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(as opposed to the correctness of the mathematical answer)
It is not enough to describe why the mathematically derived solution is correct. To resolve the paradox to the satisfaction of all, one must also describe why the intuitive solution is wrong. I think this requires three steps.
1) an understanding that the "intuitive solution" is just a dismissive term for a first analysis that turned out to have a flaw
2) a description of the logical steps that were employed in coming to the intuitive solution
3) an analysis of that logic, to find its flaw(s)
(1) I not argue the point, but rather just hope that people agree with it.
(2) I think that the intuitive solution takes the following steps:
Here is the game that I believe is used intuitively (I’ll call it the Silly game):
Should he change it or not?
Clearly the answer to this is that the chance is 50% either way, and it does not matter whether the contestant changes his mind.
Now compare the Silly game to an original Monty Hall game at a point half way through, in which the contestant has guessed door B and Monty has opened door A.
It seems that the Silly game is exactly the same as the original Monty Hall game at this point. To a viewer who has just ‘tuned in’ and does not know what has previously happened, the games look identical. Therefore, logic would dictate that the answer to the original question is that it doesn't matter whether the contestant changes their mind; the probability is 50% either way.
(3) It turns out, of course because the two games are not completely isomorphic. There is a crucial piece of information missing from the Silly game that is needed to make it isomorphic with the Monty Hall game, as follows: Monty (who knows where the car is)
Monty has thus said something concrete about door C (“I didn’t choose it, perhaps because I couldn’t choose it”), but nothing about door B. This is the source of the asymmetry between doors B and C, and the reason that door C is more likely to not have a goat behind it. Happyharris 20:57, 25 July 2005 (UTC)
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)
kudos to those contributors to this article. i have been thinking about this for days (despite the fact that i "got it" ... after about 20 minutes or so). i find the "two sets" explanation the most clear, though i'm sure some people will love the bayes' theorem explanation. i also found the talk page very entertaining. some contributors are clearly manifesting what kahneman and tversky (writers on cognitive biases - the former won a nobel prize) refer to as "belief perseverance." geez, sit down with a friend and three cards for 10 minutes and you'd realize you're wrong. anyway... i thought you might find this page interesting:
http://econwpa.wustl.edu:8089/eps/exp/papers/9906/9906001.html
besides elaborating on many of the key assumptions (often unstated) underlying the MHP, the paper presents an interesting (and supposedly earlier) problem of the same form, the "three prisoners problem." here's the upshot...
There are three prisoners, you and Prisoners A and B. Two of you are to be executed, while one will be pardoned. The prison warden knows who will be executed and who will be pardoned (like monty must know where the goats and car are). According to policy, the warden is NOT allowed to tell any prisoner if he/she is to be pardoned. You point out to the warden that if he tells you if A or B will be executed, he will not be violating any rule. The warden says C is to be executed. What is the chance that you will be pardoned?
the answer presented, like that to the the MHP, is that your chances of being pardoned remain at 1/3, while the the chances of B being pardoned, GIVEN WHAT THE WARDEN HAS TOLD YOU, is 2/3.
I found looking at this problem reinforced how *$&%^@# hard it is to get one's head around the MHP, because despite having read about the latter for days, i still had to think about the three prisoners problem for several minutes to get my head around IT, despite knowing that it was essentially the same as the MHP. K-razy.
someone mentioned the idea of blocking edits to featured articles for some period of time. while apparently there is a policy against such things, i agree with the suggestion, at least where articles that are sources of dispute. the featuring of an article doubtless attracts many potential editors, some of who may simply not know enough about the subject to edit appropriately. for example, the first time i read the monty hall page (when it was featured), someone had altered the solution section to express the misguided (given certain assumptions, which too often are unstated) 50/50 approach. needless to say, i was confused by the fact that all other sections of the article contradicted the solution.
good job. Contributed by 24.89.202.141.
I copied this back in from the archive:
Hecatonchires originally changed 'probability' into 'game theory', and then did it again two times. I've changed it back and posted a notice on his/her Talk page, pointing him/her to this discussion. I've invited him/her to start a discussion here if s/he wants it changed to 'game theory' again. Phaunt 10:54, 10 August 2005 (UTC)
Ahem, don't we have an encyclopedia to write here. It seem that you are arguing about the distance between two points on a beach ... jeesh. Well, if it is really important for one of you to prevail here then fine, go at it. Every once in a while however, look around and notice what you are spending your valuable time doing. ;-( hydnjo talk 02:15, 11 August 2005 (UTC)
Back and forth, the edits revert again and again. Can we get some consensus on this? It's two words... I don't mind the current compromise of mentioning both terms in the same sentence, but would prefer only mentioning probability. But is there any way we can get people to stop changing it every week or so? Fieari 05:22, August 26, 2005 (UTC)
My own understanding is that within the context of game theory, "Monty Hall Problem" refers to some modification of the problem from the one currently given in the introduction to this article. Specifically, some range of behaviors is permitted to the host. In this case, each of the two players chooses a strategy from within their ranges of possible behaviors, and the task is to identify a Nash equilibrium. The Mueser and Granberg paper describes this approach to the question. -- Wmarkham 21:05, 5 September 2005 (UTC)
I just reverted the additions by User:62.99.223.20 under Increasing the number of doors, because they were copied verbatim from [1]. This page was linked to, but that doesn't change the copyvio. Also, it didn't really belong there (under aids to understanding) but rather under Variants -> n doors, where a shorter discussion of this variant already exists. I invite User:62.99.223.20 to expand on this if he feels it's too short (but without violating copyright, of course).
