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Archive 1 | ← | Archive 4 | Archive 5 | Archive 6 | Archive 7 | Archive 8 | → | Archive 10 |
I've moved the existing talk page to Talk:Monty Hall problem/Archive2, so the edit history is now with the archive page. I've copied back a few recent threads. Older discussions are in Talk:Monty Hall problem/Archive1. Hope this helps, Wile E. Heresiarch 15:28, 28 July 2005 (UTC)
I'm all for racial harmony and all, but why do the images include a bald black man as the player? Could we be using a more abstract, symbolic face? (I know I know, I'm a racist because I wouldn't have noticed if it were a white guy. Granted, but I think the question is still valid.) -- Doradus 02:00, 3 May 2007 (UTC)
The two mentions of Deal or No Deal contradict one another: under 'Sequential Doors' is an explanation of the critical difference between the NBC Prime Time SMASH HIT and the Monty Hall problem as stated, but under the 'History of the problem' heading, Deal or No Deal is essentially considered to be a minor variant with the same sorts of conclusions. The numbers directly contradict one another as well. Though I'm fairly positive the first description is accurate, I'll refrain from editing for now so people can argue and yell and scream and then somebody smarter than myself can fix it. I also kinda like the fact that both are in this article, as it seems the Monty Hall problem's best contribution to society is watching people flailing about entirely confused yet certain that they're right in the face of contradicting evidence. 66.188.124.133 17:32, 16 May 2007 (UTC)
The problem with this is that the 3 scenarios are actually 4.
For the first scenario where the person picks the car it is listed as host showing "either Goat A or B". Actually these are two different scenarios:
Scenario 1: Contestant picks car, Host shows Goat A, Contestant switches, Contestant LOSES
Scenario 2: Contestant picks car, Host shows Goat B, Contestant switches, Contestant LOSES
Scenario 3: Contestant picks Goat A, Host must show Goat B, Contestant switches, Contestant WINS
Scenario 4: Contestant picks Goat B, Host must show Goat A, contestant switches, Contestant WINS
You can see that switching yields the expected 50% success.
It is rather alarming to me that this is missed by experts. I believe Quantum computing falls into this same "smoke and mirror" science. —The preceding unsigned comment was added by 143.182.124.2 ( talk • contribs).
Might I ask for more specificity as to what this article needs in order to become an FA? It's not clear how to act upon the suggestions left in the Mathematics rating box.-- Father Goose 02:31, 23 June 2007 (UTC)
From my talk page, but it makes more sense to comment here:
The current lead is a teaser, which is great for a magazine, but not for an encyclopedia. I'm sorry that my other concerns are vague, but it seems that other editors believe that this article meets FA standards, in which case please will someone replace my comment and signature by their own. Thank you! Geometry guy 10:56, 23 June 2007 (UTC)
The Monty Hall problem is a puzzle involving probability loosely based on the American game show Let's Make a Deal. The name comes from the show's host, Monty Hall. A widely known statement of the Monty Hall problem appeared in a letter to Marilyn vos Savant's Ask Marilyn column in Parade (vos Savant 1990):
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
Since there is no way for the player to know which of the two unopened doors is the winning door, many people assume that each door has an equal probability and conclude that switching does not matter. However, as long as the host knows what is behind each door, always opens a door revealing a goat, and always makes the offer to switch, opening a losing door does not change the probability of 1/3 that the car is behind the player's initially chosen door. As there is only one other unopened door, the probability that this door conceals the car must be 2/3.
The problem is also called the Monty Hall paradox; it is a veridical paradox in the sense that the solution is counterintuitive. For example, when Marilyn vos Savant offered the problem and the correct solution in her Ask Marilyn column in Parade, approximately 10,000 readers, including several hundred mathematics professors, wrote to tell her she was wrong. Some of the controversy was because the Parade statement of the problem fails to fully specify the host's behavior and is thus technically ambiguous. However, even when given completely unambiguous problem statements, explanations, simulations, and formal mathematical proofs, many people still meet the correct answer with disbelief.
I found another exchange in the FAR which resonates a little bit with my reaction to the article and may help clarify the vaguer part of my comment. I quote:
I wouldn't go as far as Abscissa, but it is not dissimilar from my reaction. Anyway, this exchange seems to have got lost in a sandwich between the arguments about inline citation that opened the review, and the impressive copyediting drive that ended it. I couldn't find it in the talk archive. Did anything come of it? Geometry guy 17:41, 23 June 2007 (UTC)
I am just wondering if it is a flaw in one of the Variations à Other host behaviors à “The host does not know what lies behind the doors, and the player loses if the host reveals the car.”
The answer is allegedly “The player loses when the car is revealed a third of the time. If the prize is still hidden, switching wins the car half of the time.” Shouldn’t switching here also result in a 2/3 probability of victory (since it is purportedly the same result (so far) as in the original Monty Hall problem)? -- 85.164.95.143 14:15, 9 July 2007 (UTC)
I believe as well there is a flaw here. Let's extend the problem to 100 doors the same way we reason the standard problem. The host luckily opens 98 goats in a row. But once this has happened, certainly you would switch, because even without the knowledge of the host, you only had a 1/100 chance of landing on the car in the first place, which has not changed. Your odds of WINNING this sort of game are of course lower because of the probability of the host opening the door to the car causing a loss. However, after he opens a goat, even if it is random, it does not affect the probability of you having picked the car in the first place, and thus the advantage of switching. Again, the 100 doors intuition here seems to support this argument. —Preceding unsigned comment added by 151.190.254.108 ( talk • contribs)
I'm thinking now I am mistaken because I read another type of explanation that I could not seem to refute, but your point #3 is still somewhat confusing because it seems to contradict #1, not show how #1 doesn't work.
Alright, I have been convinced. Perhaps my confusion can lead to something positive, like some clearer wording about why this is true. What I gather now is that when the host picks randomly, 1/2 of the time that a car is not revealed randomly, the reason is not that you got lucky picks by the host but that the car is behind the door you picked. I followed this by a tree diagram. The other explanation I found online was just a different wording of the one you used to refute my 100 door argument. Thanks!
Okay, I've spent entirely too much brainjuice on this. It probably has one or more glaring errors. But it's the algebra-ized version of the way I view the problem in my head (which has more to do with colors than with symbols; go figure). That said, I'm not so good at this whole color→algebra thing, so please fix it up:
You select a winning door | Your door is now a winner | Remaining Door(s) contain a Winner | You've already lost | |
Host picking randomly | ||||
Host picking only losing doors |
Jouster ( whisper) 23:48, 18 July 2007 (UTC)
You originally select a winning door | Your door is now a winner | All d remaining Door(s) contain a Winner | Each d remaining door contains a winner | You've already lost | |
Host picking randomly (absolute) | |||||
Host picking randomly (conditional) | |||||
Host picking only losing doors |
I'm not sure what the final word here is, but I think it's wrong. Anyway, the article is still in error, stating
Possible host behaviors in unspecified problem | |
---|---|
Host behavior | Result |
The host does not know what lies behind the doors, and the player loses if the host reveals the car. | The player loses when the car is revealed a third of the time. If the prize is still hidden, switching wins the car half of the time. |
I claim, as did whoever initiated this discussion, that
1. If the host does not know where the prize is, and
2. The host randomly opens one of the two doors not chosen,
then given the above, the stategy of not switching doors yeilds wins with probablity 1/3, exactly as it did when the host knew where the prize was!
This can be shown in a number of ways. The most convincing, using Bayes' theorem, appears last.
1. Ad absurdum: If you believe that there's a difference between the cases where the host knows where the prize is or guesses where the prize is (tosses a coin), then give some thought to the interim possibilities: The host knows that the prize had an uneven probability of being placed, say 1/4, 1/4, 1/2. Work out the probability of winning if you don't change your choice (host opens the door least likely to show the prize). Now consider what happens if the host thought he knew the probability ditribution of the prize placement, but was wrong - they used a different distribution! How can the contestant's strategy be affected by all this?!
2. If the host knows nothing, we don't need him. State the game as follows: Contestant picks a door (may as well be at random), and now opens one of the other two doors at random. Given that the opened door has no prize, what's the chance that the initial door does? I claim 1/3. If you still think that the answer is 1/2, consider the following variant:
Contestant opens one door at random. It has no prize. Now the contestant picks one of the remaining doors. Clearly the chance of winning is 1/2. Do you really believe that the two situations yeild the same probability of winning?
3. Bayes' theorem: I use the notation from the article: In Bayesian terms, probabilities are associated to propositions, and express a degree of belief in their truth, subject to whatever background information happens to be known. For this problem the background is the set of game rules, and the propositions of interest are:
For example, denotes the proposition the car is behind Door 1, and denotes the proposition the host opens Door 2 after the player has picked Door 1. Indicating the background information with , the assumptions are formally stated as follows.
First, the car can be behind any door, and all doors are a priori equally likely to hide the car. In this context a priori means before the game is played, or before seeing the goat. Hence, the prior probability of a proposition is:
Second, the host will pick a door from the remaining two at random. This rule determines the conditional probability of a proposition subject to where the car is, i.e. conditioned on a proposition . Specifically, it is:
if i = j, (the host cannot open the door picked by the player) | |||
if i ≠ j and j = k, (the host can open a door with a car behind it) | |||
if i ≠j and j ≠ k, (the host can open a door without a car behind it) |
The problem can now be solved by scoring each strategy with its associated posterior probability of winning, that is with its probability subject to the host's opening of one of the doors. Without loss of generality assume, by re-numbering the doors if necessary, that the player picks Door 1, and the host then opens Door 3, showing him or her a goat. In other words, the host makes proposition true.
The posterior probability of winning by not switching doors, subject to the game rules and , is then . Using Bayes' theorem this is expressed as:
By the assumptions stated above, the numerator of the right-hand side is:
The normalizing constant at the denominator is simply:
as can be seen in the table above. Dividing the numerator by the normalizing constant yields:
This can be stated differently:
since the host doesn't know where the prize is! Therefore
This shows that the probability that your initial choice was right remains 1/3 even after the host opens a door at random, given that the random door did not show the prize. You should switch doors, exactly as in the case where the host knew where the prize was. —Preceding unsigned comment added by Jasoncoop ( talk • contribs) 13:24, 25 October 2007 (UTC)
Wrong! of course and are each 1/3. These are the cases where your initial choice was wrong. The problem you're describing is not the right one! The question is what's the probablitity that you were initially right given that the third door is wrong. This posteriori information affects the probability in the way I described using Bayes' theorem. Can you find fault with the way I used it? Jasoncoop 14:47, 25 October 2007 (UTC)
Rick, thanks for your detailed and patient response. I think you must be a pretty good teacher. I substituted host opens door 3 and prize is not there for and came up with 1/2 as you promised.
I'm still a bit concerned about the question of what happens when the host has some knowlege about the way the prize door was chosen, but not certain information (some non uniform distribution). Maybe I'll work it out some time.
So in fact the two following situations are similar:
1. First pick a door at random and open it. Given that it's empty, each of the closed doors has probability 1/2 of containing the prize.
2. Pick a door at random, open one of the other two doors at random. Given that it's empty, each of the closed doors has probability 1/2 of containing the prize.
Somehow I find this a bit disturbing, but hopefully I'll get used to it...
