Talk:Monty Hall problem/Archive 3

message: *This question and response have been moved to down to its chronological place.  hydnjo talk 16:00, 22 November 2005 (UTC)
thread_title: Actual rules for the gameshow

message: For the people who still find it hard to see why the probability is not 50/50 when there are two doors left, I hope the following illustration may help.  I'm going to show that the Monty Hall problem is a specific case of a more general game; I'll call all the games that differ in their parameters "Hall games" for ease of use.
The basic idea behind a Hall game is this:  We start with one large set of secret-hiding items -- they can be doors to be opened, or cards to be turned over, it doesn't actually matter.  What matters is that there are n cards, but only one of them is the Prize card.  The rest are Null cards.
Step One: The cards are divided into two hands.  Each hand must have at least one card.  For simplicity's sake, we call the number of cards in the first and second hands h1 and h2.
Step Two: Someone who can see which cards are Nulls can discard some number of Nulls from one hand, the other, or both.  We call the number of cards remaining in each hand after the discarding of Nulls r1 and r2 (like h1 and h2, they must be at least one.)
Step Three: The player makes a guess at which of the two hands contains the Prize.
Step Four: The player makes a guess at which card out of the hand he selected is the Prize.
It's clear that to win the game, the player has to make correct guesses in both Step Three and Step Four.  What are the chances of picking the correct hand?  The first hand is correct h1/n of the time; the second hand is correct h2/n of the time.  If we want, we could enumerate the cases:  if h1=2 and h2=5, then the Prize could be the first card of the first hand, the second card of the first hand, the first card of the second hand ... et cetera, et cetera.
Now this is the part that many people find counter-intuitive.  If we enumerate the cases, and then we reduce Nulls to meet any legal value of r1 and r2, we find that in no case can the removal of Nulls switch the Prize from one hand to the other.  This means that the chance of the card being in the first hand or the second always stays at h1/n and h2/n -- even if the sizes of the hands do not stay at h1 and h2.  If it's dealt to that hand, it stays in that hand; therefore the chance of it being in a particular hand is always equal to the chance that it was dealt to that hand.
What about the removal of Nulls?  Does it affect anything after all?  Yes, it does -- it affects the player's chances in Step Four.  The chance that the Prize is in a particular hand is dependent upon h1 and h2 -- how many cards each hand started with.  But the chance of finding the Prize in a hand (assuming it's the right hand) is based on how many cards that hand contains after Nulls have been removed -- since there's only one Prize, the chance of finding it in a hand of r1 cards is 1/r1.
Now, what are the chances of making both guesses correctly?  If no Nulls get removed from either hand, then the chances of picking the Prize are either h1/n x 1/h1 (since r1 is equal to h1 when no Nulls have been removed) or by similar logic h2/n x 1/h2, which also multiplies out to 1/n.  If, however, one of the hands -- say, hand 1 -- has been reduced down to one card (r1 = 1), then the chances of finding the Prize in that hand is h1/n x 1/1 -- if you've correctly guessed that the Prize is in that hand, you have a 100% chance of finding it in that hand when it's the only card in the hand.
With this being the general structure of the Hall game, we can see that the Monty Hall problem is really just the case where n=3, h1=1 and h2=2, r1=1 and r2=1.  The chance that the Prize is in the player's hand is h1/n -- 1/3.  The chance that it's in Monty's hand is 2/3.  Before one Null is removed from Monty's hand, the chance of finding the Prize in his hand is 2/3 x 1/2 -- i.e., 1/3.  But when the Null is removed, the chance is now 2/3 x 1/1 -- i.e., 2/3! -- Antaeus Feldspar 03:44, 26 July 2005 (UTC)
thread_title:  Breaking it down into steps 

message: It's a bit late, but I want to congratulate and thank everyone who worked on this and helped it become a featured article. Good job! Phaunt 11:53, 27 August 2005 (UTC)
thread_title:  Congratulations 