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)
The intro refers to "the assumptions explicitly stated below" that do not appear (to me) to exist in the article.
From the introduction to the article:
The description of the puzzle appears to be quite clear on the constrants on the host's behavior, and I believe that any mention of assumtions is unnecessary here. I happen to be someone who does quibble over the statement of the problem, and I find this one to be quite clear, so this is probably not problematic.
Unfortunately, there are references to "the assumptions" throughout the article. Further unfortunately, my own, possibly nonstandard, position is that the "Monty Hall problem" is really a related family of problems. The ones in Selvin's letter and the Parade article each pose slightly different problems than the one stated above. The stated result (or any clear result) can only be obtained in those cases if additional assumptions are made. In my opinion, understanding the nature of some of the confusion surrounding the Monty Hall problem is made easier if the existence of these (nontrivial, IMO) assumptions is made clearer.
Since the article is already quite good, and my editions could be construed as having an agenda, my hope is that one of the perennial maintainers of the article is willing to adjust it in order to reflect my concerns, in a manner that is true to what the article describes. My guess is that the only changes needed are to move the existing comments about assumptions from the "Anecdotes" section to the introduction of the Parade article, and to eliminate the reference to assumptions from the intro. In fact, after consideration, I think it is safe for me to make the latter change myself. My observation is that these words currently serve no purpose one way or another, and hopefully I am a good representative of the view that the existence of assumptions is important. I will do this shortly.
-- Wmarkham 19:44, 5 September 2005 (UTC)
When you're given the choice to switch or not, it still doesn't matter, because switching or not switching is a binary decision. To give the probability of 1/3 to the scenario where you decide not to switch doesn't make any sense; when the host gives you the option, the scenario and therefore the rules change. You're no longer choosing between three doors, you're choosing between two.
It'd be different though, if you were not given the choice to switch. When one door is revealed to not be winning, you couldn't say "now my chances are 50/50" though from an intuitive standpoint you could say that, because without any ability to act on the events that occur, the probability really doesn't change from the standpoint of the start (though if we were weather forecasters we would say the chances were now 50%, but this isn't a forecast that continually updates, gamblers want to know what their overall chances are from the start, because that's where they're stuck making their decisions)
The thing here is when someone decides to base chance on when one assesses the situation or base the chance on when a choice was made. Gamblers will want to base it on when they made their choice, but weather forecasters will want to assess the situation continuously.
Jesus Christ, why are people so thick. It's like this..
I think this reasoning is too simplistic, and would give the wrong answer in some cases. For instance, would this reasoning apply to #paragraph_about_Who_Wants_to_be_a_Millionaire? I think it would lead to the wrong answer. (The right answer is that your odds are 50-50 in that case.) -- Doradus 21:31, 18 December 2005 (UTC)
I wonder if somewhere in the article it should be pointed out that under the actual rules for the “Lets Make a Deal” game show that this problem seems to be named after, switching doors didn’t actually increase your chances of winning. On the game show Monty would only offer the chance to switch ½ of the time if the player initially picked incorrectly, but would always offer the choice if the player initially picked correctly. This throws off the normal analysis in which the choice is always offered, since simply being offered the choice to switch increases the chances that your initial door pick was correct. —The preceding unsigned comment was added by 128.227.7.193 ( talk • contribs) .
I deleted the following paragraph about the Who Wants to be a Millionaire show:
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)
Agreed. No amount of meditation before the removal of two choices will affect the probability of the final outcomes. You need to tell Monty your pick and have that affect his actions. -- Doradus 21:29, 18 December 2005 (UTC)
Note that, (In the English version at least), the player can tell the presenter which of the 4 answers they think it is before the 50:50 takes one away, but it is still possible that the one they picked could be taken away. Leaving two, and a true 50:50 chance of picking at random the correct answer. -- JP Godfrey 21:10, 23 January 2006 (UTC)
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)
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!
Why?
Because the problem is mis-stated. When the contestant has chosen Goat 1 the quizmaster reveals Goat 2 - he doesn't have a choice. When the contestant has chosen Goat 2 the quizmaster reveals Goat 1 - he doesn't have a choice. When the contestant has chosen the car the quizmaster has to choose whether to reveal Goat 1 or Goat 2. These are independent possibilities and should not be selectively aggregated for the purposes of determining probabilities.
If you dispute this and believe that revealing Goat 1 or Goat 2 are aspects of the same event because the quizmaster only reveals one of them then you don't understand how probability works. The quizmaster also has to make a decision if the contestant picks the car and this decision must be included in the calculation of probabilities, as the decision tree correctly shows. The numbers allocated to the various decisions are meaningless, it is the counts that count.
Wherever you find a paradox, there's a fallacy lurking. (Hodgson's law - you read it here first) 80.47.80.51 01:45, 18 January 2006 (UTC) Graham Hodgson, 17 January 2006
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, I may be thick, but I certainly don't get it. After the door has been opened, I am left with two doors. The story so far has been very entertaining, but in fact it has given me no information which will indicate if my first choice was right or not.
That means it is now 1:2. How we got to this situation is irrelevant, unless the process of getting there gives me information which is relevant, but I don't see how it does.
The article correctly says that for some statistical calculations the past can be ignored, while for others it cannot. The major fault of the article is that it does not go on to say why in this case the past is relevant. The article cites card counting as an example where the past cannot be ignored - if I know that some cards have already gone (and I know WHICH) then I have information which affects the probability of the next card being an ace, so the past is relevant for future probability. But if we have a sequence of events which are separate events, then we have a
Markov chain, and previous events in the chain do not affect the next one: for example a series of coin tosses.
