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I've moved the existing talk page to Talk:Monty Hall problem/Archive1, so the edit history is now with the archive page. I've copied back the most recent thread. Hope this helps, Wile E. Heresiarch 04:40, 10 Aug 2004 (UTC)
Recall that the host has opened a door, and we know that door contains a goat. The contestant was either RIGHT or WRONG in the first place. In the former case, we switch to a goat and lose. In the latter case, switching guarantees a victory: we picked Goat 1 originally, and Monty has released Goat 2. So the only probability we need to know is that of being right or wrong in the first place, which are 1/3 and 2/3 respectively. 1/3 chance you will lose if you switch, 2/3 chance you will win.
Just a stupid question. Suppose you have the three doors
Door 1: Selected Door 2: Opened(goat of course) Door 3: Unknown
Since the contestant knows behind 2 doors is a goat, the Goat One from the opened door could be romoved. That leaves a fifty-fifty percent chance that he/she is correct in thier guess, so switching would not increase it by 2/3 because there are only 2 unkown doors left. I've proabably not followed this enough so tell me why I'm wrong soon please.
How does it leave 2 doors with unequal probabilities? Of two doors left, one has to have the goat and one has to have the car, becuase the host always opens up the goat.
I still don't really understand because this is different from a deck of cards. A deck of cards cannot be compared to this because a deck of cards has:
a) more choices
b) be more then one, say door.
If I put up fifty doors, and I had, lets just say, ten variations, four suites, then there is a much different probabilty then knowing that threre are only two doors, of which on is the goat and one is the car. The problem does not need the third door any more. If, say you had 34 aces of spades, and 17 aces of hearts you still would not get the same equation because it is mroe variables. You know there is a fifty fifty percent chance the door that you first seleceted has the car because the host opens a goat. You either have or don't have the goat, so fifty-fifty chance. Now if you siwtched, it would not imrvoe your chances.
But wait... if you discard a card forever, now there will be only two cards left making it one out of two.
What's wrong?
Am I right or Am I right or Am I right?
try to do it later.
-- 213.148.18.178 10:57, 17 Jan 2005 (UTC)
Except do not forget, Monty's choice is constrained, since he MUST open a door revealing a goat, in the above P(O3) = 1, not 1/2. I have deleted the sequel, as it is superfluous given my previous statement. -Dwee I have restored that text, since it's easier to see where the mistakes in a text are when you can actually look at the text. -- Antaeus Feldspar 02:54, 7 Jun 2005 (UTC)
I was talking to a friend about the Monty Hall problem today and he told me about a similar problem, and I don't think there is a Wikipedia article on it and I'm not sure how to present the solution either, but anyway here is the problem:
You are on a gameshow and the host holds out two envelopes for you to choose from A and B. So you choose an envelope (A) and it's got $2000 in it. The presenter then says that one of the envelopes has twice as much money in it as the other one and offers you the chance to switch. So you think about it this way... "If I switch I will go home with either $4000 or $1000, by not switching I will go home with $2000. There is a 50/50 chance that I will double my money by switching. A normal 50/50 bet results in me either doubling my money or losing it all, whereas here I will only lose half. Therefore this is a better than evens bet so I will make the swap." You are just about to swap envelopes when you think about the problem some more - "Surely this can't be right... ". Mintguy (T) 16:13, 15 Jul 2004 (UTC)
[Tired of nesting the blocks deeper and deeper, like bad C code.] Let's pause to list assumptions. My choice of an envelope is not correlated with the loading of the envelopes, so that I'm equally likely to have the good or the bad envelope. The host also is statistically unbiased: his telling me the news is not correeated with my initial choice of good or bad envelope. Oh, and he's telling the truth.
This is not to say that the expectation argument is right. This is the paradoxical part: that the argument has no apparent flaw in itself, but it gives the nonsensical result that one should choose and then change, even though no new information has come along to cause a change. The host's information is new, or seems to be, but how would my course of action be different if I had known it all along? Bottom line: the expectation argument leads to a silly result, but I don't believe that its flaw has been shown. Maybe this paradox deserves its own article. Dandrake 19:35, Aug 26, 2004 (UTC)
I think I'm not understanding this problem correctly. The way I read it, you either pick an envelope with X or 2X dollars in it with equal probability. Given the option to switch, this expands into four cases
Picked X | Picked 2X | |
Keep | X | 2X |
Switch | 2X | X |
Doesn't this mean that each of these events occurs with equal probablity, and that it doesn't matter? I understand the expectation argument, but I can't reconcile it with this simple grid. Cvaneg 23:13, 26 Aug 2004 (UTC)
This problem now has its own page at Envelope paradox Mintguy (T)
I'm copying this to talk:Envelope paradox. Further discussion of this subject should continue there, but I will also leave the text above so that people can follow on from here. Mintguy (T)
I decided to be a little bold and remove the assumptions section. If you read the problem as stated, then you don't need to make any of the assumptions listed. E.g., it doesn't matter whether or not you assume "Monty always opens a door," or that there is "always a goat behind the door Monty opens," because the problem clearly states that Monty opens a door and reveals a goat. That is all you need to know in order to determine the correct answer. Other points in this section also showed an incorrect understanding of the problem. -- Simoes 15:53, 24 Jul 2004 (UTC)
It's critically important to the problem that Monty knows what's behind each door. In the actual correct problem situation, there are three scenarios for what's behind each door:
Scenario | Door 1 | Door 2 | Door 3 |
---|---|---|---|
Scenario A | Car | Goat | Goat |
Scenario B | Goat | Car | Goat |
Scenario C | Goat | Goat | Car |
Let's presume that the player always first chooses Door 1 (a safe assumption, since we can always just choose how the doors are numbered after the door is chosen.) If we assume that Monty knows what's behind all doors, and opens the door he does because it has a goat behind it, then our three scenarios look like this (with the revealed goat crossed out because the player knows to avoid it):
Scenario | Player's first choice | Door 2 | Door 3 |
---|---|---|---|
Scenario A | Car | Goat | |
Scenario B | Goat | Car | |
Scenario C | Goat | Car |
If on the other hand we leave the problem too vague, by just saying "Monty opens a door that has a goat behind it", we may create the impression that it is just chance that, this time, the door had a goat behind it -- and this changes the problem completely. Because it simply makes it look as if some of our original scenarios could not have happened:
Scenario | Player's first choice | Door Monty opens | Door 3 |
---|---|---|---|
Scenario A | Car | Goat | |
Scenario C | Goat | Car |
The faulty logic thus goes "If Scenario B had been correct, there would have been a car behind the door Monty opened; but there wasn't, so therefore Scenario B is not correct. The only remaining scenarios are A, where the player's first choice contains the car, and C, where the player's first choice contains the goat; therefore the chances are 50/50 whether the player switches or stays." -- Antaeus Feldspar 20:44, 21 Nov 2004 (UTC)
The standard solution seems illogical for 2 reasons. First, probabilities describe chances, and since Monty has removed one of the uncertainty by opening up a door, he changes the conditions which the original set of probabilities described. This is to say, the 2/3 probability should not be transferrable, and the new probability designed to describe the new conditions should state that the car has a 50-50 chance of being behind either of the unopened door. Second, say the standard solution is logical, and say the car is behind door 1, and the player will always choose between door 1 or 2, and Monty will always open up door 3: if the player had originally chosen door 2, then he should change his selection to door 1; if the player had originally chosen door 1, then he should change his choice to door 2: the results of the 1st situation contradicts that of the 2nd, and vice versa; the contradiction shows that the standard solution is wrong, and it doesn't make a difference whether to stick to the original choice or not.
I found the section on "Assumptions" to be a little confusing and to obscure the true spirit of the puzzle/problem. I have therefore removed that section and reworded the problem in the introduction to make it clearer that the host knows what is behind the doors and that he always opens a door with a goat. Then the assumptions are redundant and the intentions of the host, benign or otherwise, are not a factor. I think it is obvious that the contestant will want a car, not a goat, and there is no need to state this. I have also reworded the answer to try to make it clearer and more concise.
P S K -- 213.122.39.201 17:00, 25 Oct 2004 (UTC)
Hi P S K, you made the much needed changes to one of our important articles. -- Sundar 04:40, Oct 26, 2004 (UTC)
I have written up what I think is a clearer explanation of the solution, at Monty Hall problem/temp1. I'd like others' feedback, to find out if it's as clear as I think it is and can go in the article. -- Antaeus Feldspar 01:38, 1 Dec 2004 (UTC)
The Monty Hall problem is not a problem in people's skill at using probability and mathematics. It is not a problem of gullible people falling prey to tricks. It is not even a paradox. It is a problem in communication and rhetoric.
As commonly stated, the Monty Hall problem is ambiguous as far as how the host chooses the second door.
Whether information about the third door can be inferred by the host's actions, depends on whether the host uses information about what's behind the third door to make his decision.
If the host chooses the second door randomly among the two doors not already selected by the player, and it has a goat behind it, then there is no advantage to the player switching.
If the host chooses the second door because it has a goat behind it, then the probability that the third door has a car behind it jumps to 2/3, and the player should switch.
If someone were to ask me the Monty Hall problem, and it were not clear whether the host used information from behind the third door (or, equivalently, information from behind both the first and second doors) to make his decision, then I would come back and say that the problem was ill-defined. Neither of the two common solutions would be acceptable.
Actually, a third scenario is possible: If the host chooses the second door because the third door contains a goat, then the player should never switch.
The point is that you cannot infer information about the third door unless you can infer information about the host's decision-making process. Unless this is spelled out clearly (which it most often isn't), the problem is ill-defined.
Neither side in this "debate" has done bad mathematics, and mathematicians who righteously scoff at those who express contrary views, calling it bad mathematics, or mathematical illiteracy, add nothing to our understanding.
I'd like a humanities professor's analysis of this problem, because it is not fundamentally mathematical. I'd like to see a detailed analysis of all of the different precise phrasings of this problem, and how they parse against accepted English grammar. What assumptions are socially "valid"?
An analysis of rhetoric is more important than an analysis of probability, in this problem.
(FWIW, I'm a computer scientist who writes mathematical software.)
Sorry, but I can't force myself to agree wtih your solutution. When the host opens the door he not only skips over the dorr you choose, but the other door. There is a 2/3 chance it was the goat...right? But no remove a goat, just remove the door the host opened. You would not, in your right mind, choose a goat against a car. Now there are two doors left. One is your door with car or goat, and other a door with same. The problem with most of the tables and charts here are that they use goat twice in two rows, making it two chance. Now, goat one and two are interchangable:
Player Chooses Goat One:Remaining Choice: Car
Plyaer Chooses Car: Remaining Choice: Car
You should not add:
Player Chooses Goat 2: Remaining Choice Car
That would not happen. There are three doors, of which one of them is open, which goat DOES NOT MATTER EVERYONE. There will always be two doors left open, not matter if the players chooses goat one or two. There will alwys be a car and a goat remaining. If the player choose goat one and goat 2 is opened, you can't make another choice be player chooses goat two and goat one is opened. With this amount of doors they result in the smae thing unless you did it more often. Try it.
Door One: Door 2: Door 3: Goat Goat Car
Now to help you if you still don't get it: Door One or Door two can be opened and reslult in the smae outcome. Which door they open does not matter. Goat
Very good explanation. Probability applies to a theoretical problem abstracted away from the Monty Hall Problem itself. But, it would be really interesting to have humanities' point of analysis. -- Sundar 09:35, Jan 25, 2005 (UTC)
Yep, it's impossible to reason about probability after the fact without some prior assumptions about the process. It's like someone who made a bet with a gambler and found that the coin came up 100 heads in a roll and claim that the coin is biased. It's entirely possible that a fair coin produced those outcomes, it's just as likely as any other sequence.
It has nothing to do with what Monty knows. Like the solution states, there's a 2/3 probability that the car is behind one of the doors you didn't pick. It doesn't matter if the doors are purple, have aliens behind them, are invisible or any of that other nonesense. There's a 2/3 probability that the car is behind one of the doors you didn't originally pick.
P | M | U |
---|---|---|
C | G1 | G2 |
C | G2 | G1 |
G1 | C | G2 |
G1 | G2 | C |
G2 | C | G1 |
G2 | G1 | C |
P | M | U |
---|---|---|
C | G1 | G2 |
C | G2 | G1 |
G1 | G2 | C |
G1 | G2 | C |
G2 | G1 | C |
G2 | G1 | C |
P | M | U |
---|---|---|
C | G1 | G2 |
C | G2 | G1 |
G1 | G2 | C |
G2 | G1 | C |
I realize that I am probably wrong. I mean, if this phenomenon weren't mathematically solid it wouldn't have survived for this long, but something just occurred to me.
How can my picking a certain door affect it's probability?
Consider the following:
It's the exact same set up, with three doors etc. etc. In my mind I pick door 3, but I tell the host I pick door 2 and he believes me. In his mind I've picked door 2. He opens door number 1 and it is a goat.
Which door has the two third probability and which has the one third?
How can the exact same scenario bring up two probabilities based on the door I "picked?" Picking out one door in my mind doesn't mathematically substantiate anything as it has no relevance to anything mathematically.
The problem is not asking us to plot out an entire course of action beforehand. At the point that the decision in question is being made, Monty has eliminated one of the initial possibilities and presented the contestant with an essentially new problem: There is a car behind one of these two doors, pick one. Anything that has gone before is irrelevant. Referring to the decision as "switching" is a red herring.
Please disprove this before reverting the article. -- El Mariachi
Here's one last attempt to explain -- not to trolls, but to anyone who still honestly doesn't see why the probability works out as it does:
Player A and Player B take the 13 diamond cards out of a standard deck of cards. The cards are shuffled, and then Player A receives one card face-down, which he may not look at. Player B gets the other 12 cards, and he may look at them. Both players are trying to wind up with the Ace of Diamonds in their hand.
Player B has twelve cards which he can look at. At least eleven of them are not the Ace. Player B takes out eleven non-Ace cards from his hand and lays them down face-up.
Player A now has an option: he can stay with his current hand, with the one card he was originally dealt, or he can switch his hand with Player B's hand, which was originally dealt twelve of the thirteen cards.
The only differences between this problem and the canonical Monty Hall problem is the scale (13 instead of 3), the objects (doors to be opened instead of cards to be turned over) and the fact that Player A has his initial card/door assigned, rather than selecting it himself. None of these factors makes a difference to the central point, that the chances are better of winning by switching. The only one that comes close is Player A being assigned his initial choice, and since it violates the problem rules to posit that Player A could somehow pick the winning door/card initially at a rate better than chance, it does not make a difference.