Or maybe this is the key to understanding the problem - if neither the contestant nor the host knows where the prize is, it doesn't matter who goes first. Jasoncoop 23:03, 25 October 2007 (UTC)
I believe that some who are confused by this problem are focused on the difference between a Strategy and a Choice. The Host presents the player with a Choice, and if it were truly acted on as a Choice then the probability would not be 2/3 winning. Only when the player examines the rules and develops a Strategy which is adhered to without fail, does the probability turn to hir favor. Those who have watched The Price is Right remember the indecision on the faces of players as they ponder what to do. In my (admittedly basic) understanding of probablilty, everything hinges upon a priori decisions. Thus, I posit that if a Player goes into the game with a Strategy of sticking to their original choice, they'll have 1/3 chance of winning. The Strategy of always switching when offered the inevitable Choice yields 2/3 Chance of winning. This has been established. But I think that it is important to note that if the Player has no Strategy, and truly decides to randomly Choose between switching and staying, then there is a 1/2 chance of success. I believe I am correct in this assessment, and I think it would make a good additional to the article to explain this, since I believe many of those who are confused are thinking in these terms. i.e. Explain the difference in probability between a true Strategy (Choice decided in advance) vs. random Choice. Of course, if I'm completely wrong on this (which is why I have placed this in discussion), I still think it might clarify the article to point out that this is a Strategy, and that the player is not really making a Choice - their outcome is predetermined when Monty opens a door. P.S. I note that one of the external links explains the exact case I present of 1/2 odds with a random Choice. Shouldn't that be mentioned somewhere in the article for completeness? BrianWilloughby 18:29, 11 July 2007 (UTC)
I agree with "Debunked" and I think the confusion here is with not distinguishing "or" versus "exclusive or" in example 1 or in the Venn diagram. Also with the Venn diagram the probability of picking the box with the two boxes in it is 50%, exactly the same as the probability of picking the first box. In effect in drawing the two boxes circled in the Venn diagram you have changed the probability by making it 50% 1st box or 50% "2 box" box. This again confuses "exclusive or" with "or", eg the probability of the car being behind one door is 1/3, the probability of the car being behind one of the last two box doors is 1/2 therefore the probability of the car being in the venn diagram second half is 50%, not 2/3, since it can either be in the box containing the last two boxes or be in the first box. 67.86.165.55 06:24, 1 August 2007 (UTC)
It's wrong because Scenario 1 in the image set in reality consists of 2 scenarios. Now what if we consider it from different perspective: We know the host will always pick the goat. The host picks Goat A, so the player choice is really between Goat B & the Car. If the host picks Goat B, the player choice again is only Goat A & the Car. Switching does nothing to increase the chance of winning. Since the beginning of the game the chance of the player winning the car is 50%, not 33%. 203.49.196.163 00:33, 2 August 2007 (UTC)
Thanks, Father Goose. I can see the absurdity of my statement now. So: Player picks 1 door out of 3 --> 1/3 chance of winning. If the host picks Goat A/B and player stays, his chance of winning obviously is still the same (1/3), but the chance that the car is behind the door becomes 1/2? If player decides to switch, there are 4 scenarios:
Which means by switching, his chance of winning indeed increases and becomes 1/2? Not 2/3?! 203.49.196.163 05:01, 2 August 2007 (UTC)
₠Thank you Father Goose, you are right and I withdraw my "I agree with Debunked" comment above. The example that makes it easier for me to understand is if there are 100 doors, you pick a door and then the host is told to remove 98 doors. The door he leaves has a 99% chance of being the Car, since the only scenario where he is not picking the cars is where you picked the car in the first place and the probability of that is only 1%, so it is much more likely when he opens the 98 doors that he has left closed the car. That to me is clearer for some reason than when only three doors are utilized in the problem. —The preceding unsigned comment was added by 70.181.21.230 ( talk)
For a different perspective on the problem, consider not switching doors. In each of the three possible cases, there is at least one unchosen goat for the host to show, but only in the first of the three cases (1/3 of the time) does not switching win the car.
This means that 1/3 of the time, the car is the contestant's first choice, therefore 2/3 of the time the car is not the contestant's first choice, and switching to the other available choice will win the car.
I've deleted the new section with the giant image (above) pending discussion here. In my opinion, the giant image basically replicates the other giant image (with different visual representations of the doors, car, and goat), and the text essentially repeats the text already in the second paragraph in the Solution section, making this new section simply redundant with material elsewhere in the (already too long) article. -- Rick Block ( talk) 13:56, 24 August 2007 (UTC)
Player's initial choice | Probability | Host reveals | Outcome if switching |
---|---|---|---|
1/3 | Either goat | ||
1/3 | Goat B | ||
1/3 | Goat A |
I like the idea of having a diagram in the solution section of the article. However, with the current diagram it looks like if you replace the words "Host must reveal" with "Host luckily reveals" in parts 2 and 3, you would have an 'explanation' that switching wins 2/3 of the time, even if the host forgets what is behind the three doors but happens to luckily open a door with a goat.
Last year I brought up the version of the problem where Monty luckily reveals the goat when I noticed that the probability of winning upon switching does not increase in this version. This observation confused the heck out of me until I saw the diagrams in the "To Switch or Not To Switch" section on the following webpage.
http://math.ucsd.edu/~crypto/Monty/montybg.html
The diagrams there are a little confusing at first but they definitely capture the difference in these two versions of the problem. Can we use the two diagrams in the article? I think they are useful for visualizing why the host's knowledge of what is behind each door matters. This alternate version is probably why many people don't feel sure about the solution to the standard version after they are shown it. Synesthetic
A sentence about a Bayesian solution to the Three Prisoner's Problem has been added, deleted, added, deleted, and now added again. I agree with the deletion (this is mentioned, appropriately, in the article on the Three Prisoner's Problem, but has no particular relevance to Monty Hall), but would like others to comment as well before deleting it yet again. -- Rick Block ( talk) 18:37, 10 September 2007 (UTC)
"The player's chances of winning the car actually double by switching to the door the host offers."
This seems to be wrong. Should we just delete that sentence?
The text of this article doesn't appear to state anywhere that the problem only works if we assume that the player would rather win a car than a goat. -- Mikeplokta 09:21, 14 October 2007 (UTC)
The discussion thus far creates the impression that the use of Bayes theorem is a safeguard against falling into the trap of the false answer or, equivalently, that people get the wrong answer because they don't use Bayes theorem in their heads. This is not the case. The added paragraph demonstrates that the use of Bayes theorem still leaves ample room for misformulating a problem. Moreover, it explains why the Monty Hall problem, by accentuating the contrast between "information revealed" and "total evidence", has become central to philosophical discussions about the adequacy of Bayesian reasoning in managing uncertainty. Kvihill 17:20, 16 October 2007 (UTC)
attacks in the past, and have been thoroughly debunked. The Glopk 20:28, 19 October 2007 (UTC)
Judea Pearl's book (1988) gives a Bayesian explanation for people's tendency to provide the (wrong) answer 1/2. After the hosts reveals that a goat is behind door 3, people tend to condition their beliefs on the revealed information "a goat is behind door 3" and obtain the answer:
The correct answer is obtained by conditioning on the total evidence available: "host revealed a goat behind door 3," giving:
The distinction between "information revealed" and "total evidence" has far reaching implications in reasoning under uncertainty [Pearl, 1990, 1992]
I get the feeling this is very reminiscent of Bell's inequality in relation to hidden variable theories in Quantum Mechanics - is the MH problem an analogy for that inequality perhaps? Fizzackerly 13:31, 19 October 2007 (UTC)
I was surprised to see that the "switch at the very end" strategy for n doors isn't better supported. Is this because there's no suitable reference, or is it really an open problem?
FWIW I think the following sketches a proof: when there are two doors left the sum of the probabilities of success must be equal to 1; maximising one is equivalent to minimising the other; not switching until the last chance minimizes the probability of the initially chosen foor being correct. —Preceding unsigned comment added by 190.160.252.137 ( talk) 19:49, 11 November 2007 (UTC)
I don't understand why the problem is different from the following situation: "Suppose you walk up to two doors, one of which has a car behind it. Someone tells you that there was a third door that didn't have a car but that it was removed. You must now choose one of the two doors." Why is this problem different? Why is "switching" different from "choose one of two doors?" RobertM525 03:35, 13 November 2007 (UTC)
I have a question. What if talk host doesn't know what is behind the doors, but is directed by some one, who knows what's behind, to open a specific door once the player selects a door? -- Venkataramana vurity ( talk) 16:45, 22 November 2007 (UTC)
I don't understand why it's important for the host to know what's behind the doors. Don't the following premises have the same odds?
Premise 1: "Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?"
Premise 2: "Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?"
I understand that it is improbable that the host could repeat premise #2 without eventually revealing a car, however that is a completely different issue than premise 2 as stated. —Preceding unsigned comment added by Banderson1962 ( talk • contribs) 22:38, 25 January 2008 (UTC)
This doesn't really answer the question "why is it important that the host knows what's behind the doors?" In fact, it makes no difference at all! Let's say Monty has no idea what door has the car and he's guessing as blindly as you are. Then the possibilities in the first round become 1. you picked the car and he opens a goat because that's all he has 2. you picked a goat, Monty has the car and a goat, but opens the goat randomly, and 3. you picked a goat, Monty has the car and a goat, and opens the door with the car. In scenario 3, the game is over and the second round, being moot, never occurs. However, in scenarios 1 and 2 the game goes on, and you are still confronted with the fact that when you went into the second round, Monty was twice as likely as you to have the car, and switching still doubles your odds of going home with that car. Drewtew ( talk) 06:33, 29 January 2008 (UTC)drewtew
An anonymous user inserted an aid to understanding by considering infinite doors. In this case, the first door picked has zero probability of hiding the car, and switching yields the car with probability 1. Though perhaps still a bit unpolished, I found it helpful. However, Mattbuck ( talk · contribs) undid the edit with the comment "doesn't help since it's not quite 0". How is the probability "not quite 0"? Phaunt ( talk) 11:50, 29 November 2007 (UTC)
Is the 2/3 probability supposed to be based on the initial door chosen, or the door chosen after there are 2 left and you're given the chance to switch? Because the diagram seems to indicate the former, while the jist I got from the article indicated the latter. Thanks. 71.127.243.28 ( talk) 01:24, 1 January 2008 (UTC)
There is a citations tag at the top of this article. If this article requires more citations, how is this a featured article? Moreover, it's references section does not use the proper reference tags.
So how can this article possibly be a Featured article? -- Son ( talk) 04:52, 2 January 2008 (UTC)
This article was more on-point when it was featured in July 2005, but even then it suffered the same flaw it exhibits today: matters of style aside, a "well written" article about a veridical paradox should successfully explain how common sense is led astray and convincingly explain the truth of the matter.
As discussion on the talk page indicates, subsequent elaborations such as detailed Bayesian analysis and digressions on variants, interesting though they may be, do not lend clarity. Perfectly intelligent people who fully understand the mathematics of probabilities can read the entire article and still be left suspecting some kind of elaborate academic hoax, like one of those fallacious proofs that 1+1=1.
The main problem is that Sources of confusion fails to address the fundamental misunderstanding that so easily ensnares common sense. The truth is in the article, but it is not clearly stated as the chief source of confusion: selective evidence.