The point about Markov is that there is a difference in perspective before and after any event in the sequence. If the chances of tossing a coin and getting heads is 1:2, obviously the chances of doing it twice is 1:4, but if I toss a coin and get heads, the chances of now doing it a second time are 1:2, because the perspective after the first toss no longer takes the past probability into account. This applies to all sequences of probability, unless there is a CAUSAL link between the past event and the current probability (e.g. the last toss dented the coin so that it now falls differently). The Monty hall problem looks to me like a Markov chain. If I am wrong, then the article has to show that. It does not address this problem, and I seriously dout any of you can.
Think about this: I have three cards, and I pick two of them and put them on the table in front of you. I tell you to pick one, and if you pick the higher of the two cards, you win. We all agree that your chances are 1:2. The fact that I have three cards doesn't affect the choice I gave you, and your chances would be the same if I had four cards or only the two. The fact that there WAS another door has no bearing on the chances of getting the car NOW.
(BTW, the article says that "hundreds of maths professors" have attested that the proabability is 1:2. Is that not something you 1:3 proponents should be worried about - this smugness is incredibly arrogant! If the article is right, it needs to give serious maths authorities as sources, not internet sites.) --
Doric Loon
11:24, 24 January 2006 (UTC)
Oh sure, I stand by the "assume good faith" principle and wouldn't be here if I didn't really want to understand it. The point is, though, that what Markov proved is that the probability of a future action depends entirely on the present situation and not on how we got here. How we got here is only relevant if it alters the present situation; the probabilities involved in getting here are not in themselves relevant for the probability of the next event. Now, I understand that the show host has no choice. What you haven't explained to me is what I learn from that which makes my next choice (switch or no switch) into an informed choice rather than an arbitrary (i.e. 50:50) one.
I'm not a mathematician, so of course I know I could easily be stumbling in the dark. But I do wonder if you (and the article) are not confusing two different things. Remember Markov and the coin: The chances of tossing a coin heads up twice are 1:4, but after I have tossed it heads up once, the chances of doing it a second time are 1:2. Now is it not possible that here too there are two different phases with different probabilities:
In other words, in terms of game theory, if we do it many times, I can optimise my chances by having a switch policy, but in the particular case, when I stand before two doors, it is 50:50. As with Markov's coin tossing, this seems intuitively wrong, but makes sense mathematically. I think. If that is true, then it explains why there are two strongly held views. They are answering different questions. In that case, though, the top of the article needs to rephrase the problem. -- Doric Loon 16:11, 24 January 2006 (UTC)
First of all, congratulations, that's the clearest presentation of your argument I have heard, and better than what is in the article. In particular, what you say about viewing both parts as one event is helpful. I am now comfortable with the 1:3 solution, provided we are talking about the probability of the whole. In my last comment I accepted that for the sake of argument, but now I accept it without difficulty. Standing at the beginning of the game I am cool about saying, my chances are improved by switching when the time comes.
But can you see my problem about coming into the thing half way through? The article begins by asking about the probability of picking the right door out of two AFTER a random choice has been made. Taking all three doors into account means we are including past events in the calculation (or past phases of the event, if you prefer). But that is selective. Perhaps, unbeknown to me, there were five doors, with three cars and two goats, and two car-doors were eliminated which I never heard about. That would reverse the probability. Taking the past into account is therefore dangerous. This is just instinct, but I sense your idea of viewing both parts as one event is only legitimate when you stand back and look at the whole thing, not when you are standing in the middle with one part done and the next part to be thought about.
But I WILL take your advice about asking a maths prof. --
Doric Loon
18:48, 24 January 2006 (UTC)
OK, I've got it. In maths I'm a plodder, but even plodders get there. Thanks both. --
Doric Loon
07:11, 25 January 2006 (UTC)
Just wanted to note something:
Umm.. let's say:
The chances are now 50-50. As it logically would be. Your chances of winning the car at all has actually increased your chance of winning from 1/3 to 1/2. Now you have 2 choices and one of them contains a car.
202.152.170.254 10:12, 13 February 2006 (UTC) Hartono Zhuang
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)
"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)
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)Tim Robinson.
I'm not sure I would switch. The Monty Hall problem states that the host allows you to switch only AFTER you have picked a door already (It was not part of the initial rules). So if the host knows that you picked a door with a goat, he COULD directly open that one. This would give you a much reduced probability of getting the car, and you could only get the car if you didn't switch.
Somebody please help - I think I understand the correct answer, but my mind is still struggling to get around the following: Imagine contestant A chooses door 1. Monty hall then opens door 3 to reveal a goat. At this point you introduce contestant B. Contestant B has no prior knowledge of the game. He is told he has been "allocated" door 1, he does not know why door 3 is open. He is effectively in the same position as contestant A, but he does not know that the game is fixed. This time both contestant A and B are offered the choice to switch or stick. Surely the percentage chance for contestant B is 50/50. If so how can the same two doors have different probabilities of a prize at the same time for two people standing in front of them? If contestant B does not have 50/50 why not?
No, B's chances of correctly choosing the winning door, when fully utilizing the information available to him, is 50/50 because he cannot differentiate one door from the other—that is, he doesn't know which of the two doors you picked. What is Monty's chance of picking the correct door, when fully utilizing the available to him? It's certain, of course. So you cannot make the case that the odds can't change, because they are distinct between differing amounts of knowledge. Of course this is the case. A's chances are 2/3 because he is fully utilizing the information available to him. In his case, the extra bit of information is that he knows that of the two doors you didn't pick, Monty will never open the winning door for you, thus his actions are constrained by the fact that he knows something and his actions tell A partly what that information is. The one thing that A doesn't know that Monty knows, is anything more about the door he initially chose than he ever did. The odds of that particular door being correct are 1/3, which they have always been, and will continue to be.