Anyone who still believes that once Player B has discarded eleven non-Ace cards from his hand, the chance that he's still holding the Ace drops from 12/13 to 1/2 -- come and see me in person, and we'll take the diamonds out of a deck and play, and every time the Ace is in Player B's hand, you pay me $100, and every time it's in Player A's hand, I'll pay you $200. And I'll be merciful and quit after I've taken $1000 off you. =) -- Antaeus Feldspar 20:45, 13 Feb 2005 (UTC)
Your attempts at attackign a correct answer with aimless little jabs of "you're just trolling!" are sad. You have failed to make any logical point. You are the troll here. The chance of switchign then picking either car or goat are 50/50 PERIOD. Yes the initial "draw" determines where each object is. Irrelevant. Human action or choice plays no part...whether a random die roll or a person chooses the doors, is also irrelevant. Nothin to do with "uintuitiveness". Just simple, correct logic & math. Anything else is garbage 7 a sidetrack & no one has ever remotely come near confronting the absolute empirical black & white fact that it IS a simple 50/50 choice. Exactly because of the initial dispersion of objects, you choice & if you switch or not make no impact. -- unsigned comment by 71.115.25.22 ( talk · contribs)
Ok, I've been reading the explanations for the probabilities for this Monty Hall Problem for quite some time, and I think I've come up with a much easier way to understand the answer than what I've heard so far. Rather than type out long, drawn-out mathematical equations and using all these other instances, let's look at it logically. The way I understand the problem, the contestant picks a door, the host opens a wrong door, and then the contestant is given a chance to switch. Ok, if the contestant picks the right door, with the car in it, and then switches, the contestant will be lose. There is 100% probability of this. If the contestant picks either wrong door, and then switches, then he/she will win. 100% probability. So let's look at the original odds. The chances of initially picking the right door are 1/3, and since this is the only way to lose if you switch, your chances of losing are indeed 1/3. In the same way, your chances of picking a wrong door initially, and therefore wi nning, are 2/3. How could anyone still not understand this problem?
You, the candidate, pick a door. Probability dictates that in 1 case out of 3, you picked the car. This case will now be called (C). In 2 cases out of 3, you will have picked a goat. This will now be called (G).
(C): The host will open one of the two remaining doors, revealing a goat. Changing doors will let you loose, as you picked the door with the car already; STAYING with the first choice will always win in case (C).
(G): The host will have no choice which of the remaining two doors two open, as only one contains another goat, which he is bound to show. So in case (G), CHANGING the door will always win, while staying with the goat is a clear loss.
Conclusion: It can be seen easily that in 1 out of 3 cases, staying with your originial choice will be a sure win, while in more probable 2 out of 3 cases changing doors will be a sure win.
790 12:52, 19 Feb 2005 (UTC)
Here, for those of you who don't get the math (or think you do but still think there's a 50/50 chance at the end): Monty Hall applet. Play the game over and over. It keeps track of your wins, and whether you won by switching or staying. Still not convinced? Try the iterative applet. Run it for a few thousand iterations. The statistics are borne out. -- Wapcaplet 00:20, 2 Mar 2005 (UTC)
Saying that this is a paradox simply because to some people it is counterintuitive, is like saying "My good friend Jim, I thought you were a natural blond, but now I understand that you have died your hair! I didn't expect that at all; what an amazing paradox." A paradox has a precise mathematical meaning; anything and everything that isn't what you expect isn't a paradox. The monty hall problem is perfectly reasonable and contains no mystery or internal contradictions whatsoever.
A paradox arises when there are two (at least partially) valid and contradictory conclusions. The intuitive answer of "50/50" is demonstrably wrong and completely indefensible. Therefore even if it could have been regarded as a paradox, it is entirely resolved for anyone who takes the time to think about it. To continue to claim that it is paradox, you either have to demonstrate that something is still a paradox even after it has been resolved to the satisfaction of all mathematicians and logicians, or claim that the two answers both have validity.
In casual usage it might be regarded as a paradox. --a
A Google Scholar search of "Monty Hall" and "Paradox" reveals that it is almost always not referred to as such in scientific papers, since nearly all the results returned talk about Monty Hall and also happen to mention something else, which provides the paradox part of the search. The most common way to refer to it in mathematical papers is "The Monty Hall dilemma." The conspicuous exception is "Increasing working memory demands improves probabilistic choice but not judgment on the Monty Hall" by T Ben-Zeev, J Stibel, M Dennis, S Sloman, the purpose of which is to argue that the chance of winning is actually 1/2. This reinforces my point: It only seems like a paradox to those who do not accept that the answer is unequivocally 2/3 and not 1/2.
And anyway, I am aware that there is a common definition which means that if something is simply surprising, then it is a paradox (although it should no longer be surprising after the mathematics behind it are demonstrated). What is your problem with the current revision? If you do revert, PLEASE for the sake of accuracy take out that utter crap about it being a paradox by the "mathematical definition," which is precisely what it is NOT a paradox by.
If you are going to cite evidence, use outside material, not wikipedia. If you'll check the discussion page for paradox you will find that there is considerable controversy about many of them (including Monty Hall) and even about the formal definition that is given in the first paragraph (the "apparently" part is bullshit, as several have pointed out). --a
Oh, now all of a sudden Feldspar wants to allow for broad usage of terms that have multiple meanings? Hmmm... I suppose it's ok then to call John Kerry's 1st 1st "wound" minor, eh? This unsigned comment is by 216.153.214.94, determined by a ArbCom ruling to be a sock puppet of Rex071404.
From The Collaborative International Dictionary of English v.0.48 [gcide]:
So that's what I found in the dictionary, for what it's worth. -- Wisq 18:41, 2005 Apr 24 (UTC)
The issue of the Marilyn vos Savant article was not that she was "wrong" but rather that she was mis-stating the problem. The key word she used was "guess" in regards to making the 2nd selection. By definition, "guessing" requires one to choose without making calculations. When one calculates before "guessing" it's no longer a mere guess, but an educated guess.
The other problem with her article was that the concept of odds has a dual layer in this "Monty Hall" problem which is often overlooked. The odds of finding the car do increase by switching. But the statistical likelyhood that it exists, does not. Additional information can always increase the likelyhood of a successful choice, but it does not affect the underlying statistical facts. If there are 100 piles of horseshit in your backyard and a single can of caviar buried in one of them, the singleness of the can of caviar never changes, only the odds of finding it. Suffice it to say, I have found that most people who argue about Monty Hall problems, overlook this point. Please see this variant [1] of the Monty2.gif from the article page. I suggest that this puzzle would be less puzzling if the location of the car was referred to as "odds of finding card here" rather than "odds of car being here". 216.153.214.94 06:04, 24 Mar 2005 (UTC)
I agree the probabilities are 1/2 when picking the second door, and it doesn't matter why monty picks his door because if he does reveal the prize then the game ends with you winning anyway.
I added the following statement, but it was removed with the hint, that though it is accurate, it would be "polarized and confusing" (quote Antaeus Feldspar).
Often it is suggested to increase the number of doors to help to understand the problem, for instance to 100. This approach is illegal, since when transforming the special case t=3 into the general case t=n, the order to the show host can be interpreted in at least two different ways:
Only for n=3, both interpretations are exactly the same.
The supporter of this approach refer to the second interpretation. In this case, the chance raises from 1/n to (n-1)/n, so for n=100 from 1% to 99% (in 99 out of 100 cases the candidate does not choose the prize, and in all of these cases a change to the one remaining door will lead to the prize).
The first interpretation is nevertheless just as legal and more than that also more evident than the second one. Even here the chance raises, but only slightly from 1/n to (n-1)/n x 1/(n-2), so for n=100 from 1% to ca. 1.01% (in 99 out of 100 the candidate does not choose the prize, and in 1 out of 98 cases the prize is then behind one of the many remaining doors).
For n=3, both terms - (3-1)/3 x 1/(3-2) and (3-1)/3 - result to 2/3.
So the change is a good choice for both general interpretations, as well for the one special case t=3. But it is illegal to refer to the alleged obviousness of just one general interpretation to illustrate the special case t=3.
I'm sorry if it seems polarized, but it refers to a very common approach, that is unfortunately absolutely wrong (please proof if I'm wrong). And unfortunately this approach is used in a very offensive way, like "you are wrong, and you can easily see it if you change the problem a little bit". If you change a problem, then you just CAN'T transfer the solution from the changed problem to the original problem. This is a very common mistake when dealing with mathematical problems. And unfortunately this whole thing IS about accuracy!!! And I am sorry if it is confusing, but mathematics is hard to understand some time. I hope I don't appear arrogant or something, but I think it would be a shame if such a prominent and good article contains a wrong explanation.
Maybe it is confusing because my English is bad. But in my opinion, it is better fpr an encyclopedia to be confusing than to be wrong!!!! So please discuss or improve my text, but PLEASE don't let wikipedia become a propagator of mathematical "superstition" or something!!!! :'-(
-- Abe Lincoln 14:49, 29 Apr 2005 (UTC)
Important addendum: Even if there would be only one interpretation, inferring from one special case to another special case is not a valid mathematical method for proofing a solution. It is only legal to infer from a general case to a special case. Though the general case has the "change is good"-solution for n>3 for both interpretations (see above), it is not legal to illustrate the special case n=3 with the more obvious special case n=100 with just one random interpretation. So this approach seems wrong in at least two directions:
-- Abe Lincoln 15:26, 29 Apr 2005 (UTC)
The following paper is one of the better explanations of this problem that I could find online: http://econwpa.wustl.edu/eps/exp/papers/9906/9906001.pdf I believe that it explains much more clearly why 1/2 can be a reasonable answer, as compared to this current article. -- Wmarkham
I agree. However, as presented in the current version of the article, that interpretation is entirely valid. In other words, the current presentation of the problem is insufficiently specific. Furthermore, the entire reason that this puzzle is a "paradox" is due to the confusion over what question is being asked, so it seems to me that this article ought to explain it clearly. Currently, I don't think that it does that. -- Wmarkham 23:47, 3 May 2005 (UTC)
I think so. Even with the statement "Monty always picks a goat", it is possible that Monty sometimes picks a goat that is behind the door that the contestant has initially picked. -- Wmarkham 02:38, 4 May 2005 (UTC)
I took a stab at rephrasing the question. I think it is unambiguous, and worded in a more natural manner. If not, feel free to revert. -- Wmarkham 02:19, 5 May 2005 (UTC)
I've done my best to incorporate some common confusions on this page, including one alternate (1/2) variant (and why it does not apply in this case), which I think also helps to express why the initial choice is so critical -- 2/3rds of the time, it eliminates one potential goat, forcing Monty to reveal the car.
I also added the 'problem summary' section, for lack of a better title -- I was hoping for something more like 'problem givens' or 'given statements' or 'known facts' or whatnot, but none really seemed to sound right. I added it because these are not assumptions, but rather, integral parts of the main problem... things a lot of people seem to miss, in whole or in part, when reading through and forming their conclusions.
I'm hoping all this will make the matter clearer. It's an interesting thought experiment, and I'm guessing a lot of people are so sure it's 1/2 until they see the right explanation for them, something clicks, and they realise why it's 2/3. For me, it was the scenario-based one. YMMV. -- Wisq 03:37, 2005 May 3 (UTC)
I think the wording is really essential. It must be distinguished between two different problems:
In case A it's better for Jane to switch for the described reasons. In case B it doesn't matter whether you switch, but the car is shown by Monty himself with the probabilty of 1/3. If Monty incidentally does not pick the car, the other both doors still have the same chance 1/3.
To make this a lesson in probability or rather in understandig a problem of probability, it must be worded in a way that the reader might think it is problem B when it indeed is problem A. So it must not be identified too obvious as either of the both.
The current wording is
Does anyone have other suggestions? -- Abe Lincoln 13:04, 5 May 2005 (UTC)
In "Letters to the Editor", Selvin provides a statement of the problem that describes a dialogue between "Monte Hall" (sic) and a contestant. The basic events are as follows:
The host explains that the keys to a new car are contained within one of three boxes, labelled A, B, and C. The other two are empty. The host explains that if the contestant chooses the box with the keys, the contestant will win the car. The host solicits a selection from the contestant, who selects box B. The host hands box B to the contestant. The host offers the contestant $100 for the box. The contestant declines. The host offers the contestant $200 for the box. The contestant declines. (The audience agrees!) The host provides the analysis that the probability that box B contains the keys is 1/3. The host offers $500 for the box. Once again, the contestant declines.
The host says "I'll do you a favor and open one of the remaining boxes". The host opens box A, revealing it to be empty. The host provides an analysis that the probability that box B contains the keys is 1/2. The host offers the contestant $1000 for the box. The contestant counters this offer with the offer, "I'll trade you my box B for the box C on the table".
Selvin asserts that "The contestant knows what he is doing!"
-- Wmarkham 18:06, 5 May 2005 (UTC)
I think that some of the recent edits have been heading in an unfortunate direction, making the initial problem statement so detailed in order to prevent misunderstandings that it's becoming hard to find the essential meat of the problem in the problem statement.
I'd like to propose instead that we start with a very simple statement of the problem situation, and the fact that the probability of winning by switching is 2/3, then follow up with something like:
-- Antaeus Feldspar 23:48, 5 May 2005 (UTC)
Section "The problem" says "Nydick." This author is listed in "References" as "Nudick, Robert L." Which is correct? Graue 01:47, 25 May 2005 (UTC)
Anyone think this article is NOT ready for WP:FAC nomination? I haven't worked on it much, but I'd be willing to nominate it if no one else is interested. -- Rick Block ( talk) 19:04, Jun 12, 2005 (UTC)
Here's how I finally understood it in my soul.
There is a 1/3 chance that the contestant's original door has the car, because he chooses randomly. There is a 0 chance that the door Monty opens has the car, because it always has a goat. The probabilities sum to 1. Therefore, the probability that the car is in the remaining door is 2/3, so switching is good. People think the probabilities are equal because they miss the fact that Monty deliberately, not randomly, avoids choosing the door with the car. By not picking a door, he shows it to be more likely to contain the car. Superm401 | Talk July 1, 2005 06:59 (UTC)
You are WRONG(or at least I think). Admitedlly, in the start there is a 1/3 percent chance. But then Monty skips the door you chose, along with the other one. He selects one, so he also skipped over the door you picked. Now, the goat has been revealed (or the first one) so it is fifty fifty, switching doesn't matter.
I believe the central points of these two sections might be replaced by the following explanation:
I think maybe this captures the essential points of those sections? -- Antaeus Feldspar 3 July 2005 23:00 (UTC)
This argument comes down to two mutually exclusive views:
1) The initial conditions of the game, (three doors, two goats, one car) have everything to do with the final choice.
2) The initial conditions of the game, (three doors, two goats, one car) have nothing to do with the final choice.
You can argue the niceties of the statement of the problem until the goats die from the heat produced by the studio lights, but proponents of either side will never convince the other. This is known as a religious war.
Never-the-less, I will add my two cents.
