct one 1/2 is incorrect. The way the problem is structured, opening the door does leave two doors to choose from but does not leave the remaining two doors with equal probability, in fact does not change the probability of the original choice being correct (which is 1/3). There are indeed two doors left, so since the probability of the originally chosen door is 1/3 the probability of the other one must be 2/3. How this is so is meant to be explained in the article. If you've read the article and still don't "get it" please let us know. -- Rick Block (talk) 13:49, 11 September 2006 (UTC)
152.23.84.168, thank you for approaching this with an open mind. Your approach was very logical, even if you did encounter an error. As Rick points out, just because there's two doors doesn't mean there are equal chances of the car being behind either. It may help you to see why the chances aren't equal to look at it a different way: the ending situation has two doors left to open, but there are three ways of getting there. First way: the player picks the first goat, Monty opens the door containing the second goat, and the last door has the car. Second way: the player picks the second goat, Monty opens the door containing the first goat, and the last door has the car. Third way: the player picks the car, Monty opens a door containing a goat (which of the two doesn't matter) and the last door contains the car. All three ways of getting to the ending situation are equally probable, but two of the three lead to the car being behind the last door. -- Antaeus Feldspar 14:13, 11 September 2006 (UTC)
It's true that after Monty opens the door, there are only two possibilities, but having two possibilities does not mean that the odds are 50-50. If I step into the ring with a world champion boxer, either he will win or I will win. But it's not 50-50, I have about a zero percent chance of victory. I had a hard time wrapping my mind around this problem too, until I tried it out for myself. You can play a game of "Three Card Monty" to test it. Take three playing cards, two jacks and one ace. Have a friend pick one card to see if they can find the ace. You then show them were one of the jacks are, and ask if they want to switch cards. You'll find that 2/3 of the time, switching wins. It's like getting to pick 2 cards instead of just one.--RLent 21:41, 28 September 2006 (UTC)

Solution is in bold

I really think that yes shouldn't be in bold letters because it is impossible not to notice it if you don't want the solution to be given right away. For me that spoiled the fun of trying to solve it. --70.70.132.42 20:19, 16 September 2006 (UTC)

On the other hand this is an encyclopedia not the puzzle page in the Dail Mail. Abtract 21:25, 16 September 2006 (UTC)
Oh, so you're saying the Spoiler Warning is just a mere courtesy, not really an effective warning. Then forget it. --70.70.132.42 (DrZeus) 00:13, 18 September 2006 (UTC)

Don't think any harm has been done to the article by removing the bold "yes".Davkal 14:32, 18 September 2006 (UTC)

Note--I am now known as DrZeus. I thank you, Davkal, as a new user! --DrZeus 01:45, 19 September 2006 (UTC) Also, I highly suggest that this page should become archived or something. What about the people with attention deficit disorders? What about browser page size limits? What about the children? What about the Homer Simpsons out there? It already has a size of over 165 Kilobytes long. --DrZeus 02:57, 22 September 2006 (UTC)

But What is The Relationship of this Problem with Deal or No Deal?

I understand the case described. However there seems to be a major difference with the actual problem in Deal or No Deal. That is, on the TV Show the player chooses the cases to exclude. This would essentially negate the paradoxical effect. There should be more discussion of this part of the original assumptions of the problem.

This problem is not based on Deal or No Deal, but an older (different) show called Let's Make a Deal. The format and assumptions of the problem do not even exactly match this older show, and have essentially nothing to do with the currently popular show. -- Rick Block (talk) 13:22, 18 September 2006 (UTC)

Possible host behaviors in unspecified problem (Table)

Should this table include the case where the door opened by the host happens to reveal a goat? This is mentioned in the text as a reasonable interpretation of some statements of the problem (it was my understanding when I first heard it). --catslash 21:02, 29 September 2006 (UTC)

Supposed assumption that host chooses randomly when he has a choice

This is a comment on the exposition; the result is of course correct.