message: Jesus Christ, why are people so thick. It's like this..
#Choose a door; you're lucky it's a goat! but you had a 2/3 chance of choosing a goat so the odds were on your side.
#Monty reveals the other goat.
#You switch; it has to be the car as the other goat is gone.
#You win!
#If you had chosen not to switch you would have lost.
#By not switching you're stuck with that 2/3 chance of getting a goat. Switching meant that you turned it into a 2/3 chance of winning the car.
#Easy peasy. End of story. Nighty Night Jooler 02:09, 8 November 2005 (UTC)
replies: text: But if we let the proles know how simple this is, they might start thinking for themselves, and then where will we be? Bonalaw 14:31, 22 November 2005 (UTC)
replies: text: OK, how's this for a paraphrase:  we do away with the numbers and use the layman's term "chances are."  You pick a door.  Chances are, it's a goat.  Then, Monty opens a door that he knows is a goat door.  Now, assuming you picked a goat, and of course Monty showed you a goat, the only door left is the car. You'd be a fool not to switch.  Of course, it's much less likely that you'd pick the car at first, in which case you'd lose by switching. Dyfsunctional 17:44, 14 December 2005 (UTC)
thread_title: Easy peasy

message: *Note: this was posted at the top of this talk page. hydnjo talk 15:54, 22 November 2005 (UTC)
thread_title: Actual rules for the gameshow

message: *Note- i'm not the original writer and i may be wrong, but it seems to me like the probability is actually 50/50. it's 2/3 for a person to lose. BUT say you're player 2. if  player 1 is eliminated, he had the goat. which means that player 2 either got the car, or it's left. Which means there's an equal chance for him to win or lose by switching... right? -- 24.196.238.213
**This question relates to the variant where there are two players, one is eliminated, and the question is "should you switch".  Assuming you're not eliminated the answer is no. and the probability is 2/3 that you have the car.  There are indeed two outcomes left, i.e. you have the car or switching gets it, but they have unequal probabilities in this variant - just like the two outcomes in the original problem have unequal probabilities.   The key is the realization that N possible outcomes doesn't mean each one must have a 1/N chance.  To make this one more obvious, consider a similar game with 10 doors, 1 car, and 9 contestants.  If none of the 9 choose the car, 8 are eliminated randomly.  If any of the 9 choose the car the other 8 are eliminated.   This game ends up in the same situation, a player and a door to potentially switch to, with the same two possible outcomes with what I hope are clearly not even probabilities.  In the 3-door, 2-person case, I assume you agree the unchosen door has a 1/3 chance of having the car at the beginning of the game.  Eliminating one of the players doesn't change this, but since the car is either behind the unchosen door (still 1/3 chance) or one of the players has it, when there's only one player left the probability is (1 - 1/3) = 2/3. -- Rick Block (talk) 14:47, 23 November 2005 (UTC)
thread_title:  Comment moved from article 

message: I deleted the following paragraph about the Who Wants to be a Millionaire show:
:The game show "Who wants to be a millionaire" has the same problem. The player is given 4 possible answers to a question. You can ask the host to remove 2 wrong answers, leaving you with 2 answers, one right and one wrong. Assuming you have no idea which of the 4 is right you can guess one (Lets say A), remove two and be left with two . If A is not removed then there is a 1/4th chance that A is right and a 3/4th chance that the other one is right.
On the millionaire show, the contestant does not get to pick an answer and then have two wrong answers removed.  Without picking, two are removed and if you then pick randomly there's a 50/50 chance.  Even if you mentally (randomly) "pick" and your pick is one of the two remaining answers, the result is a 50/50 chance because your pick is not related to the process by which the other answers are removed.  -- Rick Block (talk) 16:28, 16 December 2005 (UTC)
thread_title:  paragraph about Who Wants to be a Millionaire 

message: Maybe I missed it ,but it seems noone has brought up the claim that this is a pseudo antiintuative paradox..
The Wierdness of the probablilty is ONLY a result when a convergance (to that specicifed probablilty) is reached over an infinte number of cases.
Per this particular singal case ,when you are in a REAL game switching the door doen't chagne AT ALL YOUR specific case chance of winning.
"there ar e3 kinds of lies in this world : lies ,damn lies and statistics"
The Procrastinator 14:14, 30 December 2005 (UTC)
replies: text: In a real game that followed the rules set down in the problem, yes, switching the door does change your chance of winning.  Picture the following situation:  instead of doors, you and Monty have cards, and instead of three cards, you have ten cards, one of which is the Ace.  Monty shuffles the cards, lets you pick one, and keeps the other nine in his hand.  What are the chances that the card is in Monty's hand?  Obviously, nine to one.  Now, Monty discards eight non-Ace cards from his hand.  Can the Ace possibly change from one hand to the other during this step?  Clearly not, so the chance that the Ace is in Monty's hand is still nine to one, even though the actual size of Monty's hand is now one. -- Antaeus Feldspar 15:34, 30 December 2005 (UTC)
thread_title:  this problem is easier than stated 