I am in the camp number 2. The statement of the problem, no matter how carefully crafted, is a classic example of a RED HERRING. Proponents of solution number one are falling prey to the "too much information" trick. Most of the explanations above are superfluous, as they just don't apply to the issue. The only purpose of the three doors stage of the game is to provide drama and indirection.
The final choice the contestant has to make is between two doors not three. Pure and simple, two doors, 50% chance either way, end of story. What has gone on to get to that point is IRRELEVANT. Monty may have just as well arranged to have one of the goat doors open when the curtain is pulled aside to reveal the doors. It doesn't matter if you had a million doors, and opend 999,998 - at the end you still have 2 doors, one with a car and one with a goat.
There is indeed one randomizing event, and the 1/3 versus 2/3 probability applies correctly to the initial red herring part of the game, however, THAT IS NOT THE GAME. As soon as Monty opens a door with a goat and offers the contestant a new choice, a choice between two doors, it is a new set of conditions. The 1/3 chance of the opened door does not 'magically' get transferred to the unpicked door, nor does it get 'magically' spread across the two remaining doors. The entire original situation is made null, and a new set of conditions apply.
Computer simulations designed to show the results of solution number one will always support solution number one. I can just as easily devise a computer simulation that will support solution number two (and I have done so - big deal). Both simulations, by their very nature, are biased to the solution they are meant to support. -- unsigned comment by 144.140.62.104 ( talk · contribs)
I agree because the intial conditions do not matter at all. After the host opens the goat, you know behind the door you choose could be one of two things, a goat or a car, and same for door 2. You know know your door, for sure is either the right or the wrong one, compared to what you knew at the start. The problem INVOLVES changing your conditions. To make it clearer, say behind a peice of paper, you had three sets of numbers, of which tow are the same. The contestant must guess the one set, not the two other identical sets, Now, the contestant in this game knows the sets of numbers, so he has a one in three getting the correct set of numbers. But know suppose one of the set of wrong numbers was revealed. It would leave the player with one right answer and one wrong answer. It is like someone asking you the distance from here to "a" and from "m" to "n" you not given any of the vasriables. You may as well take a random guess. It is not like the host skipped over yours, even if you choose the goat the host would simply open the other. Get it?
Sorry, but you still miss the fundmental, unchanging, unarguable point. The contestant is finally and uncontrovertably given a choice between two doors. Not three. Two. I'll repeat that once again. Two Doors. There are only two doors. There is only one goat (unless the car is a Pontiac GTO ;-) ). The contestant has to make a choice between two doors. Do you get it yet? Three doors is a red herring. There are only two doors in the game.
I don't have to supply you with my computer simulation to demonstrate the point, you would point out where my simulation is biased to my solution by makeing the decision that only the last choice is relevant, while I would point out where your simulation is biased to your solution because it makes the decision that the first choice is relevant.
Consider the computer game MineSweeper (available on Windows and every Linux distribution I have seen. Fundmentally you have to 'step' on a square which may or may not contain a 'mine'. If you miss the mine, a number is exposed which tells you how many mines are in the adjacent 8 squares (or 5 for an edge, or 3 for a corner). As you progress through the game, if you step on a square that does not have any mines adjacent to it, the game will automatically clear the surrounding squares until the exposed space is bounded by squares that do have adjacent mines.
Suppose at the beginning of the game, we choose a non-edge square, there is no mine, and the number '1' is exposed. Clearly, each of the adjacent 8 squares has a 1/8 chance of hiding the mine. If we randomly choose one of the 8 and it is clear, each of the remaining 7 squares has a 1/7 chance of hiding the mine. So far I think this description is matching your arguments above, because we are eliminating the clear squares not the computer or Monty Hall.
Now suppose that as we progress the computer clears some blank space that it knows does not contain the mine. We may find situations where there is a number '1' surrounded by 7 cleared spaces and one hidden space. The chance that the remaining one space contains the mine is 100% by my rekoning and I doubt that you would argue otherwise. Now we may also find situations where the computer has exposed a number '1' with 6 cleared spaces and two uncleared spaces. Disallowing overriding information from other squares on the board (in a real game you have other clues to decide which square to pick most of the time), you have two squares to choose from. I argue that the odds are 50/50 on each one.
I know this does not describe the process of pre-picking one of the squares, so there is no room here for consideration of the alternate solution of 1/8 versus 7/8 on the two squares. But that is my point: the pre-picking of the door is irrelevant to the final choice the contestant has to make, and the MineSweeper example exactly describes the conditions and choice the contestant has to finally make. The contestant does NOT have to pick between 3 doors. End of story. (Sorry I haven't figured out how to put in a proper signature yet - Keith Latham)
I'll just add one more half cent then I'm through. In a previous edit, Antaeus said "what Monty knows is crucial to the problem". Well of course he does have to know where the goats are to safely remove one of the losing doors from the game. But what is more crucial is what the contestant knows. And what the contestant knows is that there are two doors, one of which hides a car. Removal of one losing door gives zero information about which of the remaining doors hides the car. -- Klatham 4 July 2005 11:13 (UTC)
P 1 2 +-++-+-+ P!C!|G|G| +-++-+-+ 1!G!|C|G| +-++-+-+ 2!G!|G|C| +-++-+-+
P 1 2 +-++-+-+ P!C!| |G| +-++-+-+ 1!G!|C| | +-++-+-+ 2!G!| |C| +-++-+-+
I'm very interested in the code for this computer simulation that provides a 50/50 result. Could you please display it here, or provide a link to it? I think it would be demonstrative in showing the differences in rules between the problem you are imagining we are talking about, and the problem we are actually talking about. In my experience, there are three kinds of people who disagree with the Monty Hall problem's solution... those who don't yet understand it (but will, usually with increasing n=3 to n=arbitrarily large number), those who misunderstand the problem and think we're talking about another problem entirely (who are usually convinced when the rules are unambiguously stated to their satisfaction), and those disagreeing for the sake of disagreeing (who can never be convinced). I suspect you lie under catagory two, and determining the differences between your problem and the problem we're talking about could be key. Fieari July 5, 2005 19:09 (UTC)
As a layman who just read all this (and it's fascinating), I would like to state that Anteus' grids are extremely informative, since I now understand what he's been trying to convey for months in statistical terms. However, it seems to me that the Monty Hall problem is paradoxical in that while it is true that switching increases the statistical chance of winning from 1/3 to 2/3, ruling out a door gives the illusion that there is a 50% chance that the car is behind either door.
Let me see if this makes sense:
Anteus' grid clearly demonstrates that the smartest decision that a player can make is to switch from the door he's chosen, since in 2 out of 3 cases, he'll win by doing so. However, at that point 1 door has a goat and 1 door has a car, thus giving the illusion that the choice is 50/50, which it really isn't.
Does any of that make sense? (JRVJ - I also don't know how to sign my name yet).
I disagree with including simulation results in the article. I don't know where people are going with this. The simulation results don't increase the value of the article since the solution is already nicely explained with probabilities. One could argue that these results are against Wikipedia's no original research policy. 64.229.142.80 02:10, 23 July 2005 (UTC)
KEEP! I didn't believe this article until I saw the computer program results of 2000 to 1000 or whatever it is. Computer program results are easier to understand then math results.
The probablility of the goat sitting still is quite small. Therefore, the contestant will have the following possibilities after selecting a door:
-Monty picks a door where there was a goat
-Monty picks a door where there are now two goats mating
-Monty picks a door with a goat that is sleeping, going potty, or is too indifferent to move
As such, the door the contestant picked and the other unopened door will have one of the following:
-A messy stage where a goat once was
-Two goats mating
-A stall with a goat that is sleeping, going potty, or is too indifferent to move
-A car that has the seats chewed, the bodywork dented, and the paint scratched by (a) rampant goat(s)
Therefore, the contestant's best choice would be to Not give up the prize they won earlier in the show and stay out of the Final Prize Round of Let's Make a Deal.
-Pf
The answer to the problem is no; the chance of winning the car is the same when the player switches to another door.
There are four possible scenarios, all with equal probability (1/4):
In the first two scenarios, the player wins by switching. The last two scenarios the player wins by staying. Since two out of four scenarios win by switching, the odds of winning by switching are 1/2.
If we make a distinction between goat one and two when the user picks a goat, we need to do the same when the user picks a car. Why would we do otherwise? They are both possibilities aren't they? User:Oyejorge (07:49, July 23, 2005)
I took the liberty of signing your message for you. In the future, you can do this yourself by putting four tildas ( ~~~~ ) after your message, which will both sign and date your message automatically. In responce though, the last two events you have listed there are actually, mathematically, the same event. There's no real difference. To rephrase one of the "Aids to Understanding" placed on the main page, consider the following slightly modified situation...
Now add one more element...
I'd like you to first answer the two questions I've posted above, and then if you still doubt the validity of the solution, tell me how the second problem differs from the Monty Hall problem. Fieari 08:08, July 23, 2005 (UTC) I just considered something more... if the problem was reworded to constrain Monty to picking ONLY Goat #1 if the car is picked, it wouldn't change the problem at all. In the card version, we'd have a Queen, the Jack of Hearts, and the Jack of Spades. Player two, when picking up the remaining two cards, MUST discard the Jack of Hearts if he has it, and the Jack of Spades otherwise. No randomly picking which one to discard. I propose that wording it this way doesn't change the problem at all, and neatly eliminates your last option of the four you listed. Fieari 08:21, July 23, 2005 (UTC)
Remember that the two last ways have just half the possibility as the two first. And are mathematically seen the same event as one and two. But one could say that the possibilities for the events are 1: 1/3, 2: 1/3, 3: 1/6,4 1/6. Gillis 08:19, 23 July 2005 (UTC)
Ok, I've been humbled. I don't think I should do this sort of thing late at night. Here's a diagram I used to understand the problem... It's the same info as what's in the article, just presented a little differently, maybe it'll help other neighsayers. Thanks for the clues Fieari and Gillis,
First Choice | Probability | Monty Hall Takes Away | Switching Wins? |
---|---|---|---|
Goat 1 | 1/3 | Goat 2 | Yes 100% of the time |
Goat 2 | 1/3 | Goat 1 | Yes 100% of the time |
Car | 1/3 | Goat 1 or Goat 2 | Yes 0% of the time |
Oyejorge 17:44, 23 July 2005 (UTC)
I fixed the last cell in the table, which used to read "No 0%." Also correct would be "No 100%," but that's not as straightforward. Old Nick 19:22, 25 July 2005 (UTC)
I remember once reading that this riddle/theorem whatever is much older then ever mentioned in the article. i understand it was atelast from the 19:th century.
I can't verify this with google though. Does anyone have any better knowledge of the issue or better googling-luck? Gillis 08:13, July 23, 2005 (UTC)
I would argue it's a problem in probability. Rich Farmbrough 12:44, 23 July 2005 (UTC)
Can showing you that one of the three candidates in a tight race is a loser make you change or question your vote?
I happen to believe that people love to feel included by voting for the popular candidate.
Let's say people were given a choice to vote for three candidates A B C. After the election, you find out that one of the candidates was a loser and the two candidates were so close that a special runoff election was required. Would you be more likely to be with the majority by switching your vote on the runoff?
Consider this election ..
Dino Rossi (R) and Christine O. Gregoire (D) in 2004 ran one of the tightest governor races in history. Christine Gregoire won by a mere 129 votes.
The Republicans wanted a runoff election to solve the inaccuracies.
Do party affiliations play in this problem?
This poorly-titled section formerly said this:
This is circular logic. In support of the idea that door A ends up with a 1/3 probability of having the car, it makes the statement that "the probability of the car being in A remains constant" without any support. One may reasonably ask, why doesn't the probability of the car being in C remain constant instead? This explanation provides no answer. (It also ends with a spectacularly muddled sentence that adds nothing but confusion.) I have remove the section. -- Doradus 16:29, July 23, 2005 (UTC)
Probability is 2/3 if all decisions are made before any doors are opened. Probability is 1/2 if a fresh decision independent of history is made after Monty reveals a goat.
This misunderstanding is why the problem seems counterintuitive to people.
So should you switch if Monty gives you the choice AFTER he reveals a goat? Doesn't matter. Should your predetermined strategy for playing these games be to always switch? Absolutely.
Yes, it is quite clear to me, and you have apparently met the first such person. You are wrong if you say that 2/3 chance is independent of knowledge prior to Monty revealing a goat. It comes down to history. Without the history of Monty revealing a goat, it is merely a decision between two doors, exactly one of which has the prize. That gives you a 50:50 chance, and that is why people intuitively think there should be a 50:50 chance. It is no different than any other 50:50 scenario. That is to say, if in your life's travels you keep coming across a pair of mysterious doors with one prize, it makes no difference which one you choose and how often you change your mind. However, if the game involves the entire history starting with 3 mysterious doors, then consistently switching will increase your chances to 2/3. It is all about the given history. Indeed, if you are given additional history that the prize is behind your first chosen door, then switching will reduce your chances to zero.
The problem is that people don't seem to understand probability is context-dependent--it varies with what information you put into it. Increased knowledge increases certainty. Of course this isn't explicity stated, because why would the entire problem even be presented if it is to be ignored? It is understood that it is to be incorporated into probability calculations. However, folks who don't understand the dependence of history see the Monty problem as equivalent to the 2-door problem. They seem to think that 2/3 is some sort of physical feature of the door, rather than a measure of information and uncertainty.
So I say again, if you make a fresh decision (one ignorant of the past), your chance is 1/2, whether or not you switch. That is why knowledge is a good thing. You can often use it to increase your chances in life.
This whole thing seems to hing around independent steps.
50/50 result. Some condition is set up; in this case chose 1 of 3; chance of selecting the correct outcome is exactly one third. BUT the next INDEPENDANT step is offered after the removal of one of the incorrect outcomes. What is left ? 2 options; of which one and only one is correct.Proberbility of picking the correct outcome is 1 out of 2 or 1/2.
Please demonstrate very clearly and precisely how any event prior to being offered the 2nd step choice can possibly influance the outcome of that choice.
Or to restate that - you are left with a choice of one out of two.
Somebody mentioned tossing coins; it doesn't matter how many sides previous coins had; or how many coins were tossed prior; the next toss of a double sided coin has what probability of producing a head... yes; exactly 1/2 as PRIOR conditions have zero effect. I presume at this stage no one would bet a million pounds/dollars/euros that following 1 million heads a tail would occur ? (and if you believe it MUST be a tail then you need to re-vise your basic proberbility theory
Please cut all the waffle and explain how any preceeding action affects the 1 out of 2 selection left at the end !
I'm stuck behind a dynamic isp so can only sign Pete-dtm
Jeesh, I can't make up my mind as to which is more interesting (revealing), the article or this talk page! hydnjo talk 18:26, 23 July 2005 (UTC)
The corrolation between gravity and time becomes evident at a black hole.
This is the clearest way to understand it. Good job! Is it in the main article? --
pippo2001
21:46, 23 July 2005 (UTC)
This was a great featured article. Congrats to those who contributed.