In the first section after the table of contents, it says that the statement contains an unstated assumption that if the host can open either remaining door, the one that is opened is chosen randomly. But notice this can't be right. For the general question of whether to switch doors after a losing door is revealed, it makes no difference which one is revealed. The section on the Bayesian treatment says that the assumption is made clearer by that treatment, and indeed it is used. The problem is, the Bayesian treatment here does not address this general question, but the two more specific questions, whether to switch to door 3 if door 2 is revealed to be a loser, and whether to switch to door 2 if door 3 is revealed to be a loser. To address the general question, one should not just calculate P(C2|O3) and P(C3|O2), but calculate P(C2|O3)P(O3) + P(C3|O2)P(O2). Then the randomness assumption is no longer needed; you get 2/3 either way.

Of course, the Bayesian treatment is really a red herring, for one can more easily calculate P(C2 or C3 | O2 or O3). Since P(O2 or O3) = 1, P(C2 or C3 | O2 or O3) = P(C2 or C3) = 2/3. 82.45.161.53 01:34, 5 October 2006 (UTC)

Assume the host is known to always open the lowest numbered door that the player did not choose and does not hide the car. If I pick door 1, the host must open either door 2 or door 3. If the host opens door 2, then I've picked the car and should not switch. If the host opens door 3, the car is behind door 2 and I should switch. Given these assumptions I get the car 100% of the time. Stating the host opens one of the two losing doors randomly neatly avoids this issue. Assuming you're a student of some sort, I'll leave resolving this in the Bayesian treatment as an exercise for the reader. -- Rick Block (talk) 04:14, 5 October 2006 (UTC)
That is certainly true; if the host specifically opens door 3 rather than door 2, we must assume that this gives no special information in order to get the standard result. That is why I only said that the general question does not require the randomness assumption. I realize now that the Mueser-Granberg statement does specify that door 3 is opened, so it does require at least epistemic randomness (the player must suppose that the opening of door 2 or 3 is equally likely). But I think this is an unfortunate defect of the Mueser-Granberg statement. If the problem were stated as follows: "...You begin by pointing to door number 1. The host opens another door to reveal a goat. Is it then to your advantage to switch your choice to the remaining door?"--then, I say, no such randomness assumption would be needed. For, our problem then is not to find the probability that we have chosen correctly given the specific door the host has opened, but just the probability that the remaining door is correct given that a goat door has been opened, and this is 2/3 no matter what pattern the host follows or what we know about it. I think this is more in the intended spirit of the problem. Notice that vos Savant says the host "opens another door, say no. 3" (my italics). We're interested in the general rule, not the specific cases where the host opens this door or that.
All this is somewhat academic; I suppose it's better to cite a published statement than to give an alternative, even if the alternative is better in some respect. But I also think the claims made for the Bayesian treatment are wrong. It does not save any verbiage or clarify the role of assumptions; it instead makes the problem more complicated than necessary and makes the randomness assumption seem essential, when in fact (I contend) that assumption is only necessitated by an unfortunate aspect of the problem statement.82.45.161.53 01:07, 16 October 2006 (UTC)

Introduction Rewrites

In the last couple of days, some effort has gone into honing the wording of the introduction. The first (anonymous (not me!)) edit clarified that extra assumptions were being introduced. The last (intentionally or incidentally) obscured this fact, even removing the word 'if'. The reader who only reads the introduction will miss this critical point, so I reckon it should be in - but I hesitate to revert a reversion. --catslash 13:05, 14 October 2006 (UTC)

I've changed the wording to "as long as" to make this more clear. You're absolutely correct that these are critical assumptions. Please feel free to make corrections like this. Even reverting a reversion is fine, but in this case you should probably make it absolutely clear in your edit summary why the change is being made. -- Rick Block (talk) 17:12, 14 October 2006 (UTC)
Between us we have made it much clearer.Abtract 20:03, 14 October 2006 (UTC)

The problem...