I would be guilty of original research if I wrote an explanation of this, as I have not found a source with a clear and direct statement of the issue as it relates to this paradox in particular. A good article on evaluating selective evidence might merit more than low-importance and low-priority. As a cautionary example, this article's significance could be elevated as well, if only the source of confusion were not obscured by the elaborate explanations. 67.130.129.135 ( talk) 03:27, 4 January 2008 (UTC)
To play "Devil's advocate", I will semi-seriously challenge one assertion of the article. In essence, I will agree that the problem, when correctly stated, yields 2/3 as the probability of winning by switching. However, I will contend that for the version stated in the Ask Marilyn column, the answer is not "technically ambiguous" but rather 1/2 - slightly muddy, perhaps, but not as ambiguous as all that. To understand why, it's important realize that "probability" is inevitably a manifestation of incomplete knowledge. Outside of the quantum realm, complete knowledge of a situation invariably yields prediction values of either 1.0 or 0. Probability, on the other hand, is based on the knowledge at hand, and so to claim that a probability estimate is ambiguous because important information is lacking is to misconstrue the meaning of probability. In the Ask Marilyn version, we really know only that the car is not behind door 3 and must therefore be behind one of the other two doors. This yields the value of 1/2. Of course, we also know that the host opened door 3 to reveal a goat, but for practical purposes, this is not usable knowledge. Based only on what is usable, we arrive at the 1/2 answer. I suppose one could take a Bayesian approach to interpreting the host's actions, in which case, one could estimate various probabilities based on guesses about the host's motivation. If so, that should be stated. It would be equally reasonable, however, to simply say switching or staying are equally likely to succeed as long as we don't know what the host is up to. Fmoolten ( talk) 01:37, 20 January 2008 (UTC)
I would argue that the entire notion of probability is based on estimates of outcome under circumstances that are only partially defined. Completely defined circumstances yield probability values of 0 or 1 (excluding uncertainties in the quantum realm). "Defined" is therefore a matter of degree rather than an all-or-none concept. To say that the probabilies are equal when we don't know what the host is up to is consistent with this principle. Clearly, these circumstances are less well defined than in the correct version of the puzzle, but they still permit probability estimates based on the information available. An analogy exists with a coin toss. If all we know is that an ordinary coin is tossed, the probability of heads is estimated at 0.5. However, if the toss is better defined in terms of angle and height of toss, speed of rotation, center of gravity of the coin, and the nature of the surface below, the probability changes based on the greater degree of definition. In the Ask Marilyn puzzle, we at least know that the choice is between two doors. The puzzle would be more or less completely undefined if we didn't know how many doors were involved, and in that case, a probability estimate would be virtually meaningless. Fmoolten ( talk) 17:06, 29 January 2008 (UTC)
The host doesn't want to lose cars, but everyone watches the show and knows his pattern of behavior. If we assume all the players are smart, then there's no use for the host to only open a door if the player has picked the car, because then people won't switch. But if he always opens a door then all the players switch and win more often. He has nothing to win or lose if he opens a door 50% of the time when the player hasn't picked the car and 100% of the time when the player has picked the car, because then the player wins the car by switching as often as he loses it by switching. Nonetheless in the real world some people will always switch by mistake when given the option. So there's a trade-off between a notorious host who only opens the door when the player picks the car, but has very few mistaken switchers, and a subtle host who opens the door 40% of the time when the player hasn't picked the car, and gets many people who think it doesn't matter or even think it helps based on a few episodes. How to resolve that trade-off I can't tell say. But in a proper game against a cunning host it should be at least somewhat harmful to switch doors when the choice is offered. 70.15.116.59 ( talk) 20:36, 24 January 2008 (UTC)
The article gives the impression that it's important that the host knew along where the car and goats are. Aren't the results the same whether he knew or not, since you've already eliminated the cases where he opens the 'car' door? Robin Johnson (talk) 14:07, 1 February 2008 (UTC)
I had to work this out for myself though... it miiiight be possible to express it more clearly in the article. Robin Johnson (talk) 18:34, 1 February 2008 (UTC)
This edit added a section titled "The Rigorous Solution" which includes a Bayesian analysis where the host's probability of picking which of the remaining two doors to open (in the event that the player originally picked the car) is allowed to vary, and shows switching is optimal regardless of the probability assigned. This analysis is claimed to include the effects of the host's strategy, thereby showing switching is optimal regardless of the host's strategy.
Rather than debate the merits of this analysis, I'd suggest we not even consider including it unless a reliable source for it can be found. -- Rick Block ( talk) 16:54, 10 February 2008 (UTC)
This elementary Bayesian calculation is not research. It would be better for the readers to see this. The probability if winning the car given switching is more general than 2/3. It is 1/(p+1), which equals 2/3 when p=1/2. It can be between 1/2 and 1, but always not less than the probabilty of winning car by not switching, which is equal to p/(p+1), and therefore between 0 and 1/2. —Preceding unsigned comment added by 70.137.168.95 ( talk) 02:00, 11 February 2008 (UTC)
The problem with this article is that the main point is cluttered by so many uninteresting generalizations, whereas the original game (where the host always opens a door other than the one picked by the player, which also does not have the car) is not analyzed rigorously. The most general stratgey of the host subject to the original rules of the game is characterized by a probability p, which is that probability with which the host opens door 2, given that the player chose 1 and the car is at 1. (There could be other probabilities for the cases of the player picking 2 and 3, but it suffices to conside one case.) To ignore the full analysis of the original problem is not wise, especially when the analysis reveals the surprising result that it is always optimal to switch. To emphasize the importance of this point, the player deos not need to know p in order to conclude that switching is optimal. This is true for every p. To assume p=1/2 is too restrictive! I do not understand why you want to censor this analysis using your admin power. —Preceding unsigned comment added by 70.137.145.250 ( talk) 04:04, 11 February 2008 (UTC)
Here is the rigorous solution:
A rigorous solution requires addressing the question of the strategy of the host, namely, how the host picks a door to open, when more than one is possible. Following is a Bayesian analysis, which proves that switching is optimal regardless of the host's strategy:
The prior belief of the player is that the car is behind each door with probability 1/3. Suppose the player chooses door 1.
Denote by p the conditional probability with which the host opens door 3, given that the car is behind door 1. Thus, 1-p is the conditional probability with which the host opens door 2, given that the car is behind door 1. In the event the car is not behind door 1, there is only one door that the host can open.
Following are the probabilities and conditional probabilities that lead to the conclusion:
Prob (Host opens door 3, given that the car is behind door 1) = p Prob (Host opens door 3, given that the car is behind door 2) = 1 Prob (Host opens door 3, given that the car is behind door 3) = 0 Prob (Host opens door 3 ) = (1/3) x p + (1/3) x 1 + (1/3) x 0 = (p+1)/3.
The following equation holds:
Prob (Car is behind door 1, given that the host opened door 3) x Prob(The host opens door 3) = Prob (Host opens door 3, given that the car is behind door 1) x Prob(The car is behind door 1)
Therefore,
Prob(Car behind door 1, given that the host opened door 3) x (p+1)/3 = p/3
and thus
Prob(Car behind door 1, given that the host opened door 3) = p/(p+1).
It follows that,
Prob(Car behind door 2, given that the host opened door 3) = 1/(p+1).
Since p is not greater than 1,
Prob(Car behind door 2, given that the host opened door 3) >= Prob(Car behind door 1, given that the host opened door 3).
Therefore, switching to door 2 is an an optimal strategy for every p. —Preceding unsigned comment added by 70.137.145.250 ( talk) 05:44, 11 February 2008 (UTC)
(Re: changes by Rick Block made in response to this thread) Does it really matter whether the host makes a random or non-random choice when the two unpicked doors both contain goats?-- Father Goose ( talk) 02:06, 13 February 2008 (UTC)
RE: Rick Block (Father Goose, Please read this entire section "Rigorous Solution" in this Discussion, or the Morgan et al. paper.) What Mr. Rick Block is doing is harming the reputation of Wikipedia. He deleted the rigorous solution and put the following text as the solution: "If the player chooses to switch, the player wins the car in the last two cases. A player choosing to stay with the initial choice wins in only the first case. Since in two out of three equally likely cases switching wins, the probability of winning by switching is 2/3. In other words, players who switch will win the car on average two times out of three." As explained in (Morgan et al. 1991), this statement is "the most appealing of the false solutions." Mr. Block's solution, which is the current solution of the article, is the same as what is referred to as "false solution F1" in Morgan et al., who say about it the following (p.285):
Mr. Block deleted a previous comment about the difference between unconditional and conditional probabilities. Mr. Block monopolizes this article and lowers the level of the article. It is not only the differece between "random" and "non-random." The player has no basis to believe that the host picks one of the two doors with probabilities 50:50. The host may do it with randomly with different probabilities, say 75:25 or even 100:0 (i.e., deterministically). The conditional probabilities are different in these cases, even though the unconditional probability is 2/3. Interestlingly, switching is optimal in every case. However, Mr. Block twice deleted this fact, which is cleverly explained in Morgan et al. What is the recourse against such behavior? 70.137.163.193 ( talk) 04:42, 13 February 2008 (UTC)
So, am I to understand that the current brouhaha is over whether the host's choice of the two "unpicked" doors is non-random? If so, either:
If it's the first case, that's such a huge omission from the basic statement of the problem, surely it cannot be treated as an inadvertent oversight or potential ambiguity.
To use a different example, if the host always told the player truthfully where the car was, that would change his odds of winning to 100% in all games. However, while the original phrasing doesn't preclude that as a possibility, the fact that such an unusual condition is not included in the phrasing makes it safe to assume that it is not in effect. I would say that "player knows host's predictable behavior" is a similarly unusual condition, and not something that could be treated as an ambiguity in the original problem.-- Father Goose ( talk) 03:11, 14 February 2008 (UTC)
(outindent) Father Goose: Dropping the constraint that the host select from among two goat doors with equal probability introduces an ambiguity in the sense that the numerical probability of switching by winning is not a constant for all players regardless of what the host does. It creates a potential difference between the overall chance of winning by switching seen by players in the aggregate (the unconditional probability our anonymous friend of many IP addresses is talking about) and the chance of winning by switching as seen by an individual player at the point of being asked to switch knowing which specific door the host has opened (the conditional probability). In the problem statement (both vos Savant's and the unambiguous version in the article) the question is whether some individual player is better off switching, not "what is the aggregate chance of winning by switching". If we drop the constraint, and call the host's potential preference for one door over another p (the constraint forces p=1/2), the probability an individual player sees is 1/(p+1) which ranges from 1/2 to 1 (depending on p) although the unconditional probability seen by players in the aggregate is 2/3 regardless of p (indeed, a whimsical host might decide to pick the "rightmost goat door" on Mondays, Wednesdays, and Fridays and the leftmost on Tuesdays and Thursdays making p alternate between 0 and 1 depending on the day of the week with an aggregate value of .6). Again, this is exactly the point user:anon-many-ips is making. Note that without the constraint, the difference is there and affects the player's chances whether the player knows p or not, in exactly the same way that the player's probability of winning is 1/2 (both conditionally and unconditionally) if the host opens a remaining door randomly (the forgetful Monty version) whether the player knows how Monty has selected or not. -- Rick Block ( talk) 16:49, 16 February 2008 (UTC)
Here is verbatim what they say:
Then they say:
The readers can verify that F1 is indeed the same as the solution currently in the Wikipedia article. 70.137.163.193 ( talk) 06:17, 14 February 2008 (UTC)
Ok. I coughed up the $14 and purchased the online Morgan et al article (from [1]). The version of the problem they initially analyze is essentially identical to the original statement of the problem from Parade (with no constraints on the host's behavior). They label as solution F1 the informal analysis that (paraphrased) "a player who doesn't switch wins 1/3 of the time, so switching wins 2/3 of the time" and call this a false solution. Indeed, it doesn't address the original problem as stated in Parade as it ignores the effects of the host's behavior. Rather than focus on ambiguities in the problem description they pursue an approach where the host's behavior is represented as a set of probabilities, leading to the possibility of different probabilities of winning by switching depending on the specific door the host opens. Although many host behavior variants are covered by their approach, not all are (they specifically exclude "host opens the player's door", "host opens the door revealing the car", and "host makes the offer to switch more or less often depending on the player's initial selection").