LegCircus 22:34, July 23, 2005 (UTC)
One point that's not clear: if you have a larger number of doors (say four for simplicity), is the new situation one where Monty Hall opens one door or two doors? Peter Grey 22:46, 23 July 2005 (UTC)
I've presented the solution as a way to "trick the host". I think beating the house appeals to most people in a way that dry mathematical explanation doesn't. — 131.230.133.185 23:47, 23 July 2005 (UTC)
I find this diagram at best confusing. Or should I say the diagram looks fine, the numbers at the bottom (which add to more than 1) are confusing. Rich Farmbrough
The current article (which is incorrect, by the way) states the following:
This is fundamentally wrong because it leaves out a scenario. It can be fixed in one of two ways:
Fix #1: There are two possible scenarios, both with equal probability (1/2):
In the first scenario, the player wins by switching; in the second, the player loses by switching.
Fix #2: There are four possible scenarios, each with equal probability (1/4):
In the first two scenarios, the player wins by switching; in the last two, the player loses by switching.
Once the host opens a door and reveals a goat, the old problem is gone. The new problem is there are two doors: one has a goat and one has a car. What are your odds in picking one? (Picking means changing doors, because the odds of winning stay at 1/3 for the first-chosen door.) The answer is one-half. — Fingers-of-Pyrex 00:01, July 24, 2005 (UTC)
Kinda looks like this talk page. I guess there's no hope, they're everywhere. I'm going to rv the link as it doesn't add anything new to the article except controversy. hydnjo talk 00:50, 24 July 2005 (UTC)
Methinks all should read Wikipedia:Resolving disputes. We may need to change the article's focus to the Monty Hall problem and then have two different sections for explaining the answer—one wrong and one right, of course :). In the meantime, we should apply the {{dispute}} tag to the article to point readers to this talk page. — Fingers-of-Pyrex 01:21, July 24, 2005 (UTC)
One of Wikipedia’s policies is No original research. So, please do not try to prove to me what the answer is with your own calculations. Show me your source.
The following web sites claim that the answer is 2/3 probability if you switch:
For those that assert that the probability is 1/2, please site your source. -- JamesTeterenko 03:01, 24 July 2005 (UTC)
Hi. It doesn't matter if there are one trillion doors, cards, or whatever. Let me pick one. You show me all the other goats until there are two doors left. I'll have one-trillionth chance of winning the car if I stick with my first choice. I'll have 50% chance of winning if I switch. — Fingers-of-Pyrex 02:53, July 24, 2005 (UTC)
Part of what messes up intuition is how to make use of the partial information you have when deciding whether or not to switch. Here's another way to look at it that might help (this is sort of a rewording of "Best, most concise explanation yet") :
Assume the contestant always switches, i.e. ignore the second decision. The contestant's first choice is either a) car (probability 1/3) or b) goat (probability 2/3). In (a), contestant loses, in (b) contestant wins. Thus 1/3 probability of losing, 2/3 probability of winning.
Now assume the contestant always stays. The contestant's first choice is either a) car (probability 1/3) or b) goat (probability 2/3). In (a), contestant wins, in (b) contestant loses. Thus 1/3 probability of winning, 2/3 probability of losing. Peter Grey 04:17, 24 July 2005 (UTC)
As far as I can see, the different in the two viewpoint of 2/3 and 1/2 comes from the interpretation of the problem.
See the "Full Mathematical Solution" below. aCute 05:22, 24 July 2005 (UTC)
In short, the probability of winning really is 2/3 if you use the right strategy. (It's 1/3 if you stick by your choice, the wrong strategy unless you really like goats.) It's 1/2 if you flip a coin to decide whether to switch. Davilla 15:59, 24 July 2005 (UTC)
P(X1) | P(M l X1) | P(X2 l X1, M) | . | P(X1, M, X2) |
P(X1=C) = 1/3 | P(M=G2 l X1=C) = 1/2 | P(X2=C l X1=C, M=G2) = 1/2 | . | P(X1=C, M=G2, X2=C) = 1/12 |
P(X2=G1 l X1=C, M=G2) = 1/2 | . | P(X1=C, M=G2, X2=G1) = 1/12 | ||
P(M=G1 l X1=C) = 1/2 | P(X2=C l X1=C, M=G1) = 1/2 | . | P(X1=C, M=G1, X2=C) = 1/12 | |
P(X2=G2 l X1=C, M=G1) = 1/2 | . | P(X1=C, M=G1, X2=G2) = 1/12 | ||
P(X1=G1) = 1/3 | P(M=G2 l X1=G1) = 1 | P(X2=C l X1=G1, M=G2) = 1/2 | . | P(X1=G1, M=G2, X2=C) = 1/6 |
P(X2=G1 l X1=G1, M=G2) = 1/2 | . | P(X1=G1, M=G2, X2=G1) = 1/6 | ||
P(X1=G2) = 1/3 | P(M=G1 l X1=G2) = 1 | P(X2=C l X1=G2, M=G1) = 1/2 | . | P(X1=G2, M=G1, X2=C) = 1/6 |
P(X2=G2 l X1=G2, M=G1) = 1/2 | . | P(X1=G2, M=G1, X2=G2) = 1/6 |
The probability of winning (not Monty Hall problem) is:
The probability of winning without switching is:
The probability of winning with switching is:
The answer to the Monty Hall problem:
aCute 05:22, 24 July 2005 (UTC)
Note: It is my observation to lead me to conclude the following:
The Monty Hall problem is NOT to determine the winning-probability of the last event or the entire game, but rather to determine which of the two strategies will maximize winning-probability. The problem is like comparing the subset cases in the manner similar to the Simpsons paradox. This is also like trying to find a winning strategy in Blackjack by counting cards already played previously as additional information (sort of like Monty Hall revealing a goat) in order to adjust subsequent choices to gain an advantage over the house. aCute 06:49, 24 July 2005 (UTC)
In "the formal probability diagram" within the project page of the
Monty Hall problem, the leaves of the diagram tree has the values "1/3 1/3 1/3 1/3 1/6 1/6 1/6 1/6" which cannot possibly exist since the sum of these values canNOT exceed 1.
Please someone change them ASAP.
aCute
05:22, 24 July 2005 (UTC)
Yes I have one of those too, and I found the diagram confusing (as noted above). Perhaps the colour of the YES and associates probabilites could be diffeent from the NO and assoc, probs. Perhaps a caption woul clairfy things. Rich Farmbrough 18:25, 24 July 2005 (UTC)
Okay, now I see how the diagram is constructed. Previously, I failed to see that BOTH situations are placed onto the SAME tree. aCute 19:41, 24 July 2005 (UTC)
"Never argue with someone who knows that he's right" hydnjo talk 05:27, 24 July 2005 (UTC)
When this doesn't work well, it's time to go on to something more productive. Buh bye now. hydnjo talk 05:45, 24 July 2005 (UTC)
Ok, so stupid me, but I have to do this. Lets take this down to such a basic level that even I understand.
Respectfully, hydnjo talk 21:18, 24 July 2005 (UTC)
The game is fixed from the moment that you pick one door and Monty gets two doors. All else is just fluff and misdirection to make the game last longer. You hold a 1/3 chance of winning and (because he has two doors) Monty has a 2/3 chance of winning. Swap with him at your earliest opportunity. Please. hydnjo talk 02:10, 25 July 2005 (UTC)
You get one of three possibilities and Monty gets two of the three possibilities. The fact that you chose first changes nothing. If given the opportunity to swap with Monty well, just jump at it. He did, after all, try to screw you at the very outset. And, if Monty tries to confuse you with more doors, well then hold fast. When he offers a swap (which he will) just take it. Don't forget, you got to pick one door and Monty gets all the others. Lucky lucky him he would like you to believe! hydnjo talk 02:48, 25 July 2005 (UTC)
(as opposed to the correctness of the mathematical answer)
It is not enough to describe why the mathematically derived solution is correct. To resolve the paradox to the satisfaction of all, one must also describe why the intuitive solution is wrong. I think this requires three steps.
1) an understanding that the "intuitive solution" is just a dismissive term for a first analysis that turned out to have a flaw
2) a description of the logical steps that were employed in coming to the intuitive solution
3) an analysis of that logic, to find its flaw(s)
(1) I not argue the point, but rather just hope that people agree with it.
(2) I think that the intuitive solution takes the following steps:
Here is the game that I believe is used intuitively (I’ll call it the Silly game):
Should he change it or not?
Clearly the answer to this is that the chance is 50% either way, and it does not matter whether the contestant changes his mind.
Now compare the Silly game to an original Monty Hall game at a point half way through, in which the contestant has guessed door B and Monty has opened door A.
It seems that the Silly game is exactly the same as the original Monty Hall game at this point. To a viewer who has just ‘tuned in’ and does not know what has previously happened, the games look identical. Therefore, logic would dictate that the answer to the original question is that it doesn't matter whether the contestant changes their mind; the probability is 50% either way.
(3) It turns out, of course because the two games are not completely isomorphic. There is a crucial piece of information missing from the Silly game that is needed to make it isomorphic with the Monty Hall game, as follows: Monty (who knows where the car is)
Monty has thus said something concrete about door C (“I didn’t choose it, perhaps because I couldn’t choose it”), but nothing about door B. This is the source of the asymmetry between doors B and C, and the reason that door C is more likely to not have a goat behind it. Happyharris 20:57, 25 July 2005 (UTC)
For the people who still find it hard to see why the probability is not 50/50 when there are two doors left, I hope the following illustration may help. I'm going to show that the Monty Hall problem is a specific case of a more general game; I'll call all the games that differ in their parameters "Hall games" for ease of use.
The basic idea behind a Hall game is this: We start with one large set of secret-hiding items -- they can be doors to be opened, or cards to be turned over, it doesn't actually matter. What matters is that there are n cards, but only one of them is the Prize card. The rest are Null cards.
Step One: The cards are divided into two hands. Each hand must have at least one card. For simplicity's sake, we call the number of cards in the first and second hands h1 and h2.
Step Two: Someone who can see which cards are Nulls can discard some number of Nulls from one hand, the other, or both. We call the number of cards remaining in each hand after the discarding of Nulls r1 and r2 (like h1 and h2, they must be at least one.)
Step Three: The player makes a guess at which of the two hands contains the Prize.
Step Four: The player makes a guess at which card out of the hand he selected is the Prize.
It's clear that to win the game, the player has to make correct guesses in both Step Three and Step Four. What are the chances of picking the correct hand? The first hand is correct h1/n of the time; the second hand is correct h2/n of the time. If we want, we could enumerate the cases: if h1=2 and h2=5, then the Prize could be the first card of the first hand, the second card of the first hand, the first card of the second hand ... et cetera, et cetera.
Now this is the part that many people find counter-intuitive. If we enumerate the cases, and then we reduce Nulls to meet any legal value of r1 and r2, we find that in no case can the removal of Nulls switch the Prize from one hand to the other. This means that the chance of the card being in the first hand or the second always stays at h1/n and h2/n -- even if the sizes of the hands do not stay at h1 and h2. If it's dealt to that hand, it stays in that hand; therefore the chance of it being in a particular hand is always equal to the chance that it was dealt to that hand.
What about the removal of Nulls? Does it affect anything after all? Yes, it does -- it affects the player's chances in Step Four. The chance that the Prize is in a particular hand is dependent upon h1 and h2 -- how many cards each hand started with. But the chance of finding the Prize in a hand (assuming it's the right hand) is based on how many cards that hand contains after Nulls have been removed -- since there's only one Prize, the chance of finding it in a hand of r1 cards is 1/r1.
Now, what are the chances of making both guesses correctly? If no Nulls get removed from either hand, then the chances of picking the Prize are either h1/n x 1/h1 (since r1 is equal to h1 when no Nulls have been removed) or by similar logic h2/n x 1/h2, which also multiplies out to 1/n. If, however, one of the hands -- say, hand 1 -- has been reduced down to one card (r1 = 1), then the chances of finding the Prize in that hand is h1/n x 1/1 -- if you've correctly guessed that the Prize is in that hand, you have a 100% chance of finding it in that hand when it's the only card in the hand.
With this being the general structure of the Hall game, we can see that the Monty Hall problem is really just the case where n=3, h1=1 and h2=2, r1=1 and r2=1. The chance that the Prize is in the player's hand is h1/n -- 1/3. The chance that it's in Monty's hand is 2/3. Before one Null is removed from Monty's hand, the chance of finding the Prize in his hand is 2/3 x 1/2 -- i.e., 1/3. But when the Null is removed, the chance is now 2/3 x 1/1 -- i.e., 2/3! -- Antaeus Feldspar 03:44, 26 July 2005 (UTC)
kudos to those contributors to this article. i have been thinking about this for days (despite the fact that i "got it" ... after about 20 minutes or so). i find the "two sets" explanation the most clear, though i'm sure some people will love the bayes' theorem explanation. i also found the talk page very entertaining. some contributors are clearly manifesting what kahneman and tversky (writers on cognitive biases - the former won a nobel prize) refer to as "belief perseverance." geez, sit down with a friend and three cards for 10 minutes and you'd realize you're wrong. anyway... i thought you might find this page interesting:
http://econwpa.wustl.edu:8089/eps/exp/papers/9906/9906001.html
besides elaborating on many of the key assumptions (often unstated) underlying the MHP, the paper presents an interesting (and supposedly earlier) problem of the same form, the "three prisoners problem." here's the upshot...
There are three prisoners, you and Prisoners A and B. Two of you are to be executed, while one will be pardoned. The prison warden knows who will be executed and who will be pardoned (like monty must know where the goats and car are). According to policy, the warden is NOT allowed to tell any prisoner if he/she is to be pardoned. You point out to the warden that if he tells you if A or B will be executed, he will not be violating any rule. The warden says C is to be executed. What is the chance that you will be pardoned?
the answer presented, like that to the the MHP, is that your chances of being pardoned remain at 1/3, while the the chances of B being pardoned, GIVEN WHAT THE WARDEN HAS TOLD YOU, is 2/3.
I found looking at this problem reinforced how *$&%^@# hard it is to get one's head around the MHP, because despite having read about the latter for days, i still had to think about the three prisoners problem for several minutes to get my head around IT, despite knowing that it was essentially the same as the MHP. K-razy.
someone mentioned the idea of blocking edits to featured articles for some period of time. while apparently there is a policy against such things, i agree with the suggestion, at least where articles that are sources of dispute. the featuring of an article doubtless attracts many potential editors, some of who may simply not know enough about the subject to edit appropriately. for example, the first time i read the monty hall page (when it was featured), someone had altered the solution section to express the misguided (given certain assumptions, which too often are unstated) 50/50 approach. needless to say, i was confused by the fact that all other sections of the article contradicted the solution.
good job. Contributed by 24.89.202.141.