message: To determine the probability of an outcome you list all the possible outcomes and then count the number of times the outcome you are interested in turns up. Look at the decision tree under the Venn diagrams in the main entry. How many possible outcomes are there under each of the possible contestant choices? How many of these outcomes result in the contestant winning the car?
50% in all cases!
:So how do you explain the observed fact that the contestant tends to win twice as often when he switches?  --Doradus 16:40, 2 March 2006 (UTC)
replies: text: Sorry, nope.  They are not independent possibilities.  There is a one-in-three chance that the contestant initially picks the car.  The host can only choose between revealing Goat 1 or revealing Goat 2 when this one-in-three chance has already happened, and he must choose one of the two; therefore the total probability of these two chances must be exactly one-in-three.  No more, no less.  If we were to display the probabilities visually on a pie chart, we would see the total size of the user-picks-the-car slice stay exactly the same size as it was divided into the (not-significant) possibilities of "host reveals Goat 1" and "host reveals Goat 2".  Are you seriously suggesting that that slice of the pie must actually get bigger because it's being divided into a larger number of slices? -- Antaeus Feldspar 02:18, 18 January 2006 (UTC)
replies: text: The purpose of this page is to demonstrate that there's a sucker born every minute. The (number of ways to explain the problem logically and correctly) divided by the (number of dissenting opinions) will always  be less than one. All you do by trying yet one more logically correct perspective in hopes of persuading only one more true believer is to perpetuate the ranks of non-believers by more than one. Thus the numerator will grow more slowly than the denominator and the ratio will continue to be less than one, (believers ÷ nonbelievers < 1), always. ;-)  hydnjo talk 05:03, 18 January 2006 (UTC)
replies: text: Of course - the chances of making a wrong initial choice are twice as great as the chances of making a correct initial choice, and therefore the chances of improving on the initial choice are also twice as great. I withdraw covered in confusion. 213.78.64.39 12:03, 18 January 2006 (UTC)

text: This has nothing to do with true-believers and non-believers, it's not a philosophical or moral dilemma, it's a simple mathematical problem whose solution is counterintuitive. Some understand it, some don't. Tailpig 19:37, 18 January 2006 (UTC)
replies: text: I was using the terms believers and non-believers metaphorically for those that do understand and those that don't.  :-)  hydnjo talk 19:48, 25 January 2006 (UTC)
thread_title:  Let's revert to the fundamentals of probability 

message: This is my own plain-language explanation & debunk of "switching increases your odds." (It's essentially the Markov Chain). I hope Tailpig will read it and become a "believer." :)  [töff's analysis]
:Well, first off, let me say that I've never really liked that diagram; I know how the problem goes and what the 'trick' of it is and I find it incredibly difficult to see how that diagram relates to it.
:With that said, however, your analysis is fundamentally flawed.  Let me quote from your analysis:  "Let's say you choose Goat1. The host shows Goat2. At that point, you have two paths: switch(Y) or not(N) ... and you have equal 50-50 chances to take either path." (emphasis in original)  Well, that's the source of your confusion right there, because that is in no way the original problem.  The essence of the problem is that the player chooses his strategy, whether to switch or stay -- he does not have it randomly selected for him with 50-50 probability!  It's no wonder that your calculations show the player's chances as 50-50; if the player has one of two strategies randomly assigned to him, that will make his overall chances 50-50 no matter what the probability is for a given strategy.
:If you doubt this point, let me illustrate.  I will roll six ten-sided dice in a row.  If and only if all six of them come up "10" will I put the prize in Box A; otherwise I'll put the prize in Box B.  Elementary analysis should confirm that the strategy of picking Box B will pay off 999,999 times out of 1,000,000.  If you are allowed to choose your strategy, you can win 999,999 times out of 1,000,000; if, however, I then randomly pick a 'strategy' and therefore a box for you, with equal probability of either, you now win the prize only 1 in 2 times.
:Now that you know that the problem you've been dealing with isn't the actual Monty Hall problem, please let us know if you have any problem seeing why the answer of the actual Monty Hall problem is that switching gives you a 2/3 chance of winning. -- Antaeus Feldspar 00:08, 26 January 2006 (UTC)
thread_title:  Töff's analysis 