I wrote the first specific web page on the MHP ten years ago, it has been referenced widely, and I have corresponded with countless people about this. No one way of presenting this problem is always effective at realizing comprehension. But one approach has had good success, and that's the million door version of the problem. If you pick a door from among a million, obviously your odds of it being the winning door are a million-to-one. However, if Monty opens all remaining 999,999 doors excepting the winning door, and presents you with two doors, you choice and the remaining door, is it better to switch or to stay? And if someone else walks along at that moment, not knowing anything about which door you picked and which doors Monty opened, their best odds for correctly picking the right door are 50-50. But the other guy is nearly assured of choosing the correct door—it's the one in almost a million doors that Monty very specifically, knowingly, did not open. There's a very small possibility that he didn't avoid any particular door because your door was the correct door. But the chances of you having correctly chosen the winning door from a million is 999,999 to one. It's extremely unlikely. (kmellis@kmellis.com) 69.254.138.180 06:27, 7 February 2006 (UTC)
(unwrapping back to left:) You state the following:
B, not having picked a door, nor knowing which door Monty opened sees only two doors that he cannot differentiate from each other in any way. There can be no "switching" or "staying" in a problem statement for this B fellow, …
But the scenario we were discussing states the following:
Contestant B has no prior knowledge of the game. He is told he has been "allocated" door 1, he does not know why door 3 is open. He is effectively in the same position as contestant A, but he does not know that the game is fixed.
Hence, there is indeed the concept of switching, due to the "allocation" of a door. If B picks to switch or stay randomly, he has 50-50 chances. If you tell B to switch, he has a two thirds chance. This is why I say the probability (of winning via switching versus via staying) does not change based on what you know. If you want to say that B doesn't even know what door has been allocated, you may, but that's a whole different problem. – Wisq 04:26, 10 February 2006 (UTC)
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
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)
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.
Oki, I agree, what I was meaning was here is another way to explain it.
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)
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)
The lesson to learn from the Monty Hall problem isn't one of mathematics, but of psychology -- that your instincts might be incorrect. With that in mind, the explanations given don't have to be complete. They have these two goals: 1) to be correct, 2) to be easy to understand. I tried wading into the Bayes theorem article and just couldn't make my way, so I don't think it meets goal #2 as a means of explanation.
I find it difficult to see in what way T.Z.K. disagrees with the article. I didn't like the chart that showed probabilities adding up to 200%, so I changed it. If T.Z.K. would like an article that actually is directed to mathematics, then I suggest expected number. I editted the grammar there, but the phrasing of the math is still weak. If he wants to revise wording then he can give that a shot too.
I like the extension he proposed to the problem. If one goat is blue and the other red, and you know that Monty always picks the red goat when he has a choice, then if he opens a door with a blue goat you must switch for it is 100% that your original pick was the red goat; if he shows a red goat it is now 50-50 whether or not to switch.
Such extensions are better off left out of the article lest they add to confusion.
Lastly, I like Antanaeus. He is remarkably patient with those who come here for help. JethroElfman 18:35, 22 February 2006 (UTC)
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.
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)
All right...
There's one point that isn't addressed by the above, and that is the question of whether x and y could take on values such that the optimal strategy when looking at Goat 1 is different from the strategy that is optimal when looking at Goat 2. If this was the case, then one could devise an overall strategy superior to any strategy that didn't take the identity of the goat into account.
It can be shown that the answer is "no". We will start by assuming the contrary: that x and y have values which make staying the best strategy for Goat 1 and switching the best strategy when looking at Goat 2; our overall strategy then would be "stay when looking at Goat 1; switch when looking at Goat 2."
In order for staying to be the best strategy, the following would have to be true: out of all of the times that the player ends up looking at Goat 1, more of them are due to the player having initially picked the car, and Monty making the choice to show Goat 1, than are due to the player having picked Goat 2. In terms of x and y, this works out to "x (the number of times that the player picked the car, and Monty picked Goat 1 to show) is greater than x+y (the number of times that Monty had to show Goat 1 because the player picked Goat 2)" -- in other words, x > x+y, which simplifies to 0 > y -- y is less than 0.
However, 0 is the lowest value possible for y; you could say "1 of every 5 times, Monty picks Goat 2" or "0 of every 5 times, Monty picks Goat 2" but "-1 of every 5 times, Monty picks Goat 2" has no meaning. The closest we can get to the situation we were trying to arrange is where y equals 0. This is the situation when Monty always picks Goat 1 if given a choice. Under these conditions, seeing Goat 2 means switching always wins; Monty must be switching because he has no choice. However, the player only sees Goat 2 1/3rd of the time; the other 2/3rds of the time, the player sees Goat 1, and the chances are exactly even that: a) Monty had to show Goat 1 because the player picked Goat 2, b) Monty had a choice of Goat 1 or Goat 2 (because the player picked the car) and showed Goat 1. There is no optimal strategy when the player is looking at Goat 1; no matter what strategy the player adopts, the chances of winning from this situation are 1 in 2. And as with every other set of values for x and y, the end result is 2/3: ((1/3)*(1/1)) + ((2/3)*(1/2)) = 1/3 + 1/3 = 2/3.