This is an archive of past discussions. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page. |
Archive 1 | Archive 2 | Archive 3 | Archive 4 | Archive 5 |
I've moved the existing talk page to Talk:Monty Hall problem/Archive1, so the edit history is now with the archive page. I've copied back the most recent thread. Hope this helps, Wile E. Heresiarch 04:40, 10 Aug 2004 (UTC)
Recall that the host has opened a door, and we know that door contains a goat. The contestant was either RIGHT or WRONG in the first place. In the former case, we switch to a goat and lose. In the latter case, switching guarantees a victory: we picked Goat 1 originally, and Monty has released Goat 2. So the only probability we need to know is that of being right or wrong in the first place, which are 1/3 and 2/3 respectively. 1/3 chance you will lose if you switch, 2/3 chance you will win.
Just a stupid question. Suppose you have the three doors
Door 1: Selected Door 2: Opened(goat of course) Door 3: Unknown
Since the contestant knows behind 2 doors is a goat, the Goat One from the opened door could be romoved. That leaves a fifty-fifty percent chance that he/she is correct in thier guess, so switching would not increase it by 2/3 because there are only 2 unkown doors left. I've proabably not followed this enough so tell me why I'm wrong soon please.
How does it leave 2 doors with unequal probabilities? Of two doors left, one has to have the goat and one has to have the car, becuase the host always opens up the goat.
I still don't really understand because this is different from a deck of cards. A deck of cards cannot be compared to this because a deck of cards has:
a) more choices
b) be more then one, say door.
If I put up fifty doors, and I had, lets just say, ten variations, four suites, then there is a much different probabilty then knowing that threre are only two doors, of which on is the goat and one is the car. The problem does not need the third door any more. If, say you had 34 aces of spades, and 17 aces of hearts you still would not get the same equation because it is mroe variables. You know there is a fifty fifty percent chance the door that you first seleceted has the car because the host opens a goat. You either have or don't have the goat, so fifty-fifty chance. Now if you siwtched, it would not imrvoe your chances.
But wait... if you discard a card forever, now there will be only two cards left making it one out of two.
What's wrong?
Am I right or Am I right or Am I right?
try to do it later.
-- 213.148.18.178 10:57, 17 Jan 2005 (UTC)
Except do not forget, Monty's choice is constrained, since he MUST open a door revealing a goat, in the above P(O3) = 1, not 1/2. I have deleted the sequel, as it is superfluous given my previous statement. -Dwee I have restored that text, since it's easier to see where the mistakes in a text are when you can actually look at the text. -- Antaeus Feldspar 02:54, 7 Jun 2005 (UTC)
I was talking to a friend about the Monty Hall problem today and he told me about a similar problem, and I don't think there is a Wikipedia article on it and I'm not sure how to present the solution either, but anyway here is the problem:
You are on a gameshow and the host holds out two envelopes for you to choose from A and B. So you choose an envelope (A) and it's got $2000 in it. The presenter then says that one of the envelopes has twice as much money in it as the other one and offers you the chance to switch. So you think about it this way... "If I switch I will go home with either $4000 or $1000, by not switching I will go home with $2000. There is a 50/50 chance that I will double my money by switching. A normal 50/50 bet results in me either doubling my money or losing it all, whereas here I will only lose half. Therefore this is a better than evens bet so I will make the swap." You are just about to swap envelopes when you think about the problem some more - "Surely this can't be right... ". Mintguy (T) 16:13, 15 Jul 2004 (UTC)
[Tired of nesting the blocks deeper and deeper, like bad C code.] Let's pause to list assumptions. My choice of an envelope is not correlated with the loading of the envelopes, so that I'm equally likely to have the good or the bad envelope. The host also is statistically unbiased: his telling me the news is not correeated with my initial choice of good or bad envelope. Oh, and he's telling the truth.
This is not to say that the expectation argument is right. This is the paradoxical part: that the argument has no apparent flaw in itself, but it gives the nonsensical result that one should choose and then change, even though no new information has come along to cause a change. The host's information is new, or seems to be, but how would my course of action be different if I had known it all along? Bottom line: the expectation argument leads to a silly result, but I don't believe that its flaw has been shown. Maybe this paradox deserves its own article. Dandrake 19:35, Aug 26, 2004 (UTC)
I think I'm not understanding this problem correctly. The way I read it, you either pick an envelope with X or 2X dollars in it with equal probability. Given the option to switch, this expands into four cases
Picked X | Picked 2X | |
Keep | X | 2X |
Switch | 2X | X |
Doesn't this mean that each of these events occurs with equal probablity, and that it doesn't matter? I understand the expectation argument, but I can't reconcile it with this simple grid. Cvaneg 23:13, 26 Aug 2004 (UTC)
This problem now has its own page at Envelope paradox Mintguy (T)
I'm copying this to talk:Envelope paradox. Further discussion of this subject should continue there, but I will also leave the text above so that people can follow on from here. Mintguy (T)
I decided to be a little bold and remove the assumptions section. If you read the problem as stated, then you don't need to make any of the assumptions listed. E.g., it doesn't matter whether or not you assume "Monty always opens a door," or that there is "always a goat behind the door Monty opens," because the problem clearly states that Monty opens a door and reveals a goat. That is all you need to know in order to determine the correct answer. Other points in this section also showed an incorrect understanding of the problem. -- Simoes 15:53, 24 Jul 2004 (UTC)
It's critically important to the problem that Monty knows what's behind each door. In the actual correct problem situation, there are three scenarios for what's behind each door:
Scenario | Door 1 | Door 2 | Door 3 |
---|---|---|---|
Scenario A | Car | Goat | Goat |
Scenario B | Goat | Car | Goat |
Scenario C | Goat | Goat | Car |
Let's presume that the player always first chooses Door 1 (a safe assumption, since we can always just choose how the doors are numbered after the door is chosen.) If we assume that Monty knows what's behind all doors, and opens the door he does because it has a goat behind it, then our three scenarios look like this (with the revealed goat crossed out because the player knows to avoid it):
Scenario | Player's first choice | Door 2 | Door 3 |
---|---|---|---|
Scenario A | Car | Goat | |
Scenario B | Goat | Car | |
Scenario C | Goat | Car |
If on the other hand we leave the problem too vague, by just saying "Monty opens a door that has a goat behind it", we may create the impression that it is just chance that, this time, the door had a goat behind it -- and this changes the problem completely. Because it simply makes it look as if some of our original scenarios could not have happened:
Scenario | Player's first choice | Door Monty opens | Door 3 |
---|---|---|---|
Scenario A | Car | Goat | |
Scenario C | Goat | Car |
The faulty logic thus goes "If Scenario B had been correct, there would have been a car behind the door Monty opened; but there wasn't, so therefore Scenario B is not correct. The only remaining scenarios are A, where the player's first choice contains the car, and C, where the player's first choice contains the goat; therefore the chances are 50/50 whether the player switches or stays." -- Antaeus Feldspar 20:44, 21 Nov 2004 (UTC)
The standard solution seems illogical for 2 reasons. First, probabilities describe chances, and since Monty has removed one of the uncertainty by opening up a door, he changes the conditions which the original set of probabilities described. This is to say, the 2/3 probability should not be transferrable, and the new probability designed to describe the new conditions should state that the car has a 50-50 chance of being behind either of the unopened door. Second, say the standard solution is logical, and say the car is behind door 1, and the player will always choose between door 1 or 2, and Monty will always open up door 3: if the player had originally chosen door 2, then he should change his selection to door 1; if the player had originally chosen door 1, then he should change his choice to door 2: the results of the 1st situation contradicts that of the 2nd, and vice versa; the contradiction shows that the standard solution is wrong, and it doesn't make a difference whether to stick to the original choice or not.
I found the section on "Assumptions" to be a little confusing and to obscure the true spirit of the puzzle/problem. I have therefore removed that section and reworded the problem in the introduction to make it clearer that the host knows what is behind the doors and that he always opens a door with a goat. Then the assumptions are redundant and the intentions of the host, benign or otherwise, are not a factor. I think it is obvious that the contestant will want a car, not a goat, and there is no need to state this. I have also reworded the answer to try to make it clearer and more concise.
P S K -- 213.122.39.201 17:00, 25 Oct 2004 (UTC)
Hi P S K, you made the much needed changes to one of our important articles. -- Sundar 04:40, Oct 26, 2004 (UTC)
I have written up what I think is a clearer explanation of the solution, at Monty Hall problem/temp1. I'd like others' feedback, to find out if it's as clear as I think it is and can go in the article. -- Antaeus Feldspar 01:38, 1 Dec 2004 (UTC)
The Monty Hall problem is not a problem in people's skill at using probability and mathematics. It is not a problem of gullible people falling prey to tricks. It is not even a paradox. It is a problem in communication and rhetoric.
As commonly stated, the Monty Hall problem is ambiguous as far as how the host chooses the second door.
Whether information about the third door can be inferred by the host's actions, depends on whether the host uses information about what's behind the third door to make his decision.
If the host chooses the second door randomly among the two doors not already selected by the player, and it has a goat behind it, then there is no advantage to the player switching.
If the host chooses the second door because it has a goat behind it, then the probability that the third door has a car behind it jumps to 2/3, and the player should switch.
If someone were to ask me the Monty Hall problem, and it were not clear whether the host used information from behind the third door (or, equivalently, information from behind both the first and second doors) to make his decision, then I would come back and say that the problem was ill-defined. Neither of the two common solutions would be acceptable.
Actually, a third scenario is possible: If the host chooses the second door because the third door contains a goat, then the player should never switch.
The point is that you cannot infer information about the third door unless you can infer information about the host's decision-making process. Unless this is spelled out clearly (which it most often isn't), the problem is ill-defined.
Neither side in this "debate" has done bad mathematics, and mathematicians who righteously scoff at those who express contrary views, calling it bad mathematics, or mathematical illiteracy, add nothing to our understanding.
I'd like a humanities professor's analysis of this problem, because it is not fundamentally mathematical. I'd like to see a detailed analysis of all of the different precise phrasings of this problem, and how they parse against accepted English grammar. What assumptions are socially "valid"?
An analysis of rhetoric is more important than an analysis of probability, in this problem.
(FWIW, I'm a computer scientist who writes mathematical software.)
Sorry, but I can't force myself to agree wtih your solutution. When the host opens the door he not only skips over the dorr you choose, but the other door. There is a 2/3 chance it was the goat...right? But no remove a goat, just remove the door the host opened. You would not, in your right mind, choose a goat against a car. Now there are two doors left. One is your door with car or goat, and other a door with same. The problem with most of the tables and charts here are that they use goat twice in two rows, making it two chance. Now, goat one and two are interchangable:
Player Chooses Goat One:Remaining Choice: Car
Plyaer Chooses Car: Remaining Choice: Car
You should not add:
Player Chooses Goat 2: Remaining Choice Car
That would not happen. There are three doors, of which one of them is open, which goat DOES NOT MATTER EVERYONE. There will always be two doors left open, not matter if the players chooses goat one or two. There will alwys be a car and a goat remaining. If the player choose goat one and goat 2 is opened, you can't make another choice be player chooses goat two and goat one is opened. With this amount of doors they result in the smae thing unless you did it more often. Try it.
Door One: Door 2: Door 3: Goat Goat Car
Now to help you if you still don't get it: Door One or Door two can be opened and reslult in the smae outcome. Which door they open does not matter. Goat
Very good explanation. Probability applies to a theoretical problem abstracted away from the Monty Hall Problem itself. But, it would be really interesting to have humanities' point of analysis. -- Sundar 09:35, Jan 25, 2005 (UTC)
Yep, it's impossible to reason about probability after the fact without some prior assumptions about the process. It's like someone who made a bet with a gambler and found that the coin came up 100 heads in a roll and claim that the coin is biased. It's entirely possible that a fair coin produced those outcomes, it's just as likely as any other sequence.
It has nothing to do with what Monty knows. Like the solution states, there's a 2/3 probability that the car is behind one of the doors you didn't pick. It doesn't matter if the doors are purple, have aliens behind them, are invisible or any of that other nonesense. There's a 2/3 probability that the car is behind one of the doors you didn't originally pick.
P | M | U |
---|---|---|
C | G1 | G2 |
C | G2 | G1 |
G1 | C | G2 |
G1 | G2 | C |
G2 | C | G1 |
G2 | G1 | C |
P | M | U |
---|---|---|
C | G1 | G2 |
C | G2 | G1 |
G1 | G2 | C |
G1 | G2 | C |
G2 | G1 | C |
G2 | G1 | C |
P | M | U |
---|---|---|
C | G1 | G2 |
C | G2 | G1 |
G1 | G2 | C |
G2 | G1 | C |
I realize that I am probably wrong. I mean, if this phenomenon weren't mathematically solid it wouldn't have survived for this long, but something just occurred to me.
How can my picking a certain door affect it's probability?
Consider the following:
It's the exact same set up, with three doors etc. etc. In my mind I pick door 3, but I tell the host I pick door 2 and he believes me. In his mind I've picked door 2. He opens door number 1 and it is a goat.
Which door has the two third probability and which has the one third?
How can the exact same scenario bring up two probabilities based on the door I "picked?" Picking out one door in my mind doesn't mathematically substantiate anything as it has no relevance to anything mathematically.
The problem is not asking us to plot out an entire course of action beforehand. At the point that the decision in question is being made, Monty has eliminated one of the initial possibilities and presented the contestant with an essentially new problem: There is a car behind one of these two doors, pick one. Anything that has gone before is irrelevant. Referring to the decision as "switching" is a red herring.
Please disprove this before reverting the article. -- El Mariachi
Here's one last attempt to explain -- not to trolls, but to anyone who still honestly doesn't see why the probability works out as it does:
Player A and Player B take the 13 diamond cards out of a standard deck of cards. The cards are shuffled, and then Player A receives one card face-down, which he may not look at. Player B gets the other 12 cards, and he may look at them. Both players are trying to wind up with the Ace of Diamonds in their hand.
Player B has twelve cards which he can look at. At least eleven of them are not the Ace. Player B takes out eleven non-Ace cards from his hand and lays them down face-up.
Player A now has an option: he can stay with his current hand, with the one card he was originally dealt, or he can switch his hand with Player B's hand, which was originally dealt twelve of the thirteen cards.
The only differences between this problem and the canonical Monty Hall problem is the scale (13 instead of 3), the objects (doors to be opened instead of cards to be turned over) and the fact that Player A has his initial card/door assigned, rather than selecting it himself. None of these factors makes a difference to the central point, that the chances are better of winning by switching. The only one that comes close is Player A being assigned his initial choice, and since it violates the problem rules to posit that Player A could somehow pick the winning door/card initially at a rate better than chance, it does not make a difference.