Given a problem statement where the host's actions are effectively described as variables requires distinguishing the player's initial chances (the "unconditional" problem) from the chances given the door the host reveals (the "conditional" problem). However given the unambiguous statement of the problem in the Problem section of this article, the specific door the host opens does not (cannot) affect the player's chances of winning. There's a fairly long standing consensus that the version of the problem discussed here should be an unambiguous version, permitting only a single solution which does not need to (and does not) distinguish "unconditional" vs. "conditional" (this difference is prohibited by the problem statement). IMO "Other host behaviors" would be a fine place to include a discussion of this issue. -- Rick Block ( talk) 03:47, 14 February 2008 (UTC)
More formally, when the player is asked whether to switch (assuming the player initially picked door 1) there are three possible situations corresponding to the location of the car, each with probability 1/3:
The host opens only one door, so only one of the first subcases with probability 1/6 and one other case with probability 1/3 apply. Staying wins in the 1/6 case where the player has initially picked the car while switching wins in the 1/3 case where the player has not. Switching wins twice as often as staying, so the probability of winning by switching is 2/3. In other words, playersa player who switches will win the car on average two times out of three.
Suppose the player picked door 1. Before the host has opened any door, there are three equally likely possible situations, namely, the car is behind each door with probability 1/3. It is assumed that if the car is behind door 1, then the host opens door 2 or door 3 with equal probabilities. If the car is behind door 2, then the host opens door 3, and if the car is behind door 3, he opens door 2. Therefore, the probability that the car is behind door 1 and the host opens door 3 is (1/3)x(1/2)=1/6. The probability that the car is behind door 2 and the host opens door 3 is (1/3) x 1 = 1/3. These are the only scenarios under which the host opens door 3. Therefore, the probability that the host opens door 3 is 1/6 + 1/3 = 1/2. Given that host has opened door 3, the conditional probability that the car is behind door 1 is (1/6)/(1/2)=1/3, and the conditional probability that the car is behind door 2 is (1/3)/(1/2)=2/3. Therfore, given that the host opened door 3, switching wins the car with probability 2/3. Likewise, given that the host has opened door 2, switching wins the car with probability 2/3. 70.137.163.193 ( talk) 16:44, 14 February 2008 (UTC)
(outindent) You keep saying this, but what exactly about it is false? You didn't respond to the suggestion (previous section) about expanding it from 3 to 6 cases (treating each possible door that can be opened separately). Even in the 3 case version it explicitly enumerates the cases in effect after the host has opened a door (either door). It doesn't use the words "conditional probability" and doesn't show the derivation of the equal probability cases, but is this necessary? How about:
I'm not overly attached to this description, but just want to understand what the objection is. -- Rick Block ( talk) 23:08, 14 February 2008 (UTC)
It is all from the point of view of the player. The figure in the current solution section is very misleading for the layperson, because it does not show what the player knows (the player knows that he has chosen door 1):
-- 198.4.83.52 ( talk) 20:06, 15 February 2008 (UTC)
So far so good.
How would the solution described in the text above "so far so good" change if the host opened one of two goat doors non-randomly?-- Father Goose ( talk) 22:17, 16 February 2008 (UTC)
More formally, when the player is asked whether to switch there are three possible situations corresponding to the location of the car, each with probability 1/3. Assuming the player initially picked door 1:
The host opens only one door, so only one of the first subcases and one other case with probability 1/3 apply. In the case where the player has initially picked the car, staying wins with probability 1/3*p or 1/3*(1-p) depending on whether the host opens door 3 or door 2. Switching wins in the 1/3 case where the player has not. Since p is a probability it ranges between 0 and 1, so switching always wins with at least the same probability as staying regardless of which door the host opens — and depending on p and which door the host opens might guarantee winning the car. In other words, a player will win the car more often by switching than by staying with the initially picked door.
If all players switch regardless of p and what door the host opens, their average probability of winning is 2/3 since 1/3 of these players will initially pick the car and lose while 2/3 will initially pick a goat and win.
Moved from User talk:Father Goose
Any idea why Talk:Monty Hall problem got so quiet all of a sudden? I thought we were almost there. -- Rick Block ( talk) 01:01, 20 February 2008 (UTC)
(reset indent)All right, I'll take a different tack. Is it not possible to formulate a unconditional conditional analysis that represents the fact that any time the player picks the car first, he will lose when switching, no matter which behavior the host adopts when choosing between two goats?--
Father Goose (
talk)
20:50, 21 February 2008 (UTC)
Excuse me, I meant to say a conditional analysis.-- Father Goose ( talk) 00:36, 22 February 2008 (UTC)
An anonymous user (editing from a variety of IP addresses) has expressed a concern about the solution presented in the article. All of the discussion above, starting with the section #Rigorous solution relates to this issue. Should the wording of the primary solution be changed and is the following suggested wording satisfactory?
More formally, when the player is asked whether to switch (assuming the player initially picked door 1) there are three possible situations corresponding to the location of the car, each with probability 1/3:
The host opens only one door, so only one of the first subcases with probability 1/6 and one other case with probability 1/3 apply. Staying wins in the 1/6 case where the player has initially picked the car while switching wins in the 1/3 case where the player has not. Switching wins twice as often as staying, so the probability of winning by switching is 2/3. In other words, a player who switches will win the car on average two times out of three.
After the player has chosen an initial door (assuming door 1) but before the host opens a door, there are three possible situations corresponding to the location of the car — each with probability 1/3:
The host opens only one door, so at the point the player is asked whether to switch only one of the first subcases with probability 1/6 and one of the other cases with probability 1/3 apply. Staying wins in the 1/6 case where the player has initially picked the car while switching wins in the 1/3 case where the player has not. Switching wins twice as often as staying, so the probability of winning by switching is 2/3. In other words, a player who switches will win the car on average two times out of three.
Replace the last paragraph by the following:
The host opens only one door. Consider the case where the host opened door 3 (the same argument applies as well to the case where the host opened door 2). The player is then given the option to switch. In this situation, the possibilities are only two: (i) The car is behind door 1 and the host picked at random door 3 (originally, this case had probability 1/6). (ii) The car is behind door 2 and the host had no choice but open door 3 (originally this case had probability 1/3). These two cases have a total initial probability of 1/6 + 1/3 = 1/2. Staying wins in case (i) where the player has initially picked the car, which amounts to a proportion of (1/6)/(1/2) = 1/3 (see
conditional probability). Switching wins in case (ii), which amounts to a proportion (conditional probability) of (1/3)/(1/2) = 2/3. Intuitively, switching wins twice as often as staying, so the probability of winning by switching.
-
70.137.165.14 (
talk)
06:57, 26 February 2008 (UTC)
Here's a way I find easy and simple to understand: When you choose a door the first time, 1/3 chance you got the car. So theres a 2/3 chance of the car being behind one of the other TWO doors. Now if we get rid of one of the other 2 doors, then its 2/3 chance of it being behind the remaining ONE door. LordRobert ( talk) 07:19, 23 February 2008 (UTC)
(outindent) We're clearly talking past each other. With our constraints (specifically including "host picks randomly between two goat doors") the answer is 2/3 in every case regardless of which door the player picks (so whether it's 1, 2, or 3 doesn't matter) and no matter which door the host subsequently opens. IMO this is a good property to have and allows the illustrations and specific problem setup (player picks door 1, host opens door 3) to be representative of all cases (it doesn't matter - they're all the same!). In addition, because it's the answer in every single possible scenario, it's also the average outcome considering all scenarios. I also think it's the (surprising) answer nearly anyone posing this problem presents as the solution. I think we all agree on this so far.
On the other hand, both problem statements we present (the one from Parade, and the one from Mueser and Granberg) present the problem from the perspective of a single player after the host has opened a door (Parade: You pick a door, say No. 1, and the host ... opens another door, say No. 3. Is it to your advantage to switch your choice? - Mueser and Granberg: You begin by pointing to door number 1. The host shows you that door number 3 has a goat. Do the player's chances of getting the car increase by switching). I think in both versions the answer is not meant to vary depending on the player's or host's actions, so it is perhaps ambiguous whether the actual question is "considering all possible scenarios what is the player's average chance of winning by switching" or "given a player has picked a specific door and the host has opened another specific door, what is this specific player's chance of winning by switching". Given the way it's worded, I think the latter interpretation is actually more sensible, but I accept we may not all agree on this. Whether we agree or not, I think we can (and should, but perhaps don't yet) agree that these questions can have different answers depending on the constraints.
If these questions might have different answers, can we agree a solution approach for one may not be appropriate for the other? Taking this a little further, can we agree that if we answer the player specific question for all possible scenarios we assuredly have the answer for the more general question but not necessarily vice versa?
There are 6 possible cases of player picks and doors the host might open (PHx, PxH, HPx, HxP, xPH, xHP). It seems we're showing the overall probability is 2/3 and wanting (without any particular justification) for the question to be "considering all possible scenarios". This leaves open the possibility that other people are reading the question as "specific player, specific door the host opens" (which is where our anonymous friend entered the picture). If we show both that the probability is 2/3 for each and every player because of our constraints and that the overall probability is 2/3, isn't this a better approach? -- Rick Block ( talk) 03:43, 28 February 2008 (UTC)
Yes, we are talking past each other; we are being repetitive; we are having great sport going round in circles. Can we come down to earth for a moment? Does anybody seriously question whether it is overwhelmingly notable that the problem was originally intended, and is most widely interpreted to be about probabilities of choice, and not about strategies in a game between two contestants? Does anybody dispute that, thusly interpreted, it is a very interesting veridical paradox, a moderately important layman's puzzle, and a perfectly notable encyclopedic topic? If not, let us keep the main article on topic by eschewing considerations of host strategy and allowing that the goats are as indistinguishable to the host as the doors are to the guest. If so, please explain.
To return to the interest, importance, and charm of the problem, it is just this:
Variant interpretations are interesting, notable, and encyclopedic digressions, but they should be presented in a way that does not detract from this central theme by unduly complicating the statement of the basic problem or its basic solution. 67.130.129.135 ( talk) 03:11, 29 February 2008 (UTC)
Let me try once more. I'm sorry if the following only indirectly answers Rick's questions.
Enough facts. With the added clause (which could be improved by deleting the word "special"), there is a simple analysis showing that the switching strategy always increases the chance from 1/3 to 2/3. Without that clause, but with another additional clause that is required then, namely that the player does not learn anything that is not a consequence of which door was opened (possibly because the host has preferential behaviour of which the player is aware), you can show that the always-switching strategy is an optimal strategy that increases the a priori chance (before the host opened the door) from 1/3 to 2/3. The proof of 2/3 is as before, but the proof that this is optimal requires, as far as I can see now, an elaborate case analysis (but maybe I'm wrong and there is an easy proof for this too). In either case, the frequentist interpretation is appropriate: you can't win two thirds of a car, but if the game is repeated often enough about 2 in every 3 players who follow the optimal strategy will go home in a car instead of on a goat.
Summing this up. Without some clause added to the M&G formulation, the problem cannot be analyzed. With an added clause, perverse interpretations may still be possible. Ignoring such perversions, one added clause (the one now in the article) allows a simple complete analysis; the other possible choice for an added clause (which is an awkward thing to formulate; an attempt is found above, which in fact rules out the perverse interpretation) still allows a simple proof – the same as before – that always switching is better on the average, but leaves the possibility open that it is not better in all situations, and a more complicated proof is needed to show this strategy is nevertheless optimal. --
Lambiam
20:39, 29 February 2008 (UTC)
A statistician is given the following problem:
The statistician is then told that he will win the car outright if he can provide a single numerical answer as to what his probability of winning by switching is. The statistician says that there are two probabilities he faces, which cannot be combined in any way to produce a single probability that would correctly describe his odds of winning by switching. He additionally asserts that there is no valid way of analyzing the problem as presented that could correctly produce a single numerical answer. He leaves the show empty-handed.