(Now, I ask others -- is there any point to addressing any part of this in the article itself?) -- Antaeus Feldspar 23:34, 1 March 2006 (UTC)
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)
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)
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):
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)
This is an archive of past discussions. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page. |
Archive 1 | Archive 2 | Archive 3 | Archive 4 | Archive 5 | → | Archive 10 |
(as opposed to the correctness of the mathematical answer)
It is not enough to describe why the mathematically derived solution is correct. To resolve the paradox to the satisfaction of all, one must also describe why the intuitive solution is wrong. I think this requires three steps.
1) an understanding that the "intuitive solution" is just a dismissive term for a first analysis that turned out to have a flaw
2) a description of the logical steps that were employed in coming to the intuitive solution
3) an analysis of that logic, to find its flaw(s)
(1) I not argue the point, but rather just hope that people agree with it.
(2) I think that the intuitive solution takes the following steps:
Here is the game that I believe is used intuitively (I’ll call it the Silly game):
Should he change it or not?
Clearly the answer to this is that the chance is 50% either way, and it does not matter whether the contestant changes his mind.
Now compare the Silly game to an original Monty Hall game at a point half way through, in which the contestant has guessed door B and Monty has opened door A.
It seems that the Silly game is exactly the same as the original Monty Hall game at this point. To a viewer who has just ‘tuned in’ and does not know what has previously happened, the games look identical. Therefore, logic would dictate that the answer to the original question is that it doesn't matter whether the contestant changes their mind; the probability is 50% either way.
(3) It turns out, of course because the two games are not completely isomorphic. There is a crucial piece of information missing from the Silly game that is needed to make it isomorphic with the Monty Hall game, as follows: Monty (who knows where the car is)
Monty has thus said something concrete about door C (“I didn’t choose it, perhaps because I couldn’t choose it”), but nothing about door B. This is the source of the asymmetry between doors B and C, and the reason that door C is more likely to not have a goat behind it. Happyharris 20:57, 25 July 2005 (UTC)
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)
kudos to those contributors to this article. i have been thinking about this for days (despite the fact that i "got it" ... after about 20 minutes or so). i find the "two sets" explanation the most clear, though i'm sure some people will love the bayes' theorem explanation. i also found the talk page very entertaining. some contributors are clearly manifesting what kahneman and tversky (writers on cognitive biases - the former won a nobel prize) refer to as "belief perseverance." geez, sit down with a friend and three cards for 10 minutes and you'd realize you're wrong. anyway... i thought you might find this page interesting:
http://econwpa.wustl.edu:8089/eps/exp/papers/9906/9906001.html
besides elaborating on many of the key assumptions (often unstated) underlying the MHP, the paper presents an interesting (and supposedly earlier) problem of the same form, the "three prisoners problem." here's the upshot...
There are three prisoners, you and Prisoners A and B. Two of you are to be executed, while one will be pardoned. The prison warden knows who will be executed and who will be pardoned (like monty must know where the goats and car are). According to policy, the warden is NOT allowed to tell any prisoner if he/she is to be pardoned. You point out to the warden that if he tells you if A or B will be executed, he will not be violating any rule. The warden says C is to be executed. What is the chance that you will be pardoned?
the answer presented, like that to the the MHP, is that your chances of being pardoned remain at 1/3, while the the chances of B being pardoned, GIVEN WHAT THE WARDEN HAS TOLD YOU, is 2/3.
I found looking at this problem reinforced how *$&%^@# hard it is to get one's head around the MHP, because despite having read about the latter for days, i still had to think about the three prisoners problem for several minutes to get my head around IT, despite knowing that it was essentially the same as the MHP. K-razy.
someone mentioned the idea of blocking edits to featured articles for some period of time. while apparently there is a policy against such things, i agree with the suggestion, at least where articles that are sources of dispute. the featuring of an article doubtless attracts many potential editors, some of who may simply not know enough about the subject to edit appropriately. for example, the first time i read the monty hall page (when it was featured), someone had altered the solution section to express the misguided (given certain assumptions, which too often are unstated) 50/50 approach. needless to say, i was confused by the fact that all other sections of the article contradicted the solution.
good job. Contributed by 24.89.202.141.
I copied this back in from the archive:
Hecatonchires originally changed 'probability' into 'game theory', and then did it again two times. I've changed it back and posted a notice on his/her Talk page, pointing him/her to this discussion. I've invited him/her to start a discussion here if s/he wants it changed to 'game theory' again. Phaunt 10:54, 10 August 2005 (UTC)
Ahem, don't we have an encyclopedia to write here. It seem that you are arguing about the distance between two points on a beach ... jeesh. Well, if it is really important for one of you to prevail here then fine, go at it. Every once in a while however, look around and notice what you are spending your valuable time doing. ;-( hydnjo talk 02:15, 11 August 2005 (UTC)
Back and forth, the edits revert again and again. Can we get some consensus on this? It's two words... I don't mind the current compromise of mentioning both terms in the same sentence, but would prefer only mentioning probability. But is there any way we can get people to stop changing it every week or so? Fieari 05:22, August 26, 2005 (UTC)
My own understanding is that within the context of game theory, "Monty Hall Problem" refers to some modification of the problem from the one currently given in the introduction to this article. Specifically, some range of behaviors is permitted to the host. In this case, each of the two players chooses a strategy from within their ranges of possible behaviors, and the task is to identify a Nash equilibrium. The Mueser and Granberg paper describes this approach to the question. -- Wmarkham 21:05, 5 September 2005 (UTC)
I just reverted the additions by User:62.99.223.20 under Increasing the number of doors, because they were copied verbatim from [1]. This page was linked to, but that doesn't change the copyvio. Also, it didn't really belong there (under aids to understanding) but rather under Variants -> n doors, where a shorter discussion of this variant already exists. I invite User:62.99.223.20 to expand on this if he feels it's too short (but without violating copyright, of course).