Anyone who still believes that once Player B has discarded eleven non-Ace cards from his hand, the chance that he's still holding the Ace drops from 12/13 to 1/2 -- come and see me in person, and we'll take the diamonds out of a deck and play, and every time the Ace is in Player B's hand, you pay me $100, and every time it's in Player A's hand, I'll pay you $200. And I'll be merciful and quit after I've taken $1000 off you. =) -- Antaeus Feldspar 20:45, 13 Feb 2005 (UTC)
Your attempts at attackign a correct answer with aimless little jabs of "you're just trolling!" are sad. You have failed to make any logical point. You are the troll here. The chance of switchign then picking either car or goat are 50/50 PERIOD. Yes the initial "draw" determines where each object is. Irrelevant. Human action or choice plays no part...whether a random die roll or a person chooses the doors, is also irrelevant. Nothin to do with "uintuitiveness". Just simple, correct logic & math. Anything else is garbage 7 a sidetrack & no one has ever remotely come near confronting the absolute empirical black & white fact that it IS a simple 50/50 choice. Exactly because of the initial dispersion of objects, you choice & if you switch or not make no impact. -- unsigned comment by 71.115.25.22 ( talk · contribs)
Ok, I've been reading the explanations for the probabilities for this Monty Hall Problem for quite some time, and I think I've come up with a much easier way to understand the answer than what I've heard so far. Rather than type out long, drawn-out mathematical equations and using all these other instances, let's look at it logically. The way I understand the problem, the contestant picks a door, the host opens a wrong door, and then the contestant is given a chance to switch. Ok, if the contestant picks the right door, with the car in it, and then switches, the contestant will be lose. There is 100% probability of this. If the contestant picks either wrong door, and then switches, then he/she will win. 100% probability. So let's look at the original odds. The chances of initially picking the right door are 1/3, and since this is the only way to lose if you switch, your chances of losing are indeed 1/3. In the same way, your chances of picking a wrong door initially, and therefore wi nning, are 2/3. How could anyone still not understand this problem?
You, the candidate, pick a door. Probability dictates that in 1 case out of 3, you picked the car. This case will now be called (C). In 2 cases out of 3, you will have picked a goat. This will now be called (G).
(C): The host will open one of the two remaining doors, revealing a goat. Changing doors will let you loose, as you picked the door with the car already; STAYING with the first choice will always win in case (C).
(G): The host will have no choice which of the remaining two doors two open, as only one contains another goat, which he is bound to show. So in case (G), CHANGING the door will always win, while staying with the goat is a clear loss.
Conclusion: It can be seen easily that in 1 out of 3 cases, staying with your originial choice will be a sure win, while in more probable 2 out of 3 cases changing doors will be a sure win.
790 12:52, 19 Feb 2005 (UTC)
Here, for those of you who don't get the math (or think you do but still think there's a 50/50 chance at the end): Monty Hall applet. Play the game over and over. It keeps track of your wins, and whether you won by switching or staying. Still not convinced? Try the iterative applet. Run it for a few thousand iterations. The statistics are borne out. -- Wapcaplet 00:20, 2 Mar 2005 (UTC)
Saying that this is a paradox simply because to some people it is counterintuitive, is like saying "My good friend Jim, I thought you were a natural blond, but now I understand that you have died your hair! I didn't expect that at all; what an amazing paradox." A paradox has a precise mathematical meaning; anything and everything that isn't what you expect isn't a paradox. The monty hall problem is perfectly reasonable and contains no mystery or internal contradictions whatsoever.
A paradox arises when there are two (at least partially) valid and contradictory conclusions. The intuitive answer of "50/50" is demonstrably wrong and completely indefensible. Therefore even if it could have been regarded as a paradox, it is entirely resolved for anyone who takes the time to think about it. To continue to claim that it is paradox, you either have to demonstrate that something is still a paradox even after it has been resolved to the satisfaction of all mathematicians and logicians, or claim that the two answers both have validity.
In casual usage it might be regarded as a paradox. --a
A Google Scholar search of "Monty Hall" and "Paradox" reveals that it is almost always not referred to as such in scientific papers, since nearly all the results returned talk about Monty Hall and also happen to mention something else, which provides the paradox part of the search. The most common way to refer to it in mathematical papers is "The Monty Hall dilemma." The conspicuous exception is "Increasing working memory demands improves probabilistic choice but not judgment on the Monty Hall" by T Ben-Zeev, J Stibel, M Dennis, S Sloman, the purpose of which is to argue that the chance of winning is actually 1/2. This reinforces my point: It only seems like a paradox to those who do not accept that the answer is unequivocally 2/3 and not 1/2.
And anyway, I am aware that there is a common definition which means that if something is simply surprising, then it is a paradox (although it should no longer be surprising after the mathematics behind it are demonstrated). What is your problem with the current revision? If you do revert, PLEASE for the sake of accuracy take out that utter crap about it being a paradox by the "mathematical definition," which is precisely what it is NOT a paradox by.
If you are going to cite evidence, use outside material, not wikipedia. If you'll check the discussion page for paradox you will find that there is considerable controversy about many of them (including Monty Hall) and even about the formal definition that is given in the first paragraph (the "apparently" part is bullshit, as several have pointed out). --a
Oh, now all of a sudden Feldspar wants to allow for broad usage of terms that have multiple meanings? Hmmm... I suppose it's ok then to call John Kerry's 1st 1st "wound" minor, eh? This unsigned comment is by 216.153.214.94, determined by a ArbCom ruling to be a sock puppet of Rex071404.
From The Collaborative International Dictionary of English v.0.48 [gcide]:
So that's what I found in the dictionary, for what it's worth. -- Wisq 18:41, 2005 Apr 24 (UTC)
The issue of the Marilyn vos Savant article was not that she was "wrong" but rather that she was mis-stating the problem. The key word she used was "guess" in regards to making the 2nd selection. By definition, "guessing" requires one to choose without making calculations. When one calculates before "guessing" it's no longer a mere guess, but an educated guess.
The other problem with her article was that the concept of odds has a dual layer in this "Monty Hall" problem which is often overlooked. The odds of finding the car do increase by switching. But the statistical likelyhood that it exists, does not. Additional information can always increase the likelyhood of a successful choice, but it does not affect the underlying statistical facts. If there are 100 piles of horseshit in your backyard and a single can of caviar buried in one of them, the singleness of the can of caviar never changes, only the odds of finding it. Suffice it to say, I have found that most people who argue about Monty Hall problems, overlook this point. Please see this variant [1] of the Monty2.gif from the article page. I suggest that this puzzle would be less puzzling if the location of the car was referred to as "odds of finding card here" rather than "odds of car being here". 216.153.214.94 06:04, 24 Mar 2005 (UTC)
I agree the probabilities are 1/2 when picking the second door, and it doesn't matter why monty picks his door because if he does reveal the prize then the game ends with you winning anyway.
I added the following statement, but it was removed with the hint, that though it is accurate, it would be "polarized and confusing" (quote Antaeus Feldspar).
Often it is suggested to increase the number of doors to help to understand the problem, for instance to 100. This approach is illegal, since when transforming the special case t=3 into the general case t=n, the order to the show host can be interpreted in at least two different ways:
Only for n=3, both interpretations are exactly the same.
The supporter of this approach refer to the second interpretation. In this case, the chance raises from 1/n to (n-1)/n, so for n=100 from 1% to 99% (in 99 out of 100 cases the candidate does not choose the prize, and in all of these cases a change to the one remaining door will lead to the prize).
The first interpretation is nevertheless just as legal and more than that also more evident than the second one. Even here the chance raises, but only slightly from 1/n to (n-1)/n x 1/(n-2), so for n=100 from 1% to ca. 1.01% (in 99 out of 100 the candidate does not choose the prize, and in 1 out of 98 cases the prize is then behind one of the many remaining doors).
For n=3, both terms - (3-1)/3 x 1/(3-2) and (3-1)/3 - result to 2/3.
So the change is a good choice for both general interpretations, as well for the one special case t=3. But it is illegal to refer to the alleged obviousness of just one general interpretation to illustrate the special case t=3.
I'm sorry if it seems polarized, but it refers to a very common approach, that is unfortunately absolutely wrong (please proof if I'm wrong). And unfortunately this approach is used in a very offensive way, like "you are wrong, and you can easily see it if you change the problem a little bit". If you change a problem, then you just CAN'T transfer the solution from the changed problem to the original problem. This is a very common mistake when dealing with mathematical problems. And unfortunately this whole thing IS about accuracy!!! And I am sorry if it is confusing, but mathematics is hard to understand some time. I hope I don't appear arrogant or something, but I think it would be a shame if such a prominent and good article contains a wrong explanation.
Maybe it is confusing because my English is bad. But in my opinion, it is better fpr an encyclopedia to be confusing than to be wrong!!!! So please discuss or improve my text, but PLEASE don't let wikipedia become a propagator of mathematical "superstition" or something!!!! :'-(
-- Abe Lincoln 14:49, 29 Apr 2005 (UTC)
Important addendum: Even if there would be only one interpretation, inferring from one special case to another special case is not a valid mathematical method for proofing a solution. It is only legal to infer from a general case to a special case. Though the general case has the "change is good"-solution for n>3 for both interpretations (see above), it is not legal to illustrate the special case n=3 with the more obvious special case n=100 with just one random interpretation. So this approach seems wrong in at least two directions:
-- Abe Lincoln 15:26, 29 Apr 2005 (UTC)
The following paper is one of the better explanations of this problem that I could find online: http://econwpa.wustl.edu/eps/exp/papers/9906/9906001.pdf I believe that it explains much more clearly why 1/2 can be a reasonable answer, as compared to this current article. -- Wmarkham
I agree. However, as presented in the current version of the article, that interpretation is entirely valid. In other words, the current presentation of the problem is insufficiently specific. Furthermore, the entire reason that this puzzle is a "paradox" is due to the confusion over what question is being asked, so it seems to me that this article ought to explain it clearly. Currently, I don't think that it does that. -- Wmarkham 23:47, 3 May 2005 (UTC)
I think so. Even with the statement "Monty always picks a goat", it is possible that Monty sometimes picks a goat that is behind the door that the contestant has initially picked. -- Wmarkham 02:38, 4 May 2005 (UTC)
I took a stab at rephrasing the question. I think it is unambiguous, and worded in a more natural manner. If not, feel free to revert. -- Wmarkham 02:19, 5 May 2005 (UTC)
I've done my best to incorporate some common confusions on this page, including one alternate (1/2) variant (and why it does not apply in this case), which I think also helps to express why the initial choice is so critical -- 2/3rds of the time, it eliminates one potential goat, forcing Monty to reveal the car.
I also added the 'problem summary' section, for lack of a better title -- I was hoping for something more like 'problem givens' or 'given statements' or 'known facts' or whatnot, but none really seemed to sound right. I added it because these are not assumptions, but rather, integral parts of the main problem... things a lot of people seem to miss, in whole or in part, when reading through and forming their conclusions.
I'm hoping all this will make the matter clearer. It's an interesting thought experiment, and I'm guessing a lot of people are so sure it's 1/2 until they see the right explanation for them, something clicks, and they realise why it's 2/3. For me, it was the scenario-based one. YMMV. -- Wisq 03:37, 2005 May 3 (UTC)
I think the wording is really essential. It must be distinguished between two different problems:
In case A it's better for Jane to switch for the described reasons. In case B it doesn't matter whether you switch, but the car is shown by Monty himself with the probabilty of 1/3. If Monty incidentally does not pick the car, the other both doors still have the same chance 1/3.
To make this a lesson in probability or rather in understandig a problem of probability, it must be worded in a way that the reader might think it is problem B when it indeed is problem A. So it must not be identified too obvious as either of the both.
The current wording is
Does anyone have other suggestions? -- Abe Lincoln 13:04, 5 May 2005 (UTC)
In "Letters to the Editor", Selvin provides a statement of the problem that describes a dialogue between "Monte Hall" (sic) and a contestant. The basic events are as follows:
The host explains that the keys to a new car are contained within one of three boxes, labelled A, B, and C. The other two are empty. The host explains that if the contestant chooses the box with the keys, the contestant will win the car. The host solicits a selection from the contestant, who selects box B. The host hands box B to the contestant. The host offers the contestant $100 for the box. The contestant declines. The host offers the contestant $200 for the box. The contestant declines. (The audience agrees!) The host provides the analysis that the probability that box B contains the keys is 1/3. The host offers $500 for the box. Once again, the contestant declines.
The host says "I'll do you a favor and open one of the remaining boxes". The host opens box A, revealing it to be empty. The host provides an analysis that the probability that box B contains the keys is 1/2. The host offers the contestant $1000 for the box. The contestant counters this offer with the offer, "I'll trade you my box B for the box C on the table".
Selvin asserts that "The contestant knows what he is doing!"
-- Wmarkham 18:06, 5 May 2005 (UTC)
I think that some of the recent edits have been heading in an unfortunate direction, making the initial problem statement so detailed in order to prevent misunderstandings that it's becoming hard to find the essential meat of the problem in the problem statement.
I'd like to propose instead that we start with a very simple statement of the problem situation, and the fact that the probability of winning by switching is 2/3, then follow up with something like:
-- Antaeus Feldspar 23:48, 5 May 2005 (UTC)
Section "The problem" says "Nydick." This author is listed in "References" as "Nudick, Robert L." Which is correct? Graue 01:47, 25 May 2005 (UTC)
Anyone think this article is NOT ready for WP:FAC nomination? I haven't worked on it much, but I'd be willing to nominate it if no one else is interested. -- Rick Block ( talk) 19:04, Jun 12, 2005 (UTC)
Here's how I finally understood it in my soul.
There is a 1/3 chance that the contestant's original door has the car, because he chooses randomly. There is a 0 chance that the door Monty opens has the car, because it always has a goat. The probabilities sum to 1. Therefore, the probability that the car is in the remaining door is 2/3, so switching is good. People think the probabilities are equal because they miss the fact that Monty deliberately, not randomly, avoids choosing the door with the car. By not picking a door, he shows it to be more likely to contain the car. Superm401 | Talk July 1, 2005 06:59 (UTC)
You are WRONG(or at least I think). Admitedlly, in the start there is a 1/3 percent chance. But then Monty skips the door you chose, along with the other one. He selects one, so he also skipped over the door you picked. Now, the goat has been revealed (or the first one) so it is fifty fifty, switching doesn't matter.
I believe the central points of these two sections might be replaced by the following explanation:
I think maybe this captures the essential points of those sections? -- Antaeus Feldspar 3 July 2005 23:00 (UTC)
This argument comes down to two mutually exclusive views:
1) The initial conditions of the game, (three doors, two goats, one car) have everything to do with the final choice.
2) The initial conditions of the game, (three doors, two goats, one car) have nothing to do with the final choice.
You can argue the niceties of the statement of the problem until the goats die from the heat produced by the studio lights, but proponents of either side will never convince the other. This is known as a religious war.
Never-the-less, I will add my two cents.