message: Coming back to Markov.  I suspect the reason most people have difficulty is because at school they were taught the Markov principle (usually not under that name) and this really does look like it at first sight.  Fooled me for long enough.  As Antaeus Feldspar says above, it is an optical illusion in this respect.  The article doesn't really help here.  At the top of the "Aids to understanding" section it points to this issue, saying that "The most common objection to the solution is the idea that, for various reasons, the past can be ignored when assessing the probability."  But I am not sure that what follows really helps most people see why it cannot be ignored: I certainly read it the first time with a sense of frustration that the key point was evading me.  For me the eurika effect came with Antaeus' pointing to the (in retrospect obvious) fact that the car can't move.  I've been mulling over how to explain the difference between Markov and Monty Hall.  I think the difference is that after I toss a coin, I don't toss it a second time from where it landed, but rather I pick it up first: the way it landed last time doesn't affect the next toss because I return it to a neutral position before the next event in the sequence.  The equivalent of returning to a neutral position would be if, after the first door has been opened, the game organisers were to remove the car and remaining goat and reallocate them by a random principle.  THEN the second phase would be unaffected by the first, and that would be Markov.      But they don't.  What everyone understands is that if we play the game three times, the first guess will be right once and wrong twice; what is so easy to miss is that that cannot change unless the car moves.  Now this all seems so obvious, but it is in fact the massive blind spot which makes the optical illusion trick people.  I wonder if it would be worth having a short paragraph on Markov in this article, and discuss the difference properly - and perhaps less chaotically than I can do it.  --Doric Loon 15:24, 31 January 2006 (UTC)
replies: text: I think you've got a good point, Doric.  I'm not even sure that really is the most common objection, either:  my experience with explaining it to people tends to be evenly split between those who think that probability automatically corresponds to the size of the sets, and those who think that Monty's removing a door has "reset" the probabilities and has generated a new random event, with probabilities independent from previous random events.  Though...  now that I typed that out, I'm wondering if they're really the same thing, after all. -- Antaeus Feldspar 15:37, 2 February 2006 (UTC)
thread_title: Markov again

message: "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 likelyhood, and 50% of being the option with 2/3 likelyhood, so they balance."
This has nothing to do with the Monty Hall problem.  The Monty Hall problem is about whether choosing a particular strategy can increase or lower your chances.  Talking about what would happen if a coin flip decided your strategy for you only confuses the issue and makes it harder for people to understand the real problem. -- Antaeus Feldspar 16:02, 2 February 2006 (UTC)
replies: text: Yes, perhaps the old diagram was just irritating me too much.  It showed the odds at 50/50 and was confusing.  Still, I think people consider choosing at random to be strategy in itself and extrapolate from that to the mistaken notion that both doors are equal.  The number of cries for help here on this talk page indicates to me that there's still edits to be done so the article makes its point better.   That's my idea with the random thing; to tell people that their gut instinct of it being 50/50 is correct, but only if they choose were choosing blindly.  If the page keeps any of the new diagrams I'll make better images to replace these quickies. -- JethroElfman 02:47, 3 February 2006 (UTC)
replies: text: I agree there's still room for improvement.  I think the card game experiment is maybe the best aid to understanding, since people can actually carry it out themselves and see why Monty opening a goat-door doesn't change any probabilities.  Would anyone object if I moved that to the top of the "Aids to understanding" section? -- Antaeus Feldspar 16:00, 3 February 2006 (UTC)
replies: text: Hey, I think that would be a good move.  In fact, I would eliminate the 3-card version (or make it the extrapolation) and begin stratight away with a 52 card pack where the objective is to find the ace of spades.  It is important to get this done early because the length of the article means people may not make it to the end. JethroElfman 16:58, 3 February 2006 (UTC)
replies: text: I would have to disagree about making the 52-card rather than the 3-card version the primary version; I've seen a lot of people respond to attempts to prove the 2/3 result through simulation with "well, if your simulation comes out with what's obviously the wrong result, it proves that you mis-programmed your simulation!"  I'd rather make the simulation correspond as precisely as possible to the actual problem so that there's less room for people to think that some significant factor differs between the door version and the car version. -- Antaeus Feldspar 18:00, 3 February 2006 (UTC)

text: Additionally, what about a "Common misperceptions" section, explaining common ways in which people either misunderstand the ground rules of the problem or misunderstand the effects of those ground rules on the probabilities? -- Antaeus Feldspar 16:02, 3 February 2006 (UTC)
thread_title:  Staying or switching at random 

message: Looking only at winning posibilites.
If you are going to swap you must pick a goat in the first place and your odds are 2/3 of doing that.
If you are not going to swap you must pick the car in the first place and your odds of doing that are 1/3.
--81.79.90.68 14:55, 4 February 2006 (UTC)
thread_title:  Simply. 