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)
The intro refers to "the assumptions explicitly stated below" that do not appear (to me) to exist in the article.
From the introduction to the article:
The description of the puzzle appears to be quite clear on the constrants on the host's behavior, and I believe that any mention of assumtions is unnecessary here. I happen to be someone who does quibble over the statement of the problem, and I find this one to be quite clear, so this is probably not problematic.
Unfortunately, there are references to "the assumptions" throughout the article. Further unfortunately, my own, possibly nonstandard, position is that the "Monty Hall problem" is really a related family of problems. The ones in Selvin's letter and the Parade article each pose slightly different problems than the one stated above. The stated result (or any clear result) can only be obtained in those cases if additional assumptions are made. In my opinion, understanding the nature of some of the confusion surrounding the Monty Hall problem is made easier if the existence of these (nontrivial, IMO) assumptions is made clearer.
Since the article is already quite good, and my editions could be construed as having an agenda, my hope is that one of the perennial maintainers of the article is willing to adjust it in order to reflect my concerns, in a manner that is true to what the article describes. My guess is that the only changes needed are to move the existing comments about assumptions from the "Anecdotes" section to the introduction of the Parade article, and to eliminate the reference to assumptions from the intro. In fact, after consideration, I think it is safe for me to make the latter change myself. My observation is that these words currently serve no purpose one way or another, and hopefully I am a good representative of the view that the existence of assumptions is important. I will do this shortly.
-- Wmarkham 19:44, 5 September 2005 (UTC)
When you're given the choice to switch or not, it still doesn't matter, because switching or not switching is a binary decision. To give the probability of 1/3 to the scenario where you decide not to switch doesn't make any sense; when the host gives you the option, the scenario and therefore the rules change. You're no longer choosing between three doors, you're choosing between two.
It'd be different though, if you were not given the choice to switch. When one door is revealed to not be winning, you couldn't say "now my chances are 50/50" though from an intuitive standpoint you could say that, because without any ability to act on the events that occur, the probability really doesn't change from the standpoint of the start (though if we were weather forecasters we would say the chances were now 50%, but this isn't a forecast that continually updates, gamblers want to know what their overall chances are from the start, because that's where they're stuck making their decisions)
The thing here is when someone decides to base chance on when one assesses the situation or base the chance on when a choice was made. Gamblers will want to base it on when they made their choice, but weather forecasters will want to assess the situation continuously.
Jesus Christ, why are people so thick. It's like this..
I think this reasoning is too simplistic, and would give the wrong answer in some cases. For instance, would this reasoning apply to #paragraph_about_Who_Wants_to_be_a_Millionaire? I think it would lead to the wrong answer. (The right answer is that your odds are 50-50 in that case.) -- Doradus 21:31, 18 December 2005 (UTC)
I wonder if somewhere in the article it should be pointed out that under the actual rules for the “Lets Make a Deal” game show that this problem seems to be named after, switching doors didn’t actually increase your chances of winning. On the game show Monty would only offer the chance to switch ½ of the time if the player initially picked incorrectly, but would always offer the choice if the player initially picked correctly. This throws off the normal analysis in which the choice is always offered, since simply being offered the choice to switch increases the chances that your initial door pick was correct. —The preceding unsigned comment was added by 128.227.7.193 ( talk • contribs) .
I deleted the following paragraph about the Who Wants to be a Millionaire show:
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)
Agreed. No amount of meditation before the removal of two choices will affect the probability of the final outcomes. You need to tell Monty your pick and have that affect his actions. -- Doradus 21:29, 18 December 2005 (UTC)
Note that, (In the English version at least), the player can tell the presenter which of the 4 answers they think it is before the 50:50 takes one away, but it is still possible that the one they picked could be taken away. Leaving two, and a true 50:50 chance of picking at random the correct answer. -- JP Godfrey 21:10, 23 January 2006 (UTC)
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)
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!
Why?
Because the problem is mis-stated. When the contestant has chosen Goat 1 the quizmaster reveals Goat 2 - he doesn't have a choice. When the contestant has chosen Goat 2 the quizmaster reveals Goat 1 - he doesn't have a choice. When the contestant has chosen the car the quizmaster has to choose whether to reveal Goat 1 or Goat 2. These are independent possibilities and should not be selectively aggregated for the purposes of determining probabilities.
If you dispute this and believe that revealing Goat 1 or Goat 2 are aspects of the same event because the quizmaster only reveals one of them then you don't understand how probability works. The quizmaster also has to make a decision if the contestant picks the car and this decision must be included in the calculation of probabilities, as the decision tree correctly shows. The numbers allocated to the various decisions are meaningless, it is the counts that count.
Wherever you find a paradox, there's a fallacy lurking. (Hodgson's law - you read it here first) 80.47.80.51 01:45, 18 January 2006 (UTC) Graham Hodgson, 17 January 2006
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, I may be thick, but I certainly don't get it. After the door has been opened, I am left with two doors. The story so far has been very entertaining, but in fact it has given me no information which will indicate if my first choice was right or not.
That means it is now 1:2. How we got to this situation is irrelevant, unless the process of getting there gives me information which is relevant, but I don't see how it does.
The article correctly says that for some statistical calculations the past can be ignored, while for others it cannot. The major fault of the article is that it does not go on to say why in this case the past is relevant. The article cites card counting as an example where the past cannot be ignored - if I know that some cards have already gone (and I know WHICH) then I have information which affects the probability of the next card being an ace, so the past is relevant for future probability. But if we have a sequence of events which are separate events, then we have a
Markov chain, and previous events in the chain do not affect the next one: for example a series of coin tosses.