I am in the camp number 2. The statement of the problem, no matter how carefully crafted, is a classic example of a RED HERRING. Proponents of solution number one are falling prey to the "too much information" trick. Most of the explanations above are superfluous, as they just don't apply to the issue. The only purpose of the three doors stage of the game is to provide drama and indirection.
The final choice the contestant has to make is between two doors not three. Pure and simple, two doors, 50% chance either way, end of story. What has gone on to get to that point is IRRELEVANT. Monty may have just as well arranged to have one of the goat doors open when the curtain is pulled aside to reveal the doors. It doesn't matter if you had a million doors, and opend 999,998 - at the end you still have 2 doors, one with a car and one with a goat.
There is indeed one randomizing event, and the 1/3 versus 2/3 probability applies correctly to the initial red herring part of the game, however, THAT IS NOT THE GAME. As soon as Monty opens a door with a goat and offers the contestant a new choice, a choice between two doors, it is a new set of conditions. The 1/3 chance of the opened door does not 'magically' get transferred to the unpicked door, nor does it get 'magically' spread across the two remaining doors. The entire original situation is made null, and a new set of conditions apply.
Computer simulations designed to show the results of solution number one will always support solution number one. I can just as easily devise a computer simulation that will support solution number two (and I have done so - big deal). Both simulations, by their very nature, are biased to the solution they are meant to support. -- unsigned comment by 144.140.62.104 ( talk · contribs)
I agree because the intial conditions do not matter at all. After the host opens the goat, you know behind the door you choose could be one of two things, a goat or a car, and same for door 2. You know know your door, for sure is either the right or the wrong one, compared to what you knew at the start. The problem INVOLVES changing your conditions. To make it clearer, say behind a peice of paper, you had three sets of numbers, of which tow are the same. The contestant must guess the one set, not the two other identical sets, Now, the contestant in this game knows the sets of numbers, so he has a one in three getting the correct set of numbers. But know suppose one of the set of wrong numbers was revealed. It would leave the player with one right answer and one wrong answer. It is like someone asking you the distance from here to "a" and from "m" to "n" you not given any of the vasriables. You may as well take a random guess. It is not like the host skipped over yours, even if you choose the goat the host would simply open the other. Get it?
Sorry, but you still miss the fundmental, unchanging, unarguable point. The contestant is finally and uncontrovertably given a choice between two doors. Not three. Two. I'll repeat that once again. Two Doors. There are only two doors. There is only one goat (unless the car is a Pontiac GTO ;-) ). The contestant has to make a choice between two doors. Do you get it yet? Three doors is a red herring. There are only two doors in the game.
I don't have to supply you with my computer simulation to demonstrate the point, you would point out where my simulation is biased to my solution by makeing the decision that only the last choice is relevant, while I would point out where your simulation is biased to your solution because it makes the decision that the first choice is relevant.
Consider the computer game MineSweeper (available on Windows and every Linux distribution I have seen. Fundmentally you have to 'step' on a square which may or may not contain a 'mine'. If you miss the mine, a number is exposed which tells you how many mines are in the adjacent 8 squares (or 5 for an edge, or 3 for a corner). As you progress through the game, if you step on a square that does not have any mines adjacent to it, the game will automatically clear the surrounding squares until the exposed space is bounded by squares that do have adjacent mines.
Suppose at the beginning of the game, we choose a non-edge square, there is no mine, and the number '1' is exposed. Clearly, each of the adjacent 8 squares has a 1/8 chance of hiding the mine. If we randomly choose one of the 8 and it is clear, each of the remaining 7 squares has a 1/7 chance of hiding the mine. So far I think this description is matching your arguments above, because we are eliminating the clear squares not the computer or Monty Hall.
Now suppose that as we progress the computer clears some blank space that it knows does not contain the mine. We may find situations where there is a number '1' surrounded by 7 cleared spaces and one hidden space. The chance that the remaining one space contains the mine is 100% by my rekoning and I doubt that you would argue otherwise. Now we may also find situations where the computer has exposed a number '1' with 6 cleared spaces and two uncleared spaces. Disallowing overriding information from other squares on the board (in a real game you have other clues to decide which square to pick most of the time), you have two squares to choose from. I argue that the odds are 50/50 on each one.
I know this does not describe the process of pre-picking one of the squares, so there is no room here for consideration of the alternate solution of 1/8 versus 7/8 on the two squares. But that is my point: the pre-picking of the door is irrelevant to the final choice the contestant has to make, and the MineSweeper example exactly describes the conditions and choice the contestant has to finally make. The contestant does NOT have to pick between 3 doors. End of story. (Sorry I haven't figured out how to put in a proper signature yet - Keith Latham)
I'll just add one more half cent then I'm through. In a previous edit, Antaeus said "what Monty knows is crucial to the problem". Well of course he does have to know where the goats are to safely remove one of the losing doors from the game. But what is more crucial is what the contestant knows. And what the contestant knows is that there are two doors, one of which hides a car. Removal of one losing door gives zero information about which of the remaining doors hides the car. -- Klatham 4 July 2005 11:13 (UTC)
P 1 2 +-++-+-+ P!C!|G|G| +-++-+-+ 1!G!|C|G| +-++-+-+ 2!G!|G|C| +-++-+-+
P 1 2 +-++-+-+ P!C!| |G| +-++-+-+ 1!G!|C| | +-++-+-+ 2!G!| |C| +-++-+-+
I'm very interested in the code for this computer simulation that provides a 50/50 result. Could you please display it here, or provide a link to it? I think it would be demonstrative in showing the differences in rules between the problem you are imagining we are talking about, and the problem we are actually talking about. In my experience, there are three kinds of people who disagree with the Monty Hall problem's solution... those who don't yet understand it (but will, usually with increasing n=3 to n=arbitrarily large number), those who misunderstand the problem and think we're talking about another problem entirely (who are usually convinced when the rules are unambiguously stated to their satisfaction), and those disagreeing for the sake of disagreeing (who can never be convinced). I suspect you lie under catagory two, and determining the differences between your problem and the problem we're talking about could be key. Fieari July 5, 2005 19:09 (UTC)
As a layman who just read all this (and it's fascinating), I would like to state that Anteus' grids are extremely informative, since I now understand what he's been trying to convey for months in statistical terms. However, it seems to me that the Monty Hall problem is paradoxical in that while it is true that switching increases the statistical chance of winning from 1/3 to 2/3, ruling out a door gives the illusion that there is a 50% chance that the car is behind either door.
Let me see if this makes sense:
Anteus' grid clearly demonstrates that the smartest decision that a player can make is to switch from the door he's chosen, since in 2 out of 3 cases, he'll win by doing so. However, at that point 1 door has a goat and 1 door has a car, thus giving the illusion that the choice is 50/50, which it really isn't.
Does any of that make sense? (JRVJ - I also don't know how to sign my name yet).
I disagree with including simulation results in the article. I don't know where people are going with this. The simulation results don't increase the value of the article since the solution is already nicely explained with probabilities. One could argue that these results are against Wikipedia's no original research policy. 64.229.142.80 02:10, 23 July 2005 (UTC)
KEEP! I didn't believe this article until I saw the computer program results of 2000 to 1000 or whatever it is. Computer program results are easier to understand then math results.
The probablility of the goat sitting still is quite small. Therefore, the contestant will have the following possibilities after selecting a door:
-Monty picks a door where there was a goat
-Monty picks a door where there are now two goats mating
-Monty picks a door with a goat that is sleeping, going potty, or is too indifferent to move
As such, the door the contestant picked and the other unopened door will have one of the following:
-A messy stage where a goat once was
-Two goats mating
-A stall with a goat that is sleeping, going potty, or is too indifferent to move
-A car that has the seats chewed, the bodywork dented, and the paint scratched by (a) rampant goat(s)
Therefore, the contestant's best choice would be to Not give up the prize they won earlier in the show and stay out of the Final Prize Round of Let's Make a Deal.
-Pf
The answer to the problem is no; the chance of winning the car is the same when the player switches to another door.
There are four possible scenarios, all with equal probability (1/4):
In the first two scenarios, the player wins by switching. The last two scenarios the player wins by staying. Since two out of four scenarios win by switching, the odds of winning by switching are 1/2.
If we make a distinction between goat one and two when the user picks a goat, we need to do the same when the user picks a car. Why would we do otherwise? They are both possibilities aren't they? User:Oyejorge (07:49, July 23, 2005)
I took the liberty of signing your message for you. In the future, you can do this yourself by putting four tildas ( ~~~~ ) after your message, which will both sign and date your message automatically. In responce though, the last two events you have listed there are actually, mathematically, the same event. There's no real difference. To rephrase one of the "Aids to Understanding" placed on the main page, consider the following slightly modified situation...
Now add one more element...
I'd like you to first answer the two questions I've posted above, and then if you still doubt the validity of the solution, tell me how the second problem differs from the Monty Hall problem. Fieari 08:08, July 23, 2005 (UTC) I just considered something more... if the problem was reworded to constrain Monty to picking ONLY Goat #1 if the car is picked, it wouldn't change the problem at all. In the card version, we'd have a Queen, the Jack of Hearts, and the Jack of Spades. Player two, when picking up the remaining two cards, MUST discard the Jack of Hearts if he has it, and the Jack of Spades otherwise. No randomly picking which one to discard. I propose that wording it this way doesn't change the problem at all, and neatly eliminates your last option of the four you listed. Fieari 08:21, July 23, 2005 (UTC)
Remember that the two last ways have just half the possibility as the two first. And are mathematically seen the same event as one and two. But one could say that the possibilities for the events are 1: 1/3, 2: 1/3, 3: 1/6,4 1/6. Gillis 08:19, 23 July 2005 (UTC)
Ok, I've been humbled. I don't think I should do this sort of thing late at night. Here's a diagram I used to understand the problem... It's the same info as what's in the article, just presented a little differently, maybe it'll help other neighsayers. Thanks for the clues Fieari and Gillis,
First Choice | Probability | Monty Hall Takes Away | Switching Wins? |
---|---|---|---|
Goat 1 | 1/3 | Goat 2 | Yes 100% of the time |
Goat 2 | 1/3 | Goat 1 | Yes 100% of the time |
Car | 1/3 | Goat 1 or Goat 2 | Yes 0% of the time |
Oyejorge 17:44, 23 July 2005 (UTC)
I fixed the last cell in the table, which used to read "No 0%." Also correct would be "No 100%," but that's not as straightforward. Old Nick 19:22, 25 July 2005 (UTC)
I remember once reading that this riddle/theorem whatever is much older then ever mentioned in the article. i understand it was atelast from the 19:th century.
I can't verify this with google though. Does anyone have any better knowledge of the issue or better googling-luck? Gillis 08:13, July 23, 2005 (UTC)
I would argue it's a problem in probability. Rich Farmbrough 12:44, 23 July 2005 (UTC)
Can showing you that one of the three candidates in a tight race is a loser make you change or question your vote?
I happen to believe that people love to feel included by voting for the popular candidate.
Let's say people were given a choice to vote for three candidates A B C. After the election, you find out that one of the candidates was a loser and the two candidates were so close that a special runoff election was required. Would you be more likely to be with the majority by switching your vote on the runoff?
Consider this election ..
Dino Rossi (R) and Christine O. Gregoire (D) in 2004 ran one of the tightest governor races in history. Christine Gregoire won by a mere 129 votes.
The Republicans wanted a runoff election to solve the inaccuracies.
Do party affiliations play in this problem?
This poorly-titled section formerly said this:
This is circular logic. In support of the idea that door A ends up with a 1/3 probability of having the car, it makes the statement that "the probability of the car being in A remains constant" without any support. One may reasonably ask, why doesn't the probability of the car being in C remain constant instead? This explanation provides no answer. (It also ends with a spectacularly muddled sentence that adds nothing but confusion.) I have remove the section. -- Doradus 16:29, July 23, 2005 (UTC)
Probability is 2/3 if all decisions are made before any doors are opened. Probability is 1/2 if a fresh decision independent of history is made after Monty reveals a goat.
This misunderstanding is why the problem seems counterintuitive to people.
So should you switch if Monty gives you the choice AFTER he reveals a goat? Doesn't matter. Should your predetermined strategy for playing these games be to always switch? Absolutely.
Yes, it is quite clear to me, and you have apparently met the first such person. You are wrong if you say that 2/3 chance is independent of knowledge prior to Monty revealing a goat. It comes down to history. Without the history of Monty revealing a goat, it is merely a decision between two doors, exactly one of which has the prize. That gives you a 50:50 chance, and that is why people intuitively think there should be a 50:50 chance. It is no different than any other 50:50 scenario. That is to say, if in your life's travels you keep coming across a pair of mysterious doors with one prize, it makes no difference which one you choose and how often you change your mind. However, if the game involves the entire history starting with 3 mysterious doors, then consistently switching will increase your chances to 2/3. It is all about the given history. Indeed, if you are given additional history that the prize is behind your first chosen door, then switching will reduce your chances to zero.
The problem is that people don't seem to understand probability is context-dependent--it varies with what information you put into it. Increased knowledge increases certainty. Of course this isn't explicity stated, because why would the entire problem even be presented if it is to be ignored? It is understood that it is to be incorporated into probability calculations. However, folks who don't understand the dependence of history see the Monty problem as equivalent to the 2-door problem. They seem to think that 2/3 is some sort of physical feature of the door, rather than a measure of information and uncertainty.
So I say again, if you make a fresh decision (one ignorant of the past), your chance is 1/2, whether or not you switch. That is why knowledge is a good thing. You can often use it to increase your chances in life.
This whole thing seems to hing around independent steps.
50/50 result. Some condition is set up; in this case chose 1 of 3; chance of selecting the correct outcome is exactly one third. BUT the next INDEPENDANT step is offered after the removal of one of the incorrect outcomes. What is left ? 2 options; of which one and only one is correct.Proberbility of picking the correct outcome is 1 out of 2 or 1/2.
Please demonstrate very clearly and precisely how any event prior to being offered the 2nd step choice can possibly influance the outcome of that choice.
Or to restate that - you are left with a choice of one out of two.
Somebody mentioned tossing coins; it doesn't matter how many sides previous coins had; or how many coins were tossed prior; the next toss of a double sided coin has what probability of producing a head... yes; exactly 1/2 as PRIOR conditions have zero effect. I presume at this stage no one would bet a million pounds/dollars/euros that following 1 million heads a tail would occur ? (and if you believe it MUST be a tail then you need to re-vise your basic proberbility theory
Please cut all the waffle and explain how any preceeding action affects the 1 out of 2 selection left at the end !
I'm stuck behind a dynamic isp so can only sign Pete-dtm
Jeesh, I can't make up my mind as to which is more interesting (revealing), the article or this talk page! hydnjo talk 18:26, 23 July 2005 (UTC)
The corrolation between gravity and time becomes evident at a black hole.