message: I edited this page some time ago to clearly state the problem and the answer. Since then it has been edited back into a state. The page is a mess because there is no single, clear problem and answer statement. It is long and rambling. There are several statements of the gameshow, alternative versions, multiple versions of the answer, anecdotes, and all manner of irrelevant nonsense, and the important information is simply lost. This is an example of an article where the lack of an authority and a team of expert researchers, as is found in a "real" encyclopaedia, leads to a polarisation of opinion and hence more and more verbose explanation in order to convince those who fail to understand the right answer to prevent them from editing incorrectly. 
By the way, the correct answer is yes, you should switch. If you don't believe it then set up the game yourself with an accomplice, try it 100 times and see what answer you get. Alternatively, use a computer simulation, or even do the probability theory from first principles (but do it properly). Any mathematical explanation that reaches a different answer has a flaw in it. Ignore your intuition. PK
:I agree, but aside from deleting most of the page and then locking it, how can you solve this?  Besides, although we know the correct answer, it can take a lot to convince someone else that their intuition is wrong — typically by explaining it to him or her in exactly the right terms.  By debating it on the talk page, explaining it different ways until the opposing party "gets it", and then putting that method on the page, we have organically come up with a system that (judging from the reduced debate on this page) convinces most people of the correctness of the answer.
:I would rather a long and overly verbose page, that expresses the truth in a way that almost anyone can understand, than a short and concise page that most people would look at and think "that's bogus" — and either edit it, or just walk away convinced that Wikipedia is full of lies. – Wisq 17:13, 10 February 2006 (UTC)
thread_title: This page is monstrous

message: On these types of problems, there are dual data sets at work. Set one is the information which allows us to determine the "odds" of finding the item being sought. Set two is the information which allows us to determine what the underlying statistical distribution of winning choices is. Originally, the "odds" of both data sets are the same, but they do deviate when additional information is acquired. The additional information is provided by the certainty that the removed choice is a loser. Because of this, the choosing party is no longer making a guess, but instead is making an informed calculation. Please look at the definition for guess "To predict (a result or an event) without sufficient information". Please take note that when one has sufficient informatiton, one is no longer guessing. The removal of one choice provides us with more information and we move from the position of a mere guess to that of an educated guess, which is not the same thing. Now as to the "statistical distribution" angle: If you have 10 shoe boxes on your desk and one of them contains and egg, the distribution percentange is 1/10 or 10%. Those numbers never change. However, when we gain more information about the contents - say by opening a few, our odds of finding the egg increase. People argue about these problems because of the tricky idea that there is true "guessing" involved when there is not. And also because they forget that the numbers regarding the original distrubution of winning choices is fixed. Only the odds of finding the item improve, not the statistical likelyhood that it actually existed. Merecat 05:48, 11 February 2006 (UTC)
replies: text: It seems that no matter how this is explained (including your logical choice of words) the argument will not end. There are some who will follow the theatrics rather than the logic no matter what. Give it a go in the article if you think it will help. hydnjo talk 01:11, 12 February 2006 (UTC)
thread_title:  Dual data sets being ignored 

message: If you can accept this following fact, it could be easier to see that switching will make the probability for getting the car 2/3.
Fact: If you choose a door with a goat behind, switching will gett you a car, if you choose a door with the car behind it, swithing door will gett you a goat.
Lets take it from the start. You choose a door, lets say door C. Its now a 1/3 chanse that you choose the door with the car behind it. The host now revals one of the door with the goat behind, lets say he revals door A with the goat behind. Its now ONE goat behind a door wich you cant see, and ONE car behind a door you cant see. You now have 2 doors witch you dont know whats behind, the only thing you know is that its one car and one goat left and they har hiding behind these two doors. (Just to make it perfectly clear, it COULD be that behind door C there is a goat, and therefore behind door B there is a car, it also COULD be that behind door C there is a car and therefore there is a goat behind door B, it CANNOT be that behind the two doors left there is one goat each, becouse the host has already revaled a door with a goat behind). Now comes the fact that will make it easier to understand: If you switch door, its now GUARANTEED that IF its a Goat behind Door C ( the door you first choose) you are swithing to the door with the car behind (door B), IF its a car behind door C then its GUARANTEED that you are switching to a goat (door B).
 