The point about Markov is that there is a difference in perspective before and after any event in the sequence. If the chances of tossing a coin and getting heads is 1:2, obviously the chances of doing it twice is 1:4, but if I toss a coin and get heads, the chances of now doing it a second time are 1:2, because the perspective after the first toss no longer takes the past probability into account. This applies to all sequences of probability, unless there is a CAUSAL link between the past event and the current probability (e.g. the last toss dented the coin so that it now falls differently). The Monty hall problem looks to me like a Markov chain. If I am wrong, then the article has to show that. It does not address this problem, and I seriously dout any of you can.
Think about this: I have three cards, and I pick two of them and put them on the table in front of you. I tell you to pick one, and if you pick the higher of the two cards, you win. We all agree that your chances are 1:2. The fact that I have three cards doesn't affect the choice I gave you, and your chances would be the same if I had four cards or only the two. The fact that there WAS another door has no bearing on the chances of getting the car NOW.
(BTW, the article says that "hundreds of maths professors" have attested that the proabability is 1:2. Is that not something you 1:3 proponents should be worried about - this smugness is incredibly arrogant! If the article is right, it needs to give serious maths authorities as sources, not internet sites.) --
Doric Loon
11:24, 24 January 2006 (UTC)
Oh sure, I stand by the "assume good faith" principle and wouldn't be here if I didn't really want to understand it. The point is, though, that what Markov proved is that the probability of a future action depends entirely on the present situation and not on how we got here. How we got here is only relevant if it alters the present situation; the probabilities involved in getting here are not in themselves relevant for the probability of the next event. Now, I understand that the show host has no choice. What you haven't explained to me is what I learn from that which makes my next choice (switch or no switch) into an informed choice rather than an arbitrary (i.e. 50:50) one.
I'm not a mathematician, so of course I know I could easily be stumbling in the dark. But I do wonder if you (and the article) are not confusing two different things. Remember Markov and the coin: The chances of tossing a coin heads up twice are 1:4, but after I have tossed it heads up once, the chances of doing it a second time are 1:2. Now is it not possible that here too there are two different phases with different probabilities:
In other words, in terms of game theory, if we do it many times, I can optimise my chances by having a switch policy, but in the particular case, when I stand before two doors, it is 50:50. As with Markov's coin tossing, this seems intuitively wrong, but makes sense mathematically. I think. If that is true, then it explains why there are two strongly held views. They are answering different questions. In that case, though, the top of the article needs to rephrase the problem. -- Doric Loon 16:11, 24 January 2006 (UTC)
First of all, congratulations, that's the clearest presentation of your argument I have heard, and better than what is in the article. In particular, what you say about viewing both parts as one event is helpful. I am now comfortable with the 1:3 solution, provided we are talking about the probability of the whole. In my last comment I accepted that for the sake of argument, but now I accept it without difficulty. Standing at the beginning of the game I am cool about saying, my chances are improved by switching when the time comes.
But can you see my problem about coming into the thing half way through? The article begins by asking about the probability of picking the right door out of two AFTER a random choice has been made. Taking all three doors into account means we are including past events in the calculation (or past phases of the event, if you prefer). But that is selective. Perhaps, unbeknown to me, there were five doors, with three cars and two goats, and two car-doors were eliminated which I never heard about. That would reverse the probability. Taking the past into account is therefore dangerous. This is just instinct, but I sense your idea of viewing both parts as one event is only legitimate when you stand back and look at the whole thing, not when you are standing in the middle with one part done and the next part to be thought about.
But I WILL take your advice about asking a maths prof. --
Doric Loon
18:48, 24 January 2006 (UTC)
OK, I've got it. In maths I'm a plodder, but even plodders get there. Thanks both. --
Doric Loon
07:11, 25 January 2006 (UTC)
Just wanted to note something:
Umm.. let's say:
The chances are now 50-50. As it logically would be. Your chances of winning the car at all has actually increased your chance of winning from 1/3 to 1/2. Now you have 2 choices and one of them contains a car.
202.152.170.254 10:12, 13 February 2006 (UTC) Hartono Zhuang
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)
"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)
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)Tim Robinson.
I'm not sure I would switch. The Monty Hall problem states that the host allows you to switch only AFTER you have picked a door already (It was not part of the initial rules). So if the host knows that you picked a door with a goat, he COULD directly open that one. This would give you a much reduced probability of getting the car, and you could only get the car if you didn't switch.
Somebody please help - I think I understand the correct answer, but my mind is still struggling to get around the following: Imagine contestant A chooses door 1. Monty hall then opens door 3 to reveal a goat. At this point you introduce contestant B. Contestant B has no prior knowledge of the game. He is told he has been "allocated" door 1, he does not know why door 3 is open. He is effectively in the same position as contestant A, but he does not know that the game is fixed. This time both contestant A and B are offered the choice to switch or stick. Surely the percentage chance for contestant B is 50/50. If so how can the same two doors have different probabilities of a prize at the same time for two people standing in front of them? If contestant B does not have 50/50 why not?
No, B's chances of correctly choosing the winning door, when fully utilizing the information available to him, is 50/50 because he cannot differentiate one door from the other—that is, he doesn't know which of the two doors you picked. What is Monty's chance of picking the correct door, when fully utilizing the available to him? It's certain, of course. So you cannot make the case that the odds can't change, because they are distinct between differing amounts of knowledge. Of course this is the case. A's chances are 2/3 because he is fully utilizing the information available to him. In his case, the extra bit of information is that he knows that of the two doors you didn't pick, Monty will never open the winning door for you, thus his actions are constrained by the fact that he knows something and his actions tell A partly what that information is. The one thing that A doesn't know that Monty knows, is anything more about the door he initially chose than he ever did. The odds of that particular door being correct are 1/3, which they have always been, and will continue to be.