This is the clearest way to understand it. Good job! Is it in the main article? --
pippo2001
21:46, 23 July 2005 (UTC)
This was a great featured article. Congrats to those who contributed.
LegCircus 22:34, July 23, 2005 (UTC)
One point that's not clear: if you have a larger number of doors (say four for simplicity), is the new situation one where Monty Hall opens one door or two doors? Peter Grey 22:46, 23 July 2005 (UTC)
I've presented the solution as a way to "trick the host". I think beating the house appeals to most people in a way that dry mathematical explanation doesn't. — 131.230.133.185 23:47, 23 July 2005 (UTC)
I find this diagram at best confusing. Or should I say the diagram looks fine, the numbers at the bottom (which add to more than 1) are confusing. Rich Farmbrough
The current article (which is incorrect, by the way) states the following:
This is fundamentally wrong because it leaves out a scenario. It can be fixed in one of two ways:
Fix #1: There are two possible scenarios, both with equal probability (1/2):
In the first scenario, the player wins by switching; in the second, the player loses by switching.
Fix #2: There are four possible scenarios, each with equal probability (1/4):
In the first two scenarios, the player wins by switching; in the last two, the player loses by switching.
Once the host opens a door and reveals a goat, the old problem is gone. The new problem is there are two doors: one has a goat and one has a car. What are your odds in picking one? (Picking means changing doors, because the odds of winning stay at 1/3 for the first-chosen door.) The answer is one-half. — Fingers-of-Pyrex 00:01, July 24, 2005 (UTC)
Kinda looks like this talk page. I guess there's no hope, they're everywhere. I'm going to rv the link as it doesn't add anything new to the article except controversy. hydnjo talk 00:50, 24 July 2005 (UTC)
Methinks all should read Wikipedia:Resolving disputes. We may need to change the article's focus to the Monty Hall problem and then have two different sections for explaining the answer—one wrong and one right, of course :). In the meantime, we should apply the {{dispute}} tag to the article to point readers to this talk page. — Fingers-of-Pyrex 01:21, July 24, 2005 (UTC)
One of Wikipedia’s policies is No original research. So, please do not try to prove to me what the answer is with your own calculations. Show me your source.
The following web sites claim that the answer is 2/3 probability if you switch:
For those that assert that the probability is 1/2, please site your source. -- JamesTeterenko 03:01, 24 July 2005 (UTC)
Hi. It doesn't matter if there are one trillion doors, cards, or whatever. Let me pick one. You show me all the other goats until there are two doors left. I'll have one-trillionth chance of winning the car if I stick with my first choice. I'll have 50% chance of winning if I switch. — Fingers-of-Pyrex 02:53, July 24, 2005 (UTC)
Part of what messes up intuition is how to make use of the partial information you have when deciding whether or not to switch. Here's another way to look at it that might help (this is sort of a rewording of "Best, most concise explanation yet") :
Assume the contestant always switches, i.e. ignore the second decision. The contestant's first choice is either a) car (probability 1/3) or b) goat (probability 2/3). In (a), contestant loses, in (b) contestant wins. Thus 1/3 probability of losing, 2/3 probability of winning.
Now assume the contestant always stays. The contestant's first choice is either a) car (probability 1/3) or b) goat (probability 2/3). In (a), contestant wins, in (b) contestant loses. Thus 1/3 probability of winning, 2/3 probability of losing. Peter Grey 04:17, 24 July 2005 (UTC)
As far as I can see, the different in the two viewpoint of 2/3 and 1/2 comes from the interpretation of the problem.
See the "Full Mathematical Solution" below. aCute 05:22, 24 July 2005 (UTC)
In short, the probability of winning really is 2/3 if you use the right strategy. (It's 1/3 if you stick by your choice, the wrong strategy unless you really like goats.) It's 1/2 if you flip a coin to decide whether to switch. Davilla 15:59, 24 July 2005 (UTC)
P(X1) | P(M l X1) | P(X2 l X1, M) | . | P(X1, M, X2) |
P(X1=C) = 1/3 | P(M=G2 l X1=C) = 1/2 | P(X2=C l X1=C, M=G2) = 1/2 | . | P(X1=C, M=G2, X2=C) = 1/12 |
P(X2=G1 l X1=C, M=G2) = 1/2 | . | P(X1=C, M=G2, X2=G1) = 1/12 | ||
P(M=G1 l X1=C) = 1/2 | P(X2=C l X1=C, M=G1) = 1/2 | . | P(X1=C, M=G1, X2=C) = 1/12 | |
P(X2=G2 l X1=C, M=G1) = 1/2 | . | P(X1=C, M=G1, X2=G2) = 1/12 | ||
P(X1=G1) = 1/3 | P(M=G2 l X1=G1) = 1 | P(X2=C l X1=G1, M=G2) = 1/2 | . | P(X1=G1, M=G2, X2=C) = 1/6 |
P(X2=G1 l X1=G1, M=G2) = 1/2 | . | P(X1=G1, M=G2, X2=G1) = 1/6 | ||
P(X1=G2) = 1/3 | P(M=G1 l X1=G2) = 1 | P(X2=C l X1=G2, M=G1) = 1/2 | . | P(X1=G2, M=G1, X2=C) = 1/6 |
P(X2=G2 l X1=G2, M=G1) = 1/2 | . | P(X1=G2, M=G1, X2=G2) = 1/6 |
The probability of winning (not Monty Hall problem) is:
The probability of winning without switching is:
The probability of winning with switching is:
The answer to the Monty Hall problem:
aCute 05:22, 24 July 2005 (UTC)
Note: It is my observation to lead me to conclude the following:
The Monty Hall problem is NOT to determine the winning-probability of the last event or the entire game, but rather to determine which of the two strategies will maximize winning-probability. The problem is like comparing the subset cases in the manner similar to the Simpsons paradox. This is also like trying to find a winning strategy in Blackjack by counting cards already played previously as additional information (sort of like Monty Hall revealing a goat) in order to adjust subsequent choices to gain an advantage over the house. aCute 06:49, 24 July 2005 (UTC)
In "the formal probability diagram" within the project page of the
Monty Hall problem, the leaves of the diagram tree has the values "1/3 1/3 1/3 1/3 1/6 1/6 1/6 1/6" which cannot possibly exist since the sum of these values canNOT exceed 1.
Please someone change them ASAP.
aCute
05:22, 24 July 2005 (UTC)
Yes I have one of those too, and I found the diagram confusing (as noted above). Perhaps the colour of the YES and associates probabilites could be diffeent from the NO and assoc, probs. Perhaps a caption woul clairfy things. Rich Farmbrough 18:25, 24 July 2005 (UTC)
Okay, now I see how the diagram is constructed. Previously, I failed to see that BOTH situations are placed onto the SAME tree. aCute 19:41, 24 July 2005 (UTC)
"Never argue with someone who knows that he's right" hydnjo talk 05:27, 24 July 2005 (UTC)
When this doesn't work well, it's time to go on to something more productive. Buh bye now. hydnjo talk 05:45, 24 July 2005 (UTC)
Ok, so stupid me, but I have to do this. Lets take this down to such a basic level that even I understand.
Respectfully, hydnjo talk 21:18, 24 July 2005 (UTC)
The game is fixed from the moment that you pick one door and Monty gets two doors. All else is just fluff and misdirection to make the game last longer. You hold a 1/3 chance of winning and (because he has two doors) Monty has a 2/3 chance of winning. Swap with him at your earliest opportunity. Please. hydnjo talk 02:10, 25 July 2005 (UTC)
You get one of three possibilities and Monty gets two of the three possibilities. The fact that you chose first changes nothing. If given the opportunity to swap with Monty well, just jump at it. He did, after all, try to screw you at the very outset. And, if Monty tries to confuse you with more doors, well then hold fast. When he offers a swap (which he will) just take it. Don't forget, you got to pick one door and Monty gets all the others. Lucky lucky him he would like you to believe! hydnjo talk 02:48, 25 July 2005 (UTC)
(as opposed to the correctness of the mathematical answer)
It is not enough to describe why the mathematically derived solution is correct. To resolve the paradox to the satisfaction of all, one must also describe why the intuitive solution is wrong. I think this requires three steps.
1) an understanding that the "intuitive solution" is just a dismissive term for a first analysis that turned out to have a flaw
2) a description of the logical steps that were employed in coming to the intuitive solution
3) an analysis of that logic, to find its flaw(s)
(1) I not argue the point, but rather just hope that people agree with it.
(2) I think that the intuitive solution takes the following steps:
Here is the game that I believe is used intuitively (I’ll call it the Silly game):
Should he change it or not?
Clearly the answer to this is that the chance is 50% either way, and it does not matter whether the contestant changes his mind.
Now compare the Silly game to an original Monty Hall game at a point half way through, in which the contestant has guessed door B and Monty has opened door A.
It seems that the Silly game is exactly the same as the original Monty Hall game at this point. To a viewer who has just ‘tuned in’ and does not know what has previously happened, the games look identical. Therefore, logic would dictate that the answer to the original question is that it doesn't matter whether the contestant changes their mind; the probability is 50% either way.
(3) It turns out, of course because the two games are not completely isomorphic. There is a crucial piece of information missing from the Silly game that is needed to make it isomorphic with the Monty Hall game, as follows: Monty (who knows where the car is)
Monty has thus said something concrete about door C (“I didn’t choose it, perhaps because I couldn’t choose it”), but nothing about door B. This is the source of the asymmetry between doors B and C, and the reason that door C is more likely to not have a goat behind it. Happyharris 20:57, 25 July 2005 (UTC)
For the people who still find it hard to see why the probability is not 50/50 when there are two doors left, I hope the following illustration may help. I'm going to show that the Monty Hall problem is a specific case of a more general game; I'll call all the games that differ in their parameters "Hall games" for ease of use.
The basic idea behind a Hall game is this: We start with one large set of secret-hiding items -- they can be doors to be opened, or cards to be turned over, it doesn't actually matter. What matters is that there are n cards, but only one of them is the Prize card. The rest are Null cards.
Step One: The cards are divided into two hands. Each hand must have at least one card. For simplicity's sake, we call the number of cards in the first and second hands h1 and h2.
Step Two: Someone who can see which cards are Nulls can discard some number of Nulls from one hand, the other, or both. We call the number of cards remaining in each hand after the discarding of Nulls r1 and r2 (like h1 and h2, they must be at least one.)
Step Three: The player makes a guess at which of the two hands contains the Prize.
Step Four: The player makes a guess at which card out of the hand he selected is the Prize.
It's clear that to win the game, the player has to make correct guesses in both Step Three and Step Four. What are the chances of picking the correct hand? The first hand is correct h1/n of the time; the second hand is correct h2/n of the time. If we want, we could enumerate the cases: if h1=2 and h2=5, then the Prize could be the first card of the first hand, the second card of the first hand, the first card of the second hand ... et cetera, et cetera.
Now this is the part that many people find counter-intuitive. If we enumerate the cases, and then we reduce Nulls to meet any legal value of r1 and r2, we find that in no case can the removal of Nulls switch the Prize from one hand to the other. This means that the chance of the card being in the first hand or the second always stays at h1/n and h2/n -- even if the sizes of the hands do not stay at h1 and h2. If it's dealt to that hand, it stays in that hand; therefore the chance of it being in a particular hand is always equal to the chance that it was dealt to that hand.
What about the removal of Nulls? Does it affect anything after all? Yes, it does -- it affects the player's chances in Step Four. The chance that the Prize is in a particular hand is dependent upon h1 and h2 -- how many cards each hand started with. But the chance of finding the Prize in a hand (assuming it's the right hand) is based on how many cards that hand contains after Nulls have been removed -- since there's only one Prize, the chance of finding it in a hand of r1 cards is 1/r1.
Now, what are the chances of making both guesses correctly? If no Nulls get removed from either hand, then the chances of picking the Prize are either h1/n x 1/h1 (since r1 is equal to h1 when no Nulls have been removed) or by similar logic h2/n x 1/h2, which also multiplies out to 1/n. If, however, one of the hands -- say, hand 1 -- has been reduced down to one card (r1 = 1), then the chances of finding the Prize in that hand is h1/n x 1/1 -- if you've correctly guessed that the Prize is in that hand, you have a 100% chance of finding it in that hand when it's the only card in the hand.
With this being the general structure of the Hall game, we can see that the Monty Hall problem is really just the case where n=3, h1=1 and h2=2, r1=1 and r2=1. The chance that the Prize is in the player's hand is h1/n -- 1/3. The chance that it's in Monty's hand is 2/3. Before one Null is removed from Monty's hand, the chance of finding the Prize in his hand is 2/3 x 1/2 -- i.e., 1/3. But when the Null is removed, the chance is now 2/3 x 1/1 -- i.e., 2/3! -- Antaeus Feldspar 03:44, 26 July 2005 (UTC)
kudos to those contributors to this article. i have been thinking about this for days (despite the fact that i "got it" ... after about 20 minutes or so). i find the "two sets" explanation the most clear, though i'm sure some people will love the bayes' theorem explanation. i also found the talk page very entertaining. some contributors are clearly manifesting what kahneman and tversky (writers on cognitive biases - the former won a nobel prize) refer to as "belief perseverance." geez, sit down with a friend and three cards for 10 minutes and you'd realize you're wrong. anyway... i thought you might find this page interesting:
http://econwpa.wustl.edu:8089/eps/exp/papers/9906/9906001.html
besides elaborating on many of the key assumptions (often unstated) underlying the MHP, the paper presents an interesting (and supposedly earlier) problem of the same form, the "three prisoners problem." here's the upshot...
There are three prisoners, you and Prisoners A and B. Two of you are to be executed, while one will be pardoned. The prison warden knows who will be executed and who will be pardoned (like monty must know where the goats and car are). According to policy, the warden is NOT allowed to tell any prisoner if he/she is to be pardoned. You point out to the warden that if he tells you if A or B will be executed, he will not be violating any rule. The warden says C is to be executed. What is the chance that you will be pardoned?
the answer presented, like that to the the MHP, is that your chances of being pardoned remain at 1/3, while the the chances of B being pardoned, GIVEN WHAT THE WARDEN HAS TOLD YOU, is 2/3.
I found looking at this problem reinforced how *$&%^@# hard it is to get one's head around the MHP, because despite having read about the latter for days, i still had to think about the three prisoners problem for several minutes to get my head around IT, despite knowing that it was essentially the same as the MHP. K-razy.
someone mentioned the idea of blocking edits to featured articles for some period of time. while apparently there is a policy against such things, i agree with the suggestion, at least where articles that are sources of dispute. the featuring of an article doubtless attracts many potential editors, some of who may simply not know enough about the subject to edit appropriately. for example, the first time i read the monty hall page (when it was featured), someone had altered the solution section to express the misguided (given certain assumptions, which too often are unstated) 50/50 approach. needless to say, i was confused by the fact that all other sections of the article contradicted the solution.
good job. Contributed by 24.89.202.141.