Was the statistician right?-- Father Goose ( talk) 01:58, 29 February 2008 (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 4 | Archive 5 | Archive 6 | Archive 7 | Archive 8 | → | Archive 10 |
I've moved the existing talk page to Talk:Monty Hall problem/Archive2, so the edit history is now with the archive page. I've copied back a few recent threads. Older discussions are in Talk:Monty Hall problem/Archive1. Hope this helps, Wile E. Heresiarch 15:28, 28 July 2005 (UTC)
I'm all for racial harmony and all, but why do the images include a bald black man as the player? Could we be using a more abstract, symbolic face? (I know I know, I'm a racist because I wouldn't have noticed if it were a white guy. Granted, but I think the question is still valid.) -- Doradus 02:00, 3 May 2007 (UTC)
The two mentions of Deal or No Deal contradict one another: under 'Sequential Doors' is an explanation of the critical difference between the NBC Prime Time SMASH HIT and the Monty Hall problem as stated, but under the 'History of the problem' heading, Deal or No Deal is essentially considered to be a minor variant with the same sorts of conclusions. The numbers directly contradict one another as well. Though I'm fairly positive the first description is accurate, I'll refrain from editing for now so people can argue and yell and scream and then somebody smarter than myself can fix it. I also kinda like the fact that both are in this article, as it seems the Monty Hall problem's best contribution to society is watching people flailing about entirely confused yet certain that they're right in the face of contradicting evidence. 66.188.124.133 17:32, 16 May 2007 (UTC)
The problem with this is that the 3 scenarios are actually 4.
For the first scenario where the person picks the car it is listed as host showing "either Goat A or B". Actually these are two different scenarios:
Scenario 1: Contestant picks car, Host shows Goat A, Contestant switches, Contestant LOSES
Scenario 2: Contestant picks car, Host shows Goat B, Contestant switches, Contestant LOSES
Scenario 3: Contestant picks Goat A, Host must show Goat B, Contestant switches, Contestant WINS
Scenario 4: Contestant picks Goat B, Host must show Goat A, contestant switches, Contestant WINS
You can see that switching yields the expected 50% success.
It is rather alarming to me that this is missed by experts. I believe Quantum computing falls into this same "smoke and mirror" science. —The preceding unsigned comment was added by 143.182.124.2 ( talk • contribs).
Might I ask for more specificity as to what this article needs in order to become an FA? It's not clear how to act upon the suggestions left in the Mathematics rating box.-- Father Goose 02:31, 23 June 2007 (UTC)
From my talk page, but it makes more sense to comment here:
The current lead is a teaser, which is great for a magazine, but not for an encyclopedia. I'm sorry that my other concerns are vague, but it seems that other editors believe that this article meets FA standards, in which case please will someone replace my comment and signature by their own. Thank you! Geometry guy 10:56, 23 June 2007 (UTC)
The Monty Hall problem is a puzzle involving probability loosely based on the American game show Let's Make a Deal. The name comes from the show's host, Monty Hall. A widely known statement of the Monty Hall problem appeared in a letter to Marilyn vos Savant's Ask Marilyn column in Parade (vos Savant 1990):
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
Since there is no way for the player to know which of the two unopened doors is the winning door, many people assume that each door has an equal probability and conclude that switching does not matter. However, as long as the host knows what is behind each door, always opens a door revealing a goat, and always makes the offer to switch, opening a losing door does not change the probability of 1/3 that the car is behind the player's initially chosen door. As there is only one other unopened door, the probability that this door conceals the car must be 2/3.
The problem is also called the Monty Hall paradox; it is a veridical paradox in the sense that the solution is counterintuitive. For example, when Marilyn vos Savant offered the problem and the correct solution in her Ask Marilyn column in Parade, approximately 10,000 readers, including several hundred mathematics professors, wrote to tell her she was wrong. Some of the controversy was because the Parade statement of the problem fails to fully specify the host's behavior and is thus technically ambiguous. However, even when given completely unambiguous problem statements, explanations, simulations, and formal mathematical proofs, many people still meet the correct answer with disbelief.
I found another exchange in the FAR which resonates a little bit with my reaction to the article and may help clarify the vaguer part of my comment. I quote:
I wouldn't go as far as Abscissa, but it is not dissimilar from my reaction. Anyway, this exchange seems to have got lost in a sandwich between the arguments about inline citation that opened the review, and the impressive copyediting drive that ended it. I couldn't find it in the talk archive. Did anything come of it? Geometry guy 17:41, 23 June 2007 (UTC)
I am just wondering if it is a flaw in one of the Variations à Other host behaviors à “The host does not know what lies behind the doors, and the player loses if the host reveals the car.”
The answer is allegedly “The player loses when the car is revealed a third of the time. If the prize is still hidden, switching wins the car half of the time.” Shouldn’t switching here also result in a 2/3 probability of victory (since it is purportedly the same result (so far) as in the original Monty Hall problem)? -- 85.164.95.143 14:15, 9 July 2007 (UTC)
I believe as well there is a flaw here. Let's extend the problem to 100 doors the same way we reason the standard problem. The host luckily opens 98 goats in a row. But once this has happened, certainly you would switch, because even without the knowledge of the host, you only had a 1/100 chance of landing on the car in the first place, which has not changed. Your odds of WINNING this sort of game are of course lower because of the probability of the host opening the door to the car causing a loss. However, after he opens a goat, even if it is random, it does not affect the probability of you having picked the car in the first place, and thus the advantage of switching. Again, the 100 doors intuition here seems to support this argument. —Preceding unsigned comment added by 151.190.254.108 ( talk • contribs)
I'm thinking now I am mistaken because I read another type of explanation that I could not seem to refute, but your point #3 is still somewhat confusing because it seems to contradict #1, not show how #1 doesn't work.
Alright, I have been convinced. Perhaps my confusion can lead to something positive, like some clearer wording about why this is true. What I gather now is that when the host picks randomly, 1/2 of the time that a car is not revealed randomly, the reason is not that you got lucky picks by the host but that the car is behind the door you picked. I followed this by a tree diagram. The other explanation I found online was just a different wording of the one you used to refute my 100 door argument. Thanks!
Okay, I've spent entirely too much brainjuice on this. It probably has one or more glaring errors. But it's the algebra-ized version of the way I view the problem in my head (which has more to do with colors than with symbols; go figure). That said, I'm not so good at this whole color→algebra thing, so please fix it up:
You select a winning door | Your door is now a winner | Remaining Door(s) contain a Winner | You've already lost | |
Host picking randomly | ||||
Host picking only losing doors |
Jouster ( whisper) 23:48, 18 July 2007 (UTC)
You originally select a winning door | Your door is now a winner | All d remaining Door(s) contain a Winner | Each d remaining door contains a winner | You've already lost | |
Host picking randomly (absolute) | |||||
Host picking randomly (conditional) | |||||
Host picking only losing doors |
I'm not sure what the final word here is, but I think it's wrong. Anyway, the article is still in error, stating
Possible host behaviors in unspecified problem | |
---|---|
Host behavior | Result |
The host does not know what lies behind the doors, and the player loses if the host reveals the car. | The player loses when the car is revealed a third of the time. If the prize is still hidden, switching wins the car half of the time. |
I claim, as did whoever initiated this discussion, that
1. If the host does not know where the prize is, and
2. The host randomly opens one of the two doors not chosen,
then given the above, the stategy of not switching doors yeilds wins with probablity 1/3, exactly as it did when the host knew where the prize was!
This can be shown in a number of ways. The most convincing, using Bayes' theorem, appears last.
1. Ad absurdum: If you believe that there's a difference between the cases where the host knows where the prize is or guesses where the prize is (tosses a coin), then give some thought to the interim possibilities: The host knows that the prize had an uneven probability of being placed, say 1/4, 1/4, 1/2. Work out the probability of winning if you don't change your choice (host opens the door least likely to show the prize). Now consider what happens if the host thought he knew the probability ditribution of the prize placement, but was wrong - they used a different distribution! How can the contestant's strategy be affected by all this?!
2. If the host knows nothing, we don't need him. State the game as follows: Contestant picks a door (may as well be at random), and now opens one of the other two doors at random. Given that the opened door has no prize, what's the chance that the initial door does? I claim 1/3. If you still think that the answer is 1/2, consider the following variant:
Contestant opens one door at random. It has no prize. Now the contestant picks one of the remaining doors. Clearly the chance of winning is 1/2. Do you really believe that the two situations yeild the same probability of winning?
3. Bayes' theorem: I use the notation from the article: In Bayesian terms, probabilities are associated to propositions, and express a degree of belief in their truth, subject to whatever background information happens to be known. For this problem the background is the set of game rules, and the propositions of interest are:
For example, denotes the proposition the car is behind Door 1, and denotes the proposition the host opens Door 2 after the player has picked Door 1. Indicating the background information with , the assumptions are formally stated as follows.
First, the car can be behind any door, and all doors are a priori equally likely to hide the car. In this context a priori means before the game is played, or before seeing the goat. Hence, the prior probability of a proposition is:
Second, the host will pick a door from the remaining two at random. This rule determines the conditional probability of a proposition subject to where the car is, i.e. conditioned on a proposition . Specifically, it is:
if i = j, (the host cannot open the door picked by the player) | |||
if i ≠ j and j = k, (the host can open a door with a car behind it) | |||
if i ≠j and j ≠ k, (the host can open a door without a car behind it) |
The problem can now be solved by scoring each strategy with its associated posterior probability of winning, that is with its probability subject to the host's opening of one of the doors. Without loss of generality assume, by re-numbering the doors if necessary, that the player picks Door 1, and the host then opens Door 3, showing him or her a goat. In other words, the host makes proposition true.
The posterior probability of winning by not switching doors, subject to the game rules and , is then . Using Bayes' theorem this is expressed as:
By the assumptions stated above, the numerator of the right-hand side is:
The normalizing constant at the denominator is simply:
as can be seen in the table above. Dividing the numerator by the normalizing constant yields:
This can be stated differently:
since the host doesn't know where the prize is! Therefore
This shows that the probability that your initial choice was right remains 1/3 even after the host opens a door at random, given that the random door did not show the prize. You should switch doors, exactly as in the case where the host knew where the prize was. —Preceding unsigned comment added by Jasoncoop ( talk • contribs) 13:24, 25 October 2007 (UTC)
Wrong! of course and are each 1/3. These are the cases where your initial choice was wrong. The problem you're describing is not the right one! The question is what's the probablitity that you were initially right given that the third door is wrong. This posteriori information affects the probability in the way I described using Bayes' theorem. Can you find fault with the way I used it? Jasoncoop 14:47, 25 October 2007 (UTC)
Rick, thanks for your detailed and patient response. I think you must be a pretty good teacher. I substituted host opens door 3 and prize is not there for and came up with 1/2 as you promised.
I'm still a bit concerned about the question of what happens when the host has some knowlege about the way the prize door was chosen, but not certain information (some non uniform distribution). Maybe I'll work it out some time.
So in fact the two following situations are similar:
1. First pick a door at random and open it. Given that it's empty, each of the closed doors has probability 1/2 of containing the prize.
2. Pick a door at random, open one of the other two doors at random. Given that it's empty, each of the closed doors has probability 1/2 of containing the prize.
Somehow I find this a bit disturbing, but hopefully I'll get used to it...
