Talk:Monty Hall problem/Archive 4

message: 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)
replies: text: 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)
replies: text: 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)
replies: text: 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)
replies: text: 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)
replies: text: OK, get it now. Thanks a lot! Ajcounter 13:33, 26 August 2006 (UTC)
replies: text: 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)
thread_title: Could this also explain the solution?

message: 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)
replies: text: 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)
replies: text: 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)
replies: text: 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)
thread_title:  More "catchy"?... sexy etc. ? 

message: 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)
replies: text: 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)
replies: text: 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)
replies: text: 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)
replies: text: And in only eight months since its Mainpage-FA exposure!  hydnjo talk 14:31, 9 March 2006 (UTC)
replies: text: 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)
replies: text: 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)
replies: text: Not sure, but perhaps this is related to Second law of thermodynamics. -- Rick Block (talk) 04:55, 10 March 2006 (UTC)

text: 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)
replies: text: 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)
thread_title:  Game theory redux 

message: 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)
replies: text: 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)
replies: text: 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)
replies: text: "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)
replies: text:  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)
replies: text: 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)

text: 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)

text: 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)
replies: text: 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)
replies: text: 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)

text: 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)
replies: text: 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)
thread_title:  Rewrite issues: reasonable vs. unreasonable interpretations 

message: 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)
replies: text: 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)
replies: text: 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)

text: And DavidNS1128 please tell us why you think tat this a paradox.  --hydnjo talk 04:32, 27 March 2006 (UTC)
replies: text: 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)
thread_title:  page name, problem or paradox 

message: 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:
# The most common objection to the solution is the idea that, for
# various reasons, the past can be ignored when assessing the probability.
# Although ignoring the past works fine for some games, like coin
# flipping, it doesn't work for all games. The most notable counterexample
# is card counting in some card games, which allows players to use
# 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)
replies: text: 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)

text: 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)

text: 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)

text: *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)
thread_title: The question is flawed and must be answered at a meta-level !

message: ... 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)
thread_title: [[The Neverending Story]]...

message: 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)
thread_title:  Simplified explaination 

message: 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?  --Doradus 19:28, 12 April 2006 (UTC)
replies: text: 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)
replies: text: 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)
replies: text: Ok, maybe I don't understand.  [here] is the article version I was talking about.  Which of those diagrams is a Venn diagram?  --Doradus 17:35, 28 April 2006 (UTC)
thread_title:  Venn diagrams? 

message: 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)
replies: text: 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)
thread_title:  To those who troll this article and repeatedly revert to these primitive explanations 

message: 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:
#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.
#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)
replies: text: 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)
thread_title:  Prior Information on the hosts bias 

message: This article is blatantly biased against goats. --Bonalaw 09:08, 24 April 2006 (UTC)
replies: text: Well it certainly gets my goat.  hydnjo talk 21:04, 24 April 2006 (UTC)
replies: text: Now, now, guys. Stops making asses of yourselves. Icey 19:11, 29 May 2006 (UTC)
thread_title: POV

message: 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)
replies: text: 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)
thread_title:  Randomly deciding whether to switch 

message: 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:  and here is the code: . Icey 19:30, 29 May 2006 (UTC)
thread_title:  Thought I'd test this out 

message: 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)
thread_title: the unambiguous statement

message: 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)
thread_title: ================================================

message: 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)
replies: text: 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)
replies: text:    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)
replies: text: 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)
replies: text: 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)
replies: text: 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)

text: 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)

text: 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.  --Doradus 03:07, 27 June 2006 (UTC)

text: 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)
replies: text: 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)

text: ... 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)
replies: text: ...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)
thread_title:  Error in reasoning 

message: 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)
replies: text: 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)
replies: text: Actually, Capite?   --hydnjo talk 04:00, 27 June 2006 (UTC)
replies: text: See Capisce. —Keenan Pepper 04:13, 27 June 2006 (UTC)
replies: text: 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)

text: 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)
replies: text: 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)

text: 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)
thread_title:  I have never been more confused in MY ENTIRE LIFE 

message: 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)
replies: text: 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)
thread_title:  Python Simulator 

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

text: 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)

text: 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)
thread_title:  "A Super Quick Explanation" 

message: 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)
thread_title: Bad

message: 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)
replies: text: 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)
thread_title:  Original Assumption 

message: 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)
thread_title: Haha, this is why I'm not a math student

message: See page 54 and further at http://www.skeptics.com.au/journal/2003/1.pdf)
User:faragon 16:39, 31 July 2006 (UTC)
replies: text: 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).  --Doradus 15:10, 31 July 2006 (UTC)
thread_title: Isn't it a hoax?

message: 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)
replies: text: 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.  --Doradus 17:03, 11 September 2006 (UTC)
thread_title: A new way to wrap your mind around this (for all of the naysayers)

message: no switch
Door 1	Door 2	Door 3	Result	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	
car   	goat  	goat  	win   	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win
goat  	car   	goat  	lose  	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose
goat  	goat  	car   	lose  	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose
 switch
Door 1	Door 2	Door 3	Result	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	    	
car   	goat  	goat  	lose  	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose	lose
goat  	car   	goat  	win   	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win
goat  	goat  	car   	win   	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	win 	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)
thread_title:  Table 

message: 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)
replies: text: Agreed. Also, the removed text is wrong.  --hydnjo talk 03:24, 20 August 2006 (UTC)
replies: text: 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)
replies: text: 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)
replies: text: 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)
replies: text: 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)

text: 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)
replies: text: 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)
thread_title:  Aids to understanding 

message: 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)
replies: text: 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)

text: 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)
thread_title:  Long-run game 

message: 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)
replies: text: 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)
thread_title: "Monty Hell" problem??

message: 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)
replies: text: 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)
thread_title: My initial additions follow (these have since been removed, along with other editorial changes)

message: 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)
replies: text: 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)
thread_title:  An additional misconseption that the original problem could give 

message: 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)
replies: text: 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)

text: 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)

text: 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)
thread_title:  Where have I gone wrong? 

message: 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)
replies: text: On the other hand this is an encyclopedia not the puzzle page in the Dail Mail. Abtract 21:25, 16 September 2006 (UTC)
replies: text:  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)
thread_title:  Solution is in bold 

message: 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)
thread_title:  But What is The Relationship of this Problem with Deal or No Deal? 

message: 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)
thread_title: Possible host behaviors in unspecified problem (Table)

message: 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)
replies: text: 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)
replies: text: Between us we have made it much clearer.Abtract 20:03, 14 October 2006 (UTC)
thread_title: Introduction Rewrites

message: 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)
replies: text: 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)
replies: text: 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)
replies: text: 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)
replies: text: 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)
thread_title: Yeah this is confusing but, well, it half makes sense

message: 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)
thread_title:  Wow . . . It's true!! 

message: 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)
replies: text: 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)
thread_title:  "Simplest explanation" deleted from intro 

message: 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)
replies: text: Done now, anyway. Chick Bowen 21:15, 31 October 2006 (UTC)
thread_title:  Images 

message: 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)
thread_title: Answer: It works...