I wrote the first specific web page on the MHP ten years ago, it has been referenced widely, and I have corresponded with countless people about this. No one way of presenting this problem is always effective at realizing comprehension. But one approach has had good success, and that's the million door version of the problem. If you pick a door from among a million, obviously your odds of it being the winning door are a million-to-one. However, if Monty opens all remaining 999,999 doors excepting the winning door, and presents you with two doors, you choice and the remaining door, is it better to switch or to stay? And if someone else walks along at that moment, not knowing anything about which door you picked and which doors Monty opened, their best odds for correctly picking the right door are 50-50. But the other guy is nearly assured of choosing the correct door—it's the one in almost a million doors that Monty very specifically, knowingly, did not open. There's a very small possibility that he didn't avoid any particular door because your door was the correct door. But the chances of you having correctly chosen the winning door from a million is 999,999 to one. It's extremely unlikely. (kmellis@kmellis.com) 69.254.138.180 06:27, 7 February 2006 (UTC)
(unwrapping back to left:) You state the following:
B, not having picked a door, nor knowing which door Monty opened sees only two doors that he cannot differentiate from each other in any way. There can be no "switching" or "staying" in a problem statement for this B fellow, …
But the scenario we were discussing states the following:
Contestant B has no prior knowledge of the game. He is told he has been "allocated" door 1, he does not know why door 3 is open. He is effectively in the same position as contestant A, but he does not know that the game is fixed.
Hence, there is indeed the concept of switching, due to the "allocation" of a door. If B picks to switch or stay randomly, he has 50-50 chances. If you tell B to switch, he has a two thirds chance. This is why I say the probability (of winning via switching versus via staying) does not change based on what you know. If you want to say that B doesn't even know what door has been allocated, you may, but that's a whole different problem. – Wisq 04:26, 10 February 2006 (UTC)
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
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)
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.
Oki, I agree, what I was meaning was here is another way to explain it.
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)
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)
The lesson to learn from the Monty Hall problem isn't one of mathematics, but of psychology -- that your instincts might be incorrect. With that in mind, the explanations given don't have to be complete. They have these two goals: 1) to be correct, 2) to be easy to understand. I tried wading into the Bayes theorem article and just couldn't make my way, so I don't think it meets goal #2 as a means of explanation.
I find it difficult to see in what way T.Z.K. disagrees with the article. I didn't like the chart that showed probabilities adding up to 200%, so I changed it. If T.Z.K. would like an article that actually is directed to mathematics, then I suggest expected number. I editted the grammar there, but the phrasing of the math is still weak. If he wants to revise wording then he can give that a shot too.
I like the extension he proposed to the problem. If one goat is blue and the other red, and you know that Monty always picks the red goat when he has a choice, then if he opens a door with a blue goat you must switch for it is 100% that your original pick was the red goat; if he shows a red goat it is now 50-50 whether or not to switch.
Such extensions are better off left out of the article lest they add to confusion.
Lastly, I like Antanaeus. He is remarkably patient with those who come here for help. JethroElfman 18:35, 22 February 2006 (UTC)
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.
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)
All right...
There's one point that isn't addressed by the above, and that is the question of whether x and y could take on values such that the optimal strategy when looking at Goat 1 is different from the strategy that is optimal when looking at Goat 2. If this was the case, then one could devise an overall strategy superior to any strategy that didn't take the identity of the goat into account.
It can be shown that the answer is "no". We will start by assuming the contrary: that x and y have values which make staying the best strategy for Goat 1 and switching the best strategy when looking at Goat 2; our overall strategy then would be "stay when looking at Goat 1; switch when looking at Goat 2."
In order for staying to be the best strategy, the following would have to be true: out of all of the times that the player ends up looking at Goat 1, more of them are due to the player having initially picked the car, and Monty making the choice to show Goat 1, than are due to the player having picked Goat 2. In terms of x and y, this works out to "x (the number of times that the player picked the car, and Monty picked Goat 1 to show) is greater than x+y (the number of times that Monty had to show Goat 1 because the player picked Goat 2)" -- in other words, x > x+y, which simplifies to 0 > y -- y is less than 0.
However, 0 is the lowest value possible for y; you could say "1 of every 5 times, Monty picks Goat 2" or "0 of every 5 times, Monty picks Goat 2" but "-1 of every 5 times, Monty picks Goat 2" has no meaning. The closest we can get to the situation we were trying to arrange is where y equals 0. This is the situation when Monty always picks Goat 1 if given a choice. Under these conditions, seeing Goat 2 means switching always wins; Monty must be switching because he has no choice. However, the player only sees Goat 2 1/3rd of the time; the other 2/3rds of the time, the player sees Goat 1, and the chances are exactly even that: a) Monty had to show Goat 1 because the player picked Goat 2, b) Monty had a choice of Goat 1 or Goat 2 (because the player picked the car) and showed Goat 1. There is no optimal strategy when the player is looking at Goat 1; no matter what strategy the player adopts, the chances of winning from this situation are 1 in 2. And as with every other set of values for x and y, the end result is 2/3: ((1/3)*(1/1)) + ((2/3)*(1/2)) = 1/3 + 1/3 = 2/3.
(Now, I ask others -- is there any point to addressing any part of this in the article itself?) -- Antaeus Feldspar 23:34, 1 March 2006 (UTC)
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)
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)
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):
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)