Or maybe this is the key to understanding the problem - if neither the contestant nor the host knows where the prize is, it doesn't matter who goes first. Jasoncoop 23:03, 25 October 2007 (UTC)
I believe that some who are confused by this problem are focused on the difference between a Strategy and a Choice. The Host presents the player with a Choice, and if it were truly acted on as a Choice then the probability would not be 2/3 winning. Only when the player examines the rules and develops a Strategy which is adhered to without fail, does the probability turn to hir favor. Those who have watched The Price is Right remember the indecision on the faces of players as they ponder what to do. In my (admittedly basic) understanding of probablilty, everything hinges upon a priori decisions. Thus, I posit that if a Player goes into the game with a Strategy of sticking to their original choice, they'll have 1/3 chance of winning. The Strategy of always switching when offered the inevitable Choice yields 2/3 Chance of winning. This has been established. But I think that it is important to note that if the Player has no Strategy, and truly decides to randomly Choose between switching and staying, then there is a 1/2 chance of success. I believe I am correct in this assessment, and I think it would make a good additional to the article to explain this, since I believe many of those who are confused are thinking in these terms. i.e. Explain the difference in probability between a true Strategy (Choice decided in advance) vs. random Choice. Of course, if I'm completely wrong on this (which is why I have placed this in discussion), I still think it might clarify the article to point out that this is a Strategy, and that the player is not really making a Choice - their outcome is predetermined when Monty opens a door. P.S. I note that one of the external links explains the exact case I present of 1/2 odds with a random Choice. Shouldn't that be mentioned somewhere in the article for completeness? BrianWilloughby 18:29, 11 July 2007 (UTC)
I agree with "Debunked" and I think the confusion here is with not distinguishing "or" versus "exclusive or" in example 1 or in the Venn diagram. Also with the Venn diagram the probability of picking the box with the two boxes in it is 50%, exactly the same as the probability of picking the first box. In effect in drawing the two boxes circled in the Venn diagram you have changed the probability by making it 50% 1st box or 50% "2 box" box. This again confuses "exclusive or" with "or", eg the probability of the car being behind one door is 1/3, the probability of the car being behind one of the last two box doors is 1/2 therefore the probability of the car being in the venn diagram second half is 50%, not 2/3, since it can either be in the box containing the last two boxes or be in the first box. 67.86.165.55 06:24, 1 August 2007 (UTC)
It's wrong because Scenario 1 in the image set in reality consists of 2 scenarios. Now what if we consider it from different perspective: We know the host will always pick the goat. The host picks Goat A, so the player choice is really between Goat B & the Car. If the host picks Goat B, the player choice again is only Goat A & the Car. Switching does nothing to increase the chance of winning. Since the beginning of the game the chance of the player winning the car is 50%, not 33%. 203.49.196.163 00:33, 2 August 2007 (UTC)
Thanks, Father Goose. I can see the absurdity of my statement now. So: Player picks 1 door out of 3 --> 1/3 chance of winning. If the host picks Goat A/B and player stays, his chance of winning obviously is still the same (1/3), but the chance that the car is behind the door becomes 1/2? If player decides to switch, there are 4 scenarios:
Which means by switching, his chance of winning indeed increases and becomes 1/2? Not 2/3?! 203.49.196.163 05:01, 2 August 2007 (UTC)
₠Thank you Father Goose, you are right and I withdraw my "I agree with Debunked" comment above. The example that makes it easier for me to understand is if there are 100 doors, you pick a door and then the host is told to remove 98 doors. The door he leaves has a 99% chance of being the Car, since the only scenario where he is not picking the cars is where you picked the car in the first place and the probability of that is only 1%, so it is much more likely when he opens the 98 doors that he has left closed the car. That to me is clearer for some reason than when only three doors are utilized in the problem. —The preceding unsigned comment was added by 70.181.21.230 ( talk)
For a different perspective on the problem, consider not switching doors. In each of the three possible cases, there is at least one unchosen goat for the host to show, but only in the first of the three cases (1/3 of the time) does not switching win the car.
This means that 1/3 of the time, the car is the contestant's first choice, therefore 2/3 of the time the car is not the contestant's first choice, and switching to the other available choice will win the car.
I've deleted the new section with the giant image (above) pending discussion here. In my opinion, the giant image basically replicates the other giant image (with different visual representations of the doors, car, and goat), and the text essentially repeats the text already in the second paragraph in the Solution section, making this new section simply redundant with material elsewhere in the (already too long) article. -- Rick Block ( talk) 13:56, 24 August 2007 (UTC)
Player's initial choice | Probability | Host reveals | Outcome if switching |
---|---|---|---|
1/3 | Either goat | ||
1/3 | Goat B | ||
1/3 | Goat A |
I like the idea of having a diagram in the solution section of the article. However, with the current diagram it looks like if you replace the words "Host must reveal" with "Host luckily reveals" in parts 2 and 3, you would have an 'explanation' that switching wins 2/3 of the time, even if the host forgets what is behind the three doors but happens to luckily open a door with a goat.
Last year I brought up the version of the problem where Monty luckily reveals the goat when I noticed that the probability of winning upon switching does not increase in this version. This observation confused the heck out of me until I saw the diagrams in the "To Switch or Not To Switch" section on the following webpage.
http://math.ucsd.edu/~crypto/Monty/montybg.html
The diagrams there are a little confusing at first but they definitely capture the difference in these two versions of the problem. Can we use the two diagrams in the article? I think they are useful for visualizing why the host's knowledge of what is behind each door matters. This alternate version is probably why many people don't feel sure about the solution to the standard version after they are shown it. Synesthetic
A sentence about a Bayesian solution to the Three Prisoner's Problem has been added, deleted, added, deleted, and now added again. I agree with the deletion (this is mentioned, appropriately, in the article on the Three Prisoner's Problem, but has no particular relevance to Monty Hall), but would like others to comment as well before deleting it yet again. -- Rick Block ( talk) 18:37, 10 September 2007 (UTC)
"The player's chances of winning the car actually double by switching to the door the host offers."
This seems to be wrong. Should we just delete that sentence?
The text of this article doesn't appear to state anywhere that the problem only works if we assume that the player would rather win a car than a goat. -- Mikeplokta 09:21, 14 October 2007 (UTC)
The discussion thus far creates the impression that the use of Bayes theorem is a safeguard against falling into the trap of the false answer or, equivalently, that people get the wrong answer because they don't use Bayes theorem in their heads. This is not the case. The added paragraph demonstrates that the use of Bayes theorem still leaves ample room for misformulating a problem. Moreover, it explains why the Monty Hall problem, by accentuating the contrast between "information revealed" and "total evidence", has become central to philosophical discussions about the adequacy of Bayesian reasoning in managing uncertainty. Kvihill 17:20, 16 October 2007 (UTC)
attacks in the past, and have been thoroughly debunked. The Glopk 20:28, 19 October 2007 (UTC)
Judea Pearl's book (1988) gives a Bayesian explanation for people's tendency to provide the (wrong) answer 1/2. After the hosts reveals that a goat is behind door 3, people tend to condition their beliefs on the revealed information "a goat is behind door 3" and obtain the answer:
The correct answer is obtained by conditioning on the total evidence available: "host revealed a goat behind door 3," giving:
The distinction between "information revealed" and "total evidence" has far reaching implications in reasoning under uncertainty [Pearl, 1990, 1992]
I get the feeling this is very reminiscent of Bell's inequality in relation to hidden variable theories in Quantum Mechanics - is the MH problem an analogy for that inequality perhaps? Fizzackerly 13:31, 19 October 2007 (UTC)
I was surprised to see that the "switch at the very end" strategy for n doors isn't better supported. Is this because there's no suitable reference, or is it really an open problem?
FWIW I think the following sketches a proof: when there are two doors left the sum of the probabilities of success must be equal to 1; maximising one is equivalent to minimising the other; not switching until the last chance minimizes the probability of the initially chosen foor being correct. —Preceding unsigned comment added by 190.160.252.137 ( talk) 19:49, 11 November 2007 (UTC)
I don't understand why the problem is different from the following situation: "Suppose you walk up to two doors, one of which has a car behind it. Someone tells you that there was a third door that didn't have a car but that it was removed. You must now choose one of the two doors." Why is this problem different? Why is "switching" different from "choose one of two doors?" RobertM525 03:35, 13 November 2007 (UTC)
I have a question. What if talk host doesn't know what is behind the doors, but is directed by some one, who knows what's behind, to open a specific door once the player selects a door? -- Venkataramana vurity ( talk) 16:45, 22 November 2007 (UTC)
I don't understand why it's important for the host to know what's behind the doors. Don't the following premises have the same odds?
Premise 1: "Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?"
Premise 2: "Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?"
I understand that it is improbable that the host could repeat premise #2 without eventually revealing a car, however that is a completely different issue than premise 2 as stated. —Preceding unsigned comment added by Banderson1962 ( talk • contribs) 22:38, 25 January 2008 (UTC)
This doesn't really answer the question "why is it important that the host knows what's behind the doors?" In fact, it makes no difference at all! Let's say Monty has no idea what door has the car and he's guessing as blindly as you are. Then the possibilities in the first round become 1. you picked the car and he opens a goat because that's all he has 2. you picked a goat, Monty has the car and a goat, but opens the goat randomly, and 3. you picked a goat, Monty has the car and a goat, and opens the door with the car. In scenario 3, the game is over and the second round, being moot, never occurs. However, in scenarios 1 and 2 the game goes on, and you are still confronted with the fact that when you went into the second round, Monty was twice as likely as you to have the car, and switching still doubles your odds of going home with that car. Drewtew ( talk) 06:33, 29 January 2008 (UTC)drewtew
An anonymous user inserted an aid to understanding by considering infinite doors. In this case, the first door picked has zero probability of hiding the car, and switching yields the car with probability 1. Though perhaps still a bit unpolished, I found it helpful. However, Mattbuck ( talk · contribs) undid the edit with the comment "doesn't help since it's not quite 0". How is the probability "not quite 0"? Phaunt ( talk) 11:50, 29 November 2007 (UTC)
Is the 2/3 probability supposed to be based on the initial door chosen, or the door chosen after there are 2 left and you're given the chance to switch? Because the diagram seems to indicate the former, while the jist I got from the article indicated the latter. Thanks. 71.127.243.28 ( talk) 01:24, 1 January 2008 (UTC)
There is a citations tag at the top of this article. If this article requires more citations, how is this a featured article? Moreover, it's references section does not use the proper reference tags.
So how can this article possibly be a Featured article? -- Son ( talk) 04:52, 2 January 2008 (UTC)
This article was more on-point when it was featured in July 2005, but even then it suffered the same flaw it exhibits today: matters of style aside, a "well written" article about a veridical paradox should successfully explain how common sense is led astray and convincingly explain the truth of the matter.
As discussion on the talk page indicates, subsequent elaborations such as detailed Bayesian analysis and digressions on variants, interesting though they may be, do not lend clarity. Perfectly intelligent people who fully understand the mathematics of probabilities can read the entire article and still be left suspecting some kind of elaborate academic hoax, like one of those fallacious proofs that 1+1=1.
The main problem is that Sources of confusion fails to address the fundamental misunderstanding that so easily ensnares common sense. The truth is in the article, but it is not clearly stated as the chief source of confusion: selective evidence.