As you first choose door C it was 1/3 chanse of selecting the door with the goat behind, when you swicth its GUARANTEED that you are swithing to a door witch the content not being the same as the inital door you choose (read that sentence twice). And therfore if you switch its 2/3 chanse of getting a car.
:That seems like a pretty convoluted way of saying "picking a goat and switching will get you a car, and you have 2/3rds chance of picking a goat". :)  Really, that's the absolute simplest explanation. – Wisq 00:10, 14 February 2006 (UTC)
thread_title:  Another way to explain it 

message: This whole section has become an attack and counter-attack on each other's arguing techniques.  If you want to continue this, you can do it on your user talk pages.  --Doradus 17:21, 27 February 2006 (UTC)
thread_title:  (deleted) Disagreement with explanations 

message: Start with the usual. You pick ONE door and the host gets the remaining TWO doors. Stop. Now would you think it a good idea to swap your ONE door for the host's TWO doors?  If you say NO, then I have nothing more for you, go away. 
But if you say YES, I want to swap my ONE door for the host's TWO doors then you're almost there. That's it, except for the host's theatrics of opening a losing (and he knows it) door. What you're doing is swapping your ONE door for his TWO doors, even if he tried to mess with your head by opening one of them (oh, sure, like he's going to open a car door) and the diversions like the audience screaming "swap" - "don't swap" and the crew saying "cue the flashing lights" and the "host's silly grin". It's all about messing with your head remember. you're swapping your ONE door for his TWO doors.  
To dramatize the situation, start with ten doors. You get ONE and the host gets NINE (they would never do this as it would expose the whole thing). Now, the host (between commercials) opens EIGHT of his NINE doors (never ever opening a car door and he knows it). Wacha think now?  hydnjo talk 02:25, 18 February 2006 (UTC)
thread_title: One more time

message: The answer to the problem is NO. The logical fallacy of the reasoning which leads to conclude in the positive solution lies in the fact of considering the second goat when there is no more second goat. 
Think of this: two doors remain. One has a goat, the other one has a car. Chances are 50/50. It does not matter as to whether my door conceals the car or the goat. My switching or not is like a new decision, a new choice. In this new choice, I chose to keep my door (which is the same as to say that I re-pick that door), or to pick the other door. The open door does no longer count. The chance to switch creates a new choice with new alternatives, erasing the old ones.
:The answer really is YES.  Even though there are only two doors left, there are three ways of getting there:  by picking goat #1, by picking goat #2, or by picking the car.  Two out of those three ways result in switching being the right answer. -- Antaeus Feldspar 21:27, 24 February 2006 (UTC)
thread_title:  Number of doors vs. number of ways of getting there 