I see this ... (although even this version contains an unstated assumption that if the player initially picked the car the host will open one of the remaining two doors randomly) ... has been put back in as though it is a quote but it doesn't seem to be, it isn't within the "...". If it is a quote, who said it? If it isn't then IMHO it is entirely irrelevant/meaningless and shou;d be removed.Abtract 22:33, 15 October 2006 (UTC)

Sorry, my summary got mangled. The "even though" bit is NOT a quote, that part of the summary refers to the change within the italicized quote (changing "I will" back to "I'll", which is within the quote from the Mueser and Granberg paper). The reason the other part is important to keep is not that it's a quote, but because the Mueser and Granberg version is not actually unambiguous (see the immediately preceding section on this talk page). The reason both the introductory problem statement and the "unambiguous" version are both quotes is to try to avoid constant tweaks to their wording. Because they're quotes, they really can't be changed. Unfortunately, the Mueser and Granberg "unambiguous statement" still permits host behaviors which lead to a different conclusion (i.e. it has a little bit of ambiguity in it). To make this statement unambiguous the "extra assumption" is specified outside the quote. I'll try a version to make this intent more clear. -- Rick Block (talk) 23:01, 15 October 2006 (UTC)
The Mueser-Granberg statement now lacks a question. It does not state what we are to find. Was that in the original quote? Also, the statement, "The problem generally meant to be considered also includes the assumption that if the player initially picked the car, the specific door the host opens reveals no more information to the player," is potentially confusing. What information could be conveyed it the player has already chosen the right car? And it's not clear what "more" refers to or what kind of information is being ruled out. I'm not sure what to say, but maybe something like "The problem generally meant to be considered also includes the assumption that the specific door the host opens reveals no special information about whether the player's initial choice is correct." 82.45.161.53 01:11, 16 October 2006 (UTC)
The paper is here (per the references in the articles), and it does not include the question. The point of the addition is that if the host has a specific algorithm for opening a "goat" door, and this algorithm is known to the player, the player can in some cases figure out whether the initial selection was correct (for example, as in the thread above, if the host is known to open the lowest numbered unselected goat door). The suggested revision sounds fine to me. -- Rick Block (talk) 02:39, 16 October 2006 (UTC)

A thoroughly honest game-show host has placed a car behind one of three doors. There is a goat behind each of the other doors. You have no prior knowledge that allows you to distinguish among the doors. "First you point toward a door," he says. "Then I'll open one of the other doors to reveal a goat. After I've shown you the goat, you make your final choice whether to stick with your initial choice of doors, or to switch to the remaining door. You win whatever is behind the door." ... Maybe I am dim but it seems to me that this statement of the problem (as currently in the article) does not need a further qualifyer because it is unambiguous as stated. Note these facts:

1) host always is "thoroughly honest"
2) host always places car behind door so he knows which door it is behind
3) host always opens a door with a goat "... I'll open ... to reveal a goat ... etc"
4) host always offers a second chance ""... you make your final choice ..."

Under these circumstances, it is surely not neccesary to indicate that the host opens a door at random if the contestant has picked the car. The method by which the host selects a goat is immaterial, the significant things is that it is known that he always picks a goat. Abtract 08:49, 16 October 2006 (UTC)