I would be guilty of original research if I wrote an explanation of this, as I have not found a source with a clear and direct statement of the issue as it relates to this paradox in particular. A good article on evaluating selective evidence might merit more than low-importance and low-priority. As a cautionary example, this article's significance could be elevated as well, if only the source of confusion were not obscured by the elaborate explanations. 67.130.129.135 ( talk) 03:27, 4 January 2008 (UTC)
To play "Devil's advocate", I will semi-seriously challenge one assertion of the article. In essence, I will agree that the problem, when correctly stated, yields 2/3 as the probability of winning by switching. However, I will contend that for the version stated in the Ask Marilyn column, the answer is not "technically ambiguous" but rather 1/2 - slightly muddy, perhaps, but not as ambiguous as all that. To understand why, it's important realize that "probability" is inevitably a manifestation of incomplete knowledge. Outside of the quantum realm, complete knowledge of a situation invariably yields prediction values of either 1.0 or 0. Probability, on the other hand, is based on the knowledge at hand, and so to claim that a probability estimate is ambiguous because important information is lacking is to misconstrue the meaning of probability. In the Ask Marilyn version, we really know only that the car is not behind door 3 and must therefore be behind one of the other two doors. This yields the value of 1/2. Of course, we also know that the host opened door 3 to reveal a goat, but for practical purposes, this is not usable knowledge. Based only on what is usable, we arrive at the 1/2 answer. I suppose one could take a Bayesian approach to interpreting the host's actions, in which case, one could estimate various probabilities based on guesses about the host's motivation. If so, that should be stated. It would be equally reasonable, however, to simply say switching or staying are equally likely to succeed as long as we don't know what the host is up to. Fmoolten ( talk) 01:37, 20 January 2008 (UTC)
I would argue that the entire notion of probability is based on estimates of outcome under circumstances that are only partially defined. Completely defined circumstances yield probability values of 0 or 1 (excluding uncertainties in the quantum realm). "Defined" is therefore a matter of degree rather than an all-or-none concept. To say that the probabilies are equal when we don't know what the host is up to is consistent with this principle. Clearly, these circumstances are less well defined than in the correct version of the puzzle, but they still permit probability estimates based on the information available. An analogy exists with a coin toss. If all we know is that an ordinary coin is tossed, the probability of heads is estimated at 0.5. However, if the toss is better defined in terms of angle and height of toss, speed of rotation, center of gravity of the coin, and the nature of the surface below, the probability changes based on the greater degree of definition. In the Ask Marilyn puzzle, we at least know that the choice is between two doors. The puzzle would be more or less completely undefined if we didn't know how many doors were involved, and in that case, a probability estimate would be virtually meaningless. Fmoolten ( talk) 17:06, 29 January 2008 (UTC)
The host doesn't want to lose cars, but everyone watches the show and knows his pattern of behavior. If we assume all the players are smart, then there's no use for the host to only open a door if the player has picked the car, because then people won't switch. But if he always opens a door then all the players switch and win more often. He has nothing to win or lose if he opens a door 50% of the time when the player hasn't picked the car and 100% of the time when the player has picked the car, because then the player wins the car by switching as often as he loses it by switching. Nonetheless in the real world some people will always switch by mistake when given the option. So there's a trade-off between a notorious host who only opens the door when the player picks the car, but has very few mistaken switchers, and a subtle host who opens the door 40% of the time when the player hasn't picked the car, and gets many people who think it doesn't matter or even think it helps based on a few episodes. How to resolve that trade-off I can't tell say. But in a proper game against a cunning host it should be at least somewhat harmful to switch doors when the choice is offered. 70.15.116.59 ( talk) 20:36, 24 January 2008 (UTC)
The article gives the impression that it's important that the host knew along where the car and goats are. Aren't the results the same whether he knew or not, since you've already eliminated the cases where he opens the 'car' door? Robin Johnson (talk) 14:07, 1 February 2008 (UTC)
I had to work this out for myself though... it miiiight be possible to express it more clearly in the article. Robin Johnson (talk) 18:34, 1 February 2008 (UTC)
This edit added a section titled "The Rigorous Solution" which includes a Bayesian analysis where the host's probability of picking which of the remaining two doors to open (in the event that the player originally picked the car) is allowed to vary, and shows switching is optimal regardless of the probability assigned. This analysis is claimed to include the effects of the host's strategy, thereby showing switching is optimal regardless of the host's strategy.
Rather than debate the merits of this analysis, I'd suggest we not even consider including it unless a reliable source for it can be found. -- Rick Block ( talk) 16:54, 10 February 2008 (UTC)
This elementary Bayesian calculation is not research. It would be better for the readers to see this. The probability if winning the car given switching is more general than 2/3. It is 1/(p+1), which equals 2/3 when p=1/2. It can be between 1/2 and 1, but always not less than the probabilty of winning car by not switching, which is equal to p/(p+1), and therefore between 0 and 1/2. —Preceding unsigned comment added by 70.137.168.95 ( talk) 02:00, 11 February 2008 (UTC)
The problem with this article is that the main point is cluttered by so many uninteresting generalizations, whereas the original game (where the host always opens a door other than the one picked by the player, which also does not have the car) is not analyzed rigorously. The most general stratgey of the host subject to the original rules of the game is characterized by a probability p, which is that probability with which the host opens door 2, given that the player chose 1 and the car is at 1. (There could be other probabilities for the cases of the player picking 2 and 3, but it suffices to conside one case.) To ignore the full analysis of the original problem is not wise, especially when the analysis reveals the surprising result that it is always optimal to switch. To emphasize the importance of this point, the player deos not need to know p in order to conclude that switching is optimal. This is true for every p. To assume p=1/2 is too restrictive! I do not understand why you want to censor this analysis using your admin power. —Preceding unsigned comment added by 70.137.145.250 ( talk) 04:04, 11 February 2008 (UTC)
Here is the rigorous solution:
A rigorous solution requires addressing the question of the strategy of the host, namely, how the host picks a door to open, when more than one is possible. Following is a Bayesian analysis, which proves that switching is optimal regardless of the host's strategy:
The prior belief of the player is that the car is behind each door with probability 1/3. Suppose the player chooses door 1.
Denote by p the conditional probability with which the host opens door 3, given that the car is behind door 1. Thus, 1-p is the conditional probability with which the host opens door 2, given that the car is behind door 1. In the event the car is not behind door 1, there is only one door that the host can open.
Following are the probabilities and conditional probabilities that lead to the conclusion:
Prob (Host opens door 3, given that the car is behind door 1) = p Prob (Host opens door 3, given that the car is behind door 2) = 1 Prob (Host opens door 3, given that the car is behind door 3) = 0 Prob (Host opens door 3 ) = (1/3) x p + (1/3) x 1 + (1/3) x 0 = (p+1)/3.
The following equation holds:
Prob (Car is behind door 1, given that the host opened door 3) x Prob(The host opens door 3) = Prob (Host opens door 3, given that the car is behind door 1) x Prob(The car is behind door 1)
Therefore,
Prob(Car behind door 1, given that the host opened door 3) x (p+1)/3 = p/3
and thus
Prob(Car behind door 1, given that the host opened door 3) = p/(p+1).
It follows that,
Prob(Car behind door 2, given that the host opened door 3) = 1/(p+1).
Since p is not greater than 1,
Prob(Car behind door 2, given that the host opened door 3) >= Prob(Car behind door 1, given that the host opened door 3).
Therefore, switching to door 2 is an an optimal strategy for every p. —Preceding unsigned comment added by 70.137.145.250 ( talk) 05:44, 11 February 2008 (UTC)
(Re: changes by Rick Block made in response to this thread) Does it really matter whether the host makes a random or non-random choice when the two unpicked doors both contain goats?-- Father Goose ( talk) 02:06, 13 February 2008 (UTC)
RE: Rick Block (Father Goose, Please read this entire section "Rigorous Solution" in this Discussion, or the Morgan et al. paper.) What Mr. Rick Block is doing is harming the reputation of Wikipedia. He deleted the rigorous solution and put the following text as the solution: "If the player chooses to switch, the player wins the car in the last two cases. A player choosing to stay with the initial choice wins in only the first case. Since in two out of three equally likely cases switching wins, the probability of winning by switching is 2/3. In other words, players who switch will win the car on average two times out of three." As explained in (Morgan et al. 1991), this statement is "the most appealing of the false solutions." Mr. Block's solution, which is the current solution of the article, is the same as what is referred to as "false solution F1" in Morgan et al., who say about it the following (p.285):
Mr. Block deleted a previous comment about the difference between unconditional and conditional probabilities. Mr. Block monopolizes this article and lowers the level of the article. It is not only the differece between "random" and "non-random." The player has no basis to believe that the host picks one of the two doors with probabilities 50:50. The host may do it with randomly with different probabilities, say 75:25 or even 100:0 (i.e., deterministically). The conditional probabilities are different in these cases, even though the unconditional probability is 2/3. Interestlingly, switching is optimal in every case. However, Mr. Block twice deleted this fact, which is cleverly explained in Morgan et al. What is the recourse against such behavior? 70.137.163.193 ( talk) 04:42, 13 February 2008 (UTC)
So, am I to understand that the current brouhaha is over whether the host's choice of the two "unpicked" doors is non-random? If so, either:
If it's the first case, that's such a huge omission from the basic statement of the problem, surely it cannot be treated as an inadvertent oversight or potential ambiguity.
To use a different example, if the host always told the player truthfully where the car was, that would change his odds of winning to 100% in all games. However, while the original phrasing doesn't preclude that as a possibility, the fact that such an unusual condition is not included in the phrasing makes it safe to assume that it is not in effect. I would say that "player knows host's predictable behavior" is a similarly unusual condition, and not something that could be treated as an ambiguity in the original problem.-- Father Goose ( talk) 03:11, 14 February 2008 (UTC)
(outindent) Father Goose: Dropping the constraint that the host select from among two goat doors with equal probability introduces an ambiguity in the sense that the numerical probability of switching by winning is not a constant for all players regardless of what the host does. It creates a potential difference between the overall chance of winning by switching seen by players in the aggregate (the unconditional probability our anonymous friend of many IP addresses is talking about) and the chance of winning by switching as seen by an individual player at the point of being asked to switch knowing which specific door the host has opened (the conditional probability). In the problem statement (both vos Savant's and the unambiguous version in the article) the question is whether some individual player is better off switching, not "what is the aggregate chance of winning by switching". If we drop the constraint, and call the host's potential preference for one door over another p (the constraint forces p=1/2), the probability an individual player sees is 1/(p+1) which ranges from 1/2 to 1 (depending on p) although the unconditional probability seen by players in the aggregate is 2/3 regardless of p (indeed, a whimsical host might decide to pick the "rightmost goat door" on Mondays, Wednesdays, and Fridays and the leftmost on Tuesdays and Thursdays making p alternate between 0 and 1 depending on the day of the week with an aggregate value of .6). Again, this is exactly the point user:anon-many-ips is making. Note that without the constraint, the difference is there and affects the player's chances whether the player knows p or not, in exactly the same way that the player's probability of winning is 1/2 (both conditionally and unconditionally) if the host opens a remaining door randomly (the forgetful Monty version) whether the player knows how Monty has selected or not. -- Rick Block ( talk) 16:49, 16 February 2008 (UTC)
Here is verbatim what they say:
Then they say:
The readers can verify that F1 is indeed the same as the solution currently in the Wikipedia article. 70.137.163.193 ( talk) 06:17, 14 February 2008 (UTC)
Ok. I coughed up the $14 and purchased the online Morgan et al article (from [1]). The version of the problem they initially analyze is essentially identical to the original statement of the problem from Parade (with no constraints on the host's behavior). They label as solution F1 the informal analysis that (paraphrased) "a player who doesn't switch wins 1/3 of the time, so switching wins 2/3 of the time" and call this a false solution. Indeed, it doesn't address the original problem as stated in Parade as it ignores the effects of the host's behavior. Rather than focus on ambiguities in the problem description they pursue an approach where the host's behavior is represented as a set of probabilities, leading to the possibility of different probabilities of winning by switching depending on the specific door the host opens. Although many host behavior variants are covered by their approach, not all are (they specifically exclude "host opens the player's door", "host opens the door revealing the car", and "host makes the offer to switch more or less often depending on the player's initial selection").