message: Recently an editor brought up an interesting point for discussion:  namely, the possibility that the answer to the problem might change depending on whether Monty picks a goat to show (when he has a choice) with equal probability for each goat, or with unequal probability.  I'd like to discuss why the answer is "no, the answer to the problem does not change."
To see why not, let's start with a discussion of how to calculate probabilities in a case where one probability determines what situation you are in, and the situation determines your chance of "winning" from that situation.  For instance, consider a game where you and an opponent each pick a card from separate decks; if your cards are both face cards (J, Q, K, A) or both non-face cards (2-10) you win; otherwise you lose.  The way to calculate the total probability is to multiply your chance of getting into each situation by the chance of winning in that situation, and add those products together.  So, in our example game, you have a 4 in 13 chance of picking a face card, which gives you a 4 in 13 chance that your opponent will pick a face card and you'll win, and similarly you have a 9 in 13 chance of getting a non-face card, which gives you a 9 in 13 chance of winning; the total probability of winning is thus ((4/13)*(4/13)) + ((9/13)*(9/13)) = 16/169 + 81/169 = 97/169.
So let us say that Monty picks a goat to show (when he has a choice) with possibly-unequal probability:  for every x times that he chooses to show Goat 1, he chooses to show Goat 2 y times.  Thus, when he has a choice of which goat to show, he shows Goat 1 x/(x+y) of the time, and Goat 2 y/(x+y) of the time.
How often does Monty show Goat 1 in total, then?  Well, Monty only has a choice about which goat to show when the player chooses the car initially.  By the conditions of the problem, the player's initial choice is equally likely to be the car, Goat 1 or Goat 2.  Therefore, if Monty has a choice of goats x+y times, then x+y times he must have to show Goat 1 because the player picked Goat 2.  Therefore, Monty shows Goat 1 2x+y times in all -- x times because he chooses to, x+y times because he has to.  By similar logic we can determine that Monty shows Goat 2 x+2y times.
The chance that the player sees Goat 1 is therefore (2x+y)/(3x+3y), and the chance that the player sees Goat 2 is likewise (x+2y)/(3x+3y).  We can now look at what the chances are of winning if we employ a switching strategy in these situations.  We already saw that Monty shows Goat 1 2x+y times in total.  x+y of those times, he had to show Goat 1 because the player had chosen Goat 2; these are cases where the player must switch to get the car.  The remaining x times, Monty had a choice of which goat to show; having a choice means that the player picked the car initially.  Therefore, when the player is looking at Goat 1, his chance of winning with a switching strategy is (x+y)/(2x+y).
Before we do the same for Goat 2, let's stay with Goat 1 a little further.  We already know we're going to multiply the probability of seeing Goat 1, (2x+y)/(3x+3y), by the probability of winning by switching when looking at Goat 1, (x+y)/(2x+y).  Since the numerator of the first number is the same as the denominator of the second, they cancel out, and the product of the two probabilities is (x+y)/(3x+3y) -- which we can see always works out to 1/3, no matter what values x and y have!  The same logic must apply to Goat 2, as well.  This accords with what we already know about the problem:  1/3 of the time, the player has chosen Goat 1, and must switch to win; 1/3 of the time, the player has chosen Goat 2, and must switch to win.  The remaining 1/3 of the time, the player chooses the car initially; no matter how Monty makes his choice about which goat to reveal, it does not change the fact that in this case, switching will lose and staying will win. 
(I want to address one more minor point on this matter, but I'm out of time at this computer; I'll be at another computer in about ninety minutes.) -- Antaeus Feldspar 21:27, 1 March 2006 (UTC)
replies: text: No!  hydnjo talk 23:39, 1 March 2006 (UTC)
thread_title:  Does it matter how Monty picks which goat to show? 

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: There is some discussion above about rewriting the article, from scratch, incorporating most of what is currently in the article but editing it heavily... (see above) but there are some strange things about the article right now that someone needs to look at. I propose the new article look something like (or at least incorporate these elements in some order, if someone can rework it to be slightly better):
* A brief statement of the problem (and why it might be considered significant) (FOUR SENTENCES MAX!)
* The origin of the problem, the history of the problem. But NOT variants of the problem. -- I would start with Marilyn. And probably finish with her.
* A statement of the problem with the express contraints.
* The solution to the problem, with subsections on the various ways of understanding the problem. Perhaps also, starting with the simplest and ending with the most complex. Like Bayes's Theorem? WTF? Is someone who does not understand the solution at this point seriously going to understand that?
* Similar problems with:
:* Further history of the problem (e.g. Gardner, &c.)
:* Links to similar problems (three card problem, boy girl problem, &c.)
In summary, I think this article could be 50% of the size its now and 200% of the quality. Are there some people who would be willing to work on this with me? Or others who think it is a bad idea and that the article is best in its current form? - Abscissa 06:27, 10 March 2006 (UTC)
replies: text: I have no argument against the need but I am somewhat troubled by the suggested process. The proper way of dealing with a seriously deteriorated FA is to list it at WP:FARC (see some of the candidates there) with the reasons that it no longer meets WP:WIAFA. Only after a consensus is achieved should the article be rewritten. After that, if we have the desire to do so, the article can be submitted for WP:PR and/or WP:FAC.  --hydnjo talk 12:34, 10 March 2006 (UTC)
replies: text: Sure, sounds good to me. - Abscissa 12:52, 10 March 2006 (UTC)
thread_title:  Article Rewrite