Given the above and that the host is known to always open the lowest numbered remaining goat door, in the problem as stated (player picks door 1, host opens door 3), the player loses by switching with 100% certainty. Over time, always switching even with this additional assumption will win 2/3 of the time (the host will open door 2 2/3 of the time with this additional assumption), but adding this knowledge about the host's actions destroys the Bayesian analysis, and most of the other explanations of the solution. To keep each iteration of the game with the same probability (with switching winning 2/3 of the time) it is necessary to add the further constraint that the host picks randomly if there are 2 remaining goat doors. This constraint is explicitly present in the Bayesian analysis. -- Rick Block (talk) 13:35, 16 October 2006 (UTC)
But nowhere does the statement above indicate "the host is known to open the lowest numbered remaining goat door". If it did, that wouldn't be an ambiguity it would be a nonsense and we would indeed need the further qualifier to make his choice random but I repeat nowhere does it say the host opens the lower goat door it simply says " ... I'll open one of the other doors to reveal a goat ...". Under these circumstances, no additional qualification is needed.IMHOAbtract 15:16, 16 October 2006 (UTC)
Exactly. It doesn't say the host behaves this way, but it doesn't say he doesn't behave this way. Hence, it's ambiguous. I agree "host opens lowest numbered goat door" is a not a reasonable interpretation, but neither is "host opens a random door" (sometimes revealing the car) or "host only offers the chance to switch if the player has initially picked the car" (in which case switching always loses). An unambiguous statement of the problem (as I think most people mean the problem to be interpreted) must prohibit all host behaviors that make the winning choice obvious. Perhaps it might be worth finding a different unambiguous statement of the problem to quote (and reference). Vos Savant has a book (that I haven't seen) with an extended discussion of this problem. Perhaps she includes a fully unambiguous description. -- Rick Block (talk) 17:47, 16 October 2006 (UTC)
In response to Abtract: As argued in an above thread, it depends on the question. Assume the host always opens an unchosen goat door. If we ask, "What is the probability that the remaining door conceals a car, given only that he has opened a goat door?", then no further randomness is needed. If we ask, "What is the conditional probability that door 2 conceals a car, given that he has opened door 3?," then we need randomness. 82.45.161.53 06:34, 17 October 2006 (UTC)

Yeah this is confusing but, well, it half makes sense

I'm somewhat confused.

I think I understand the point - you have a 2/3 chance of choosing a goat, the present MUST choose a goat and so he is limited to choosing only 2 of the 3 boxes. IF you have chosen the car he can abitrarily pick either of the boxes, but IF you have chosen one of the goats, he is forced to choose the other.

Since there is a higher chance you have picked a goat, and he HAS to pick a goat, there is a higher chance that you have both chosen a goat, than there is that only he has. As a result more often than not you must switch.

Is that about right?

It is very confusing though. 50/50 was my initial understanding, but like many have said, when you know the rules the odds are no longer 50/50. Even so my brain is still struggling to accept it (even though the other part of my brain says "look dude, these guys know much more than you, just accept you're wrong!!!") ny156uk 20:04, 25 October 2006 (UTC)

Yes, that's about right. Since he always has to pick a goat the chances are really only related to your initial pick. You have a 2/3 chance of picking a goat (in which case switching wins the car) and a 1/3 chance picking the car (in which case switching gets you a goat). It sometimes help to actually do this with playing cards (ace of spaces and two more cards) where you play the role of the host. Deal two to yourself, one to the "player". Look at the two, and pitch one card that is not the ace. You can see whether switching wins for the player by what card you have left. After you do this a few times, you'll probably realize you don't even have to bother picking a card to pitch. "Switching" wins if the ace is in your hand (2/3 chance), loses if it isn't (1/3 chance). -- Rick Block (talk) 21:53, 25 October 2006 (UTC)
This phrase should be removed: "You begin by pointing to door number 1. The host shows you that door number 3 has a goat." Under those circumstances (knowing for sure that door #3 has a goat) leads to the correct conclusion of a 50/50 chance with doors 1 and 2. The host's choice of door needs to be "free" for the switch to give a 2/3 chance of winning. Ajapted 11:59, 31 October 2006 (UTC)
This phrase is part of a quote, and surprisingly, even after the door is opened showing you the goat, you have a 2/3 of winning by switching. I'm not sure what you mean by "free". Can you explain? -- Rick Block (talk) 14:36, 31 October 2006 (UTC)
Actually, the host's pick cannot be "free" but must be constrained. He must pick a door that he knows has a goat behind it. He has to know or there is no real game as otherwise, he could open the door with the car (assuming you did not pick it) and then he will have ruined the show. No, he has to know and has to pick a goat. Nothing free about it. --Justanother 15:04, 31 October 2006 (UTC)

Wow . . . It's true!!