Given a problem statement where the host's actions are effectively described as variables requires distinguishing the player's initial chances (the "unconditional" problem) from the chances given the door the host reveals (the "conditional" problem). However given the unambiguous statement of the problem in the Problem section of this article, the specific door the host opens does not (cannot) affect the player's chances of winning. There's a fairly long standing consensus that the version of the problem discussed here should be an unambiguous version, permitting only a single solution which does not need to (and does not) distinguish "unconditional" vs. "conditional" (this difference is prohibited by the problem statement). IMO "Other host behaviors" would be a fine place to include a discussion of this issue. -- Rick Block ( talk) 03:47, 14 February 2008 (UTC)
More formally, when the player is asked whether to switch (assuming the player initially picked door 1) there are three possible situations corresponding to the location of the car, each with probability 1/3:
The host opens only one door, so only one of the first subcases with probability 1/6 and one other case with probability 1/3 apply. Staying wins in the 1/6 case where the player has initially picked the car while switching wins in the 1/3 case where the player has not. Switching wins twice as often as staying, so the probability of winning by switching is 2/3. In other words, playersa player who switches will win the car on average two times out of three.
Suppose the player picked door 1. Before the host has opened any door, there are three equally likely possible situations, namely, the car is behind each door with probability 1/3. It is assumed that if the car is behind door 1, then the host opens door 2 or door 3 with equal probabilities. If the car is behind door 2, then the host opens door 3, and if the car is behind door 3, he opens door 2. Therefore, the probability that the car is behind door 1 and the host opens door 3 is (1/3)x(1/2)=1/6. The probability that the car is behind door 2 and the host opens door 3 is (1/3) x 1 = 1/3. These are the only scenarios under which the host opens door 3. Therefore, the probability that the host opens door 3 is 1/6 + 1/3 = 1/2. Given that host has opened door 3, the conditional probability that the car is behind door 1 is (1/6)/(1/2)=1/3, and the conditional probability that the car is behind door 2 is (1/3)/(1/2)=2/3. Therfore, given that the host opened door 3, switching wins the car with probability 2/3. Likewise, given that the host has opened door 2, switching wins the car with probability 2/3. 70.137.163.193 ( talk) 16:44, 14 February 2008 (UTC)
(outindent) You keep saying this, but what exactly about it is false? You didn't respond to the suggestion (previous section) about expanding it from 3 to 6 cases (treating each possible door that can be opened separately). Even in the 3 case version it explicitly enumerates the cases in effect after the host has opened a door (either door). It doesn't use the words "conditional probability" and doesn't show the derivation of the equal probability cases, but is this necessary? How about:
I'm not overly attached to this description, but just want to understand what the objection is. -- Rick Block ( talk) 23:08, 14 February 2008 (UTC)
It is all from the point of view of the player. The figure in the current solution section is very misleading for the layperson, because it does not show what the player knows (the player knows that he has chosen door 1):
-- 198.4.83.52 ( talk) 20:06, 15 February 2008 (UTC)
So far so good.
How would the solution described in the text above "so far so good" change if the host opened one of two goat doors non-randomly?-- Father Goose ( talk) 22:17, 16 February 2008 (UTC)
More formally, when the player is asked whether to switch there are three possible situations corresponding to the location of the car, each with probability 1/3. Assuming the player initially picked door 1:
The host opens only one door, so only one of the first subcases and one other case with probability 1/3 apply. In the case where the player has initially picked the car, staying wins with probability 1/3*p or 1/3*(1-p) depending on whether the host opens door 3 or door 2. Switching wins in the 1/3 case where the player has not. Since p is a probability it ranges between 0 and 1, so switching always wins with at least the same probability as staying regardless of which door the host opens — and depending on p and which door the host opens might guarantee winning the car. In other words, a player will win the car more often by switching than by staying with the initially picked door.
If all players switch regardless of p and what door the host opens, their average probability of winning is 2/3 since 1/3 of these players will initially pick the car and lose while 2/3 will initially pick a goat and win.
Moved from User talk:Father Goose
Any idea why Talk:Monty Hall problem got so quiet all of a sudden? I thought we were almost there. -- Rick Block ( talk) 01:01, 20 February 2008 (UTC)
(reset indent)All right, I'll take a different tack. Is it not possible to formulate a unconditional conditional analysis that represents the fact that any time the player picks the car first, he will lose when switching, no matter which behavior the host adopts when choosing between two goats?--
Father Goose (
talk)
20:50, 21 February 2008 (UTC)
Excuse me, I meant to say a conditional analysis.-- Father Goose ( talk) 00:36, 22 February 2008 (UTC)
An anonymous user (editing from a variety of IP addresses) has expressed a concern about the solution presented in the article. All of the discussion above, starting with the section #Rigorous solution relates to this issue. Should the wording of the primary solution be changed and is the following suggested wording satisfactory?
More formally, when the player is asked whether to switch (assuming the player initially picked door 1) there are three possible situations corresponding to the location of the car, each with probability 1/3:
The host opens only one door, so only one of the first subcases with probability 1/6 and one other case with probability 1/3 apply. Staying wins in the 1/6 case where the player has initially picked the car while switching wins in the 1/3 case where the player has not. Switching wins twice as often as staying, so the probability of winning by switching is 2/3. In other words, a player who switches will win the car on average two times out of three.
After the player has chosen an initial door (assuming door 1) but before the host opens a door, there are three possible situations corresponding to the location of the car — each with probability 1/3:
The host opens only one door, so at the point the player is asked whether to switch only one of the first subcases with probability 1/6 and one of the other cases with probability 1/3 apply. Staying wins in the 1/6 case where the player has initially picked the car while switching wins in the 1/3 case where the player has not. Switching wins twice as often as staying, so the probability of winning by switching is 2/3. In other words, a player who switches will win the car on average two times out of three.
Replace the last paragraph by the following:
The host opens only one door. Consider the case where the host opened door 3 (the same argument applies as well to the case where the host opened door 2). The player is then given the option to switch. In this situation, the possibilities are only two: (i) The car is behind door 1 and the host picked at random door 3 (originally, this case had probability 1/6). (ii) The car is behind door 2 and the host had no choice but open door 3 (originally this case had probability 1/3). These two cases have a total initial probability of 1/6 + 1/3 = 1/2. Staying wins in case (i) where the player has initially picked the car, which amounts to a proportion of (1/6)/(1/2) = 1/3 (see
conditional probability). Switching wins in case (ii), which amounts to a proportion (conditional probability) of (1/3)/(1/2) = 2/3. Intuitively, switching wins twice as often as staying, so the probability of winning by switching.
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70.137.165.14 (
talk)
06:57, 26 February 2008 (UTC)
Here's a way I find easy and simple to understand: When you choose a door the first time, 1/3 chance you got the car. So theres a 2/3 chance of the car being behind one of the other TWO doors. Now if we get rid of one of the other 2 doors, then its 2/3 chance of it being behind the remaining ONE door. LordRobert ( talk) 07:19, 23 February 2008 (UTC)
(outindent) We're clearly talking past each other. With our constraints (specifically including "host picks randomly between two goat doors") the answer is 2/3 in every case regardless of which door the player picks (so whether it's 1, 2, or 3 doesn't matter) and no matter which door the host subsequently opens. IMO this is a good property to have and allows the illustrations and specific problem setup (player picks door 1, host opens door 3) to be representative of all cases (it doesn't matter - they're all the same!). In addition, because it's the answer in every single possible scenario, it's also the average outcome considering all scenarios. I also think it's the (surprising) answer nearly anyone posing this problem presents as the solution. I think we all agree on this so far.
On the other hand, both problem statements we present (the one from Parade, and the one from Mueser and Granberg) present the problem from the perspective of a single player after the host has opened a door (Parade: You pick a door, say No. 1, and the host ... opens another door, say No. 3. Is it to your advantage to switch your choice? - Mueser and Granberg: You begin by pointing to door number 1. The host shows you that door number 3 has a goat. Do the player's chances of getting the car increase by switching). I think in both versions the answer is not meant to vary depending on the player's or host's actions, so it is perhaps ambiguous whether the actual question is "considering all possible scenarios what is the player's average chance of winning by switching" or "given a player has picked a specific door and the host has opened another specific door, what is this specific player's chance of winning by switching". Given the way it's worded, I think the latter interpretation is actually more sensible, but I accept we may not all agree on this. Whether we agree or not, I think we can (and should, but perhaps don't yet) agree that these questions can have different answers depending on the constraints.
If these questions might have different answers, can we agree a solution approach for one may not be appropriate for the other? Taking this a little further, can we agree that if we answer the player specific question for all possible scenarios we assuredly have the answer for the more general question but not necessarily vice versa?
There are 6 possible cases of player picks and doors the host might open (PHx, PxH, HPx, HxP, xPH, xHP). It seems we're showing the overall probability is 2/3 and wanting (without any particular justification) for the question to be "considering all possible scenarios". This leaves open the possibility that other people are reading the question as "specific player, specific door the host opens" (which is where our anonymous friend entered the picture). If we show both that the probability is 2/3 for each and every player because of our constraints and that the overall probability is 2/3, isn't this a better approach? -- Rick Block ( talk) 03:43, 28 February 2008 (UTC)
Yes, we are talking past each other; we are being repetitive; we are having great sport going round in circles. Can we come down to earth for a moment? Does anybody seriously question whether it is overwhelmingly notable that the problem was originally intended, and is most widely interpreted to be about probabilities of choice, and not about strategies in a game between two contestants? Does anybody dispute that, thusly interpreted, it is a very interesting veridical paradox, a moderately important layman's puzzle, and a perfectly notable encyclopedic topic? If not, let us keep the main article on topic by eschewing considerations of host strategy and allowing that the goats are as indistinguishable to the host as the doors are to the guest. If so, please explain.
To return to the interest, importance, and charm of the problem, it is just this:
Variant interpretations are interesting, notable, and encyclopedic digressions, but they should be presented in a way that does not detract from this central theme by unduly complicating the statement of the basic problem or its basic solution. 67.130.129.135 ( talk) 03:11, 29 February 2008 (UTC)
Let me try once more. I'm sorry if the following only indirectly answers Rick's questions.
Enough facts. With the added clause (which could be improved by deleting the word "special"), there is a simple analysis showing that the switching strategy always increases the chance from 1/3 to 2/3. Without that clause, but with another additional clause that is required then, namely that the player does not learn anything that is not a consequence of which door was opened (possibly because the host has preferential behaviour of which the player is aware), you can show that the always-switching strategy is an optimal strategy that increases the a priori chance (before the host opened the door) from 1/3 to 2/3. The proof of 2/3 is as before, but the proof that this is optimal requires, as far as I can see now, an elaborate case analysis (but maybe I'm wrong and there is an easy proof for this too). In either case, the frequentist interpretation is appropriate: you can't win two thirds of a car, but if the game is repeated often enough about 2 in every 3 players who follow the optimal strategy will go home in a car instead of on a goat.
Summing this up. Without some clause added to the M&G formulation, the problem cannot be analyzed. With an added clause, perverse interpretations may still be possible. Ignoring such perversions, one added clause (the one now in the article) allows a simple complete analysis; the other possible choice for an added clause (which is an awkward thing to formulate; an attempt is found above, which in fact rules out the perverse interpretation) still allows a simple proof – the same as before – that always switching is better on the average, but leaves the possibility open that it is not better in all situations, and a more complicated proof is needed to show this strategy is nevertheless optimal. --
Lambiam
20:39, 29 February 2008 (UTC)
A statistician is given the following problem:
The statistician is then told that he will win the car outright if he can provide a single numerical answer as to what his probability of winning by switching is. The statistician says that there are two probabilities he faces, which cannot be combined in any way to produce a single probability that would correctly describe his odds of winning by switching. He additionally asserts that there is no valid way of analyzing the problem as presented that could correctly produce a single numerical answer. He leaves the show empty-handed.
Was the statistician right?-- Father Goose ( talk) 01:58, 29 February 2008 (UTC)