I have to admit when I first saw this I thought it was a crock. And after thinking about it; I still thought it was a crock. I thought the question was confusing the odds of picking the right door with the odds of the car being behind a specific door (that is not supposed to make sense, it was a confusion and, by definition, confusions don't make sense). Then the light went off (or on). The 100-door example did it for me. The host knows where the car is. Out of 100 doors, he picked 2. And he only picked yours because it was the one you picked. Had you picked another then that would have been the second door. Switch, for God's sake, switch. I should say that this is a probabilistic answer and assumes always that you picked the wrong door and the host is showing you what might be a better choice. Many years ago I hung out with serious gamblers and I discovered that they do not believe in probability as governing whether they win or lose at gambling, at least most of the ones I met. They believe in things like luck and intuition (gut feeling). So while a mathematician would switch, a gambler would likely not switch and would probably consider switching "bad mojo". And I found this cause I have another editor's talk page on my watch list but I will remove it as our conversation there is finished. But my thanks to that editor for helping me discover something cool --Justanother 21:15, 25 October 2006 (UTC)

"Simplest explanation" deleted from intro

I just deleted the following from the intro:

Let’s compare the two possible strategies:
First strategy: you select a door and keep whatever is in that door disregarding whatever the host does afterwards.
Second strategy: you select a door, the host opens a door showing a goat, and you accept the change and select the remaining door.
When you selct your door the probability to pick the car is 1/3 and a goat is 2/3. If you follow strategy #1, you win the car when you first select a car (p=1/3), but if you use strategy #2 you win the car when you first select a goat (p=2/3). Hence, you always should use strategy #2.

This basically repeats what's in the previous paragraph using an "address the reader" style (which should be avoided, see WP:MOS#Avoid second-person pronouns), and isn't strictly correct. The correctness issue is that staying with the first choice does not end up with a 1/3 chance regardless of what the host does. If the host opens a random remaining door (exposing the car, presumably ending the game 1/3 of time), the chances after this are 50/50. -- Rick Block (talk) 03:18, 27 October 2006 (UTC)

Well. Sorry Rick, the asumptions are that the host knows what is behind the doors and allways opens a door (allways showing a goat). Am I wrong? —The preceding unsigned comment was added by 88.16.192.169 (talk • contribs) .

Yes, these are the assumptions but the issue is these assumptions do not appear in this simplified explanation. They do appear in the paragraph that preceded this explanation, so could be construed to be understood, However, the previous paragraph also includes this very same argument. -- Rick Block (talk) 13:31, 27 October 2006 (UTC)

Images

Are there any objections to my deleting Image:Monty.gif and Image:Monty2.gif? They're unused and appear to be redundant to the images on this page. Thanks. Chick Bowen 04:58, 31 October 2006 (UTC)

Done now, anyway. Chick Bowen 21:15, 31 October 2006 (UTC)

Answer: It works...

For anyone with doubts, like me at first, here is why: i) you have a 2/3 chance of picking a goat first ii) if you pick a goat at first, the host must choose the other goat, so that both goats have been chosen iii) i) + ii) mean there is a 2/3 chance that both goats have been chosen. iv) if both goats have been chosen, the remaining door has the car, so 2/3 of the time, the remaining door will have the car. xD

i think othre people have said this already...but i didnt understand them, so i wrote my own :P HeroInTraining 06:53, 2 November 2006 (UTC)


Simulation Listed On Article Is Incorrect

I just read over the article, and was looking at the card simulation to see if it was correctly run. Here is the order of events in the card simulation:

  1. Take 3 cards from a deck (one ace, and 2 red twos)
  2. Distribute one card at random (face down) to the 'player'
  3. You as the 'simulator' (Monty) looks at the two remaining cards, and discards a red two (you will have at least one)
  4. If the card remaining in the simulator's (Monty's) hand is an ace, record that as a round where the player would have won by switching
  5. If the card remaining in the simulator's (Monty's) hand is a red two, record that as a round where the player would have won by staying

Here is the tricky part, and the flaw in this experiment. The experiment does not take into account that the player has a 2nd opportunity to make a choice (the switching/staying part). If, in the card game, choosing is represented by dealing one card at random, then one step is missing. After step 3, the simulator (Monty) should take the card that was given to the player, and the card that is left in Monty's hand, shuffle them, and distribute one randomly back to the player (to simulate the player's 2nd choice). This is why there is a 50/50 chance of getting the ace.

It is easy to see that yes, there is a 2/3 chance of an ace being among the two cards that the player does not choose. But there is a completely new experiment (scenario) as soon as the player is given a second choice. That experiment has different initial conditions (in that one card is discarded, to simulate that a person won't intentionally choose the goat). It is the matter of choice that restructures the game each and every time a choice is given.

In simple terms, the player, before making a choice, absorbs all information from the environment. In the case of Monty Hall, the environment (initial condition), is 3 unopened doors. So the first choice a player makes will have a 1/3 chance of getting a car. But the fact is that no matter what the player chooses, the game must go on to part 2 (Monty reveals a door with a goat). Once priority is passed back to the player, that player must make a second conscious decision. That player absorbs the information from the environment (2nd initial condition), and sees that there are now only 2 unopened doors. Now that player knows there is a 50/50 chance that a car will be behind one of the two doors. This is because of two things. First, the player's initial selection of one out of three doors has no affect on the new decision he or she has to make (one out of two doors). The second thing (related to the first), is that the door revealed by Monty is removed from the situation, and when the player is given priority to choose again, the environment must be looked at (new initial conditions).

Also, if you will notice, it does not matter in any way what Monty knows or thinks. All that matters is what he does, which is that he reveals a door (not picked by the player) that has a goat behind it.

In conclusion, the original Monty Hall solution of 2/3 cannot be proven if the experiment is itself based on the flawed Monty Hall solution. This is shown in the card experiment, which contains the same exact flaw that the Monty Hall paradox has. This is why every experiment done to prove Monty Hall must take into account that a player has 2 choices, and that he or she bases those 2 choices on 2 separate initial conditions (two separate environments). In the card experiment, by not properly representing a player's 2nd choice, you end up proving that 2/3 of the time the ace will be in the two cards the player did not choose. That is absolutely correct. But that is a completely separate experiment than when the player has a second choice to make, between two cards. This is why in Monty Hall, the player still has an equal chance to get the car once the environment has changed (by in effect removing one door), and the player is given priority to make a second choice. AFpilot157 13:35, 2 November 2006


This part of your analysis:
"after step 3, the simulator (Monty) should take the card that was given to the player, and the card that is left in Monty's hand, shuffle them, and distribute one randomnly back to the player (to simulate the player's 2nd choice)."
This is precisely what does not happen in the puzzle as stated and explains the change from a 2/3 chance of winning by switching in the original game to a 1/2 chance of winning in the new game (your version). There being no counterpart to the random shuffling and second distribution in the original game. That is, after Monty has opened a door he does not change the places of the remaining goat and car in some shuffling manoeuvre.Davkal 19:58, 2 November 2006 (UTC)

I understand what you mean, but how can the card simulation use random distribution to simulate choice 1, but not random distribution to simulate choice 2? Yes its true that 2/3 of the time the ace will be in the pile that represents 'switching', but it no longer becomes 2/3 once the 'switching' pile is reduced from two cards to one (the one revealed card is discarded). You cannot redefine the attributes of the 'switching' pile while maintaining the original probability, because changing the attributes ends up changing the probability.

For example, let us say that I pick door number 1. Monty opens door 3 and shows me the goat. He now asks if I want to switch. There are now two doors (the third could be blown up for all that matters, including the goat). Behind one of two doors is a car. I go ahead and flip a coin. If it comes up heads I choose door number 1. If it comes up tails I choose door number 2. There is a 50/50 chance for heads and tails, therefore a 50/50 chance for door number 1 and door number 2, and in conclusion a 50/50 chance for what those doors directly represent (car or goat).

By the way, it would not matter whether Monty randomly shuffled the remaining car and goat, because to you as the player, you don't know what is behind the doors. If Monty shuffles the goat and car, there is a 50/50 chance they will go back behind the same doors they originally were behind, and you are left again with the same choice, door number 1 or door number 2. Is it totally true that revealing door number 3 gives you no information about door number 1? Not directly, but indirectly it does. Once you reveal a goat behind door number 3, the chances of a goat being behind the door you picked decreases to 50/50.

For all that it matters, Monty might as well not even give players a first choice. He might as well show them 3 doors, reveal one as a goat, then ask them to pick one of the remaining two doors. Is that not a 50/50 chance? The trick is that the player's first choice is meaningless, and has no real bearing on the second scenario (environment) involving 2 doors, 1 goat, and 1 car. AFpilot157 15:35, 2 November 2006

Random dealing can simulate choice one because there is supposed to be no way for the person to tell the difference between the doors so this is simply shorthand for "pick a door, any door" - this step could be left out of the simulation and an actual choice made to represent the first choice. After this though the choice is whether to stick to the original card/door or switch to the one Monty leaves and this is not random anymore inasmuch as the originally picked card/door has been picked and can't change, and the other is determined by Monty's knowledge of what is behind the cards/doors. That is, what constitute the stick and the switch cards/doors are completely determined at the outset (except where the first choice is the car and Monty can randomly choose which goat to get rid of). The simplest way, I have found, to think about it is to note that the second choice really is between your orignal choice and the other two - Monty opening a door (given that he knows what is behind them and always shows a goat) is simply an empty move always made possible by the fact that there will be at least one goat behind the two non-chosen doors.Davkal 22:04, 2 November 2006 (UTC)
Sure, you can eliminate the first choosing. To do that in the card experiment, get two boxes. Shuffle and deal one card into box A, and two cards into box B. Next, remove one card from box B but it can't be the desired card. Okay, now choose a box. Would you rather have box A or B? Which is more likely to have the desired card? JethroElfman 22:57, 2 November 2006 (UTC)
I didn't mean eliminate the first choice, I meant eliminate the random deal part that was representing the first choice in the simulator and actually choose the card - just like the door is chosen in the game. That's what seemed to be causing the confusion here. So, for example, take the three cards and pretend your the guest and pick one, then put your Monty hat on and remove one of the other two...Davkal 23:02, 2 November 2006 (UTC)
"Here is the tricky part, and the flaw in this experiment. The experiment does not take into account that the player has a 2nd opportunity to make a choice (the switching/staying part). If, in the card game, choosing is represented by dealing one card at random, then one step is missing. After step 3, the simulator (Monty) should take the card that was given to the player, and the card that is left in Monty's hand, shuffle them, and distribute one randomly back to the player (to simulate the player's 2nd choice). This is why there is a 50/50 chance of getting the ace." I understand what you're saying here, AFpilot157, but the problem you're talking about is not the Monty Hall problem. In the Monty Hall problem, the player has his choice of strategies, and the question is, does any strategy markedly outperform the other strategies? (And the answer is "Yes" -- the strategy of switching will win 2/3rds of the time.) You are instead talking about the arrangement of the cards/doors being randomized, and the player's "choice" of strategies also being randomized -- which does not happen in the original problem.
Now you might be saying that the player has no more information about one card than the other, so choosing a card for the player simulates the player's lack of knowledge about the cards. However, it is not the case that the player has no more information about one card than the other -- that is in fact circularly assuming exactly the premise which you are trying to prove. The player, if he knows the theory of the game, knows that the door he picked only has a 1/3 chance of containing the car, and he knows that 2/3 of the time, the other door is the remaining door from a set consisting of one goat and one car, from which the goat has been removed. Those two doors are not equal. -- Antaeus Feldspar 01:46, 3 November 2006 (UTC)