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Ok, the main problem I see is with the invalid grouping of outcomes, which causes the number of possibilities to be reduced. The following 2 are syntactically identical ways of stating the same thing, both of which illustrate the number of possible outcomes:
OR
In the first example, if the player chooses to switch, the car is won in the first two cases and lost in the second two. A player choosing to stay with the initial choice loses in the first two and wins in the third and fourth cases. In the second example, if the player stays they lose in one and win in the other.
The original question is a classic case of riddle misdirection, nothing more. The fact that it is as old as it is and people still do not see it astounds me. If one of the incorrect doors is eliminated, and the player is then forced to make a decision, it changes to a straight 50/50 chance of getting the correct door. The initial choice can then be discarded. This article seriously needs to be revised.
Mvandemar 05:30, 8 November 2006 (UTC)
How can the probability magically change??? What if we assume their are only two doors to choose from at the start of the game, the third door being an attempt to persuade participants into falsely believing three choices exist. The question of the zero probability door being revealed in the middle of the game should be irrelevant because we know at the onset of the game that a door with zero probability will be revealed before the end. -- Nbritton 17:05, 9 March 2007 (UTC)
It has a 2 in 3 chance of being wrong.
-- 220.237.67.125 15:58, 8 November 2006 (UTC)
The explanation above is only true for external obser who knows the host choice is not random. However in the perspective of the contestant who assumes the choice of the host is random the chance is still 50/50. I suggest you emphasise this fact in the article. Nice and clean explanation though from user above.
Actually this explanation seems perfectly correct, whether or not the host choice is random. If the contestant's first choice is incorrect then opening the second door must indicate the correct choice. Either Monty Hall opens the second door and reveals a goat, in which case the contestant should switch to the last remaining door, or else Monty Hall opens a door and reveals the car, in which case the contestant should obviously switch to that door.-- Lorenzo Traldi 10:41, 20 December 2006 (UTC)
ok so a lot of people have been saying that if Monty forgets where the car is but luckily picks a door with a goat in to "eliminate" this makes the remaining choice 50/50. i am sorry but i thought i was understanding this thing up until that. how does it? how can odds change based on the host who is surely peripheral to the experiment? maybe i've missed it but i consider this to be unexplained and SEVERLY confusing 28/11
A simple solution to the "Monty doesn't know" problem is that if Monty doesn't know, then his choice is between two doors that each hav a 1 in 3 chance (the other 1 in 3 chance door being already picked by the contestant). Assuming Monty picks the goat door (as we must) this simply removes one of the 1 in 3 chance doors and we are left with two doors that each had a 1 in 3 chance (the same chance)when there were three and so now as equal probabilities have a 1 in 2 chance since there are only two. Davkal 18:09, 28 November 2006 (UTC)
yes but how is the immediate above different from the original problem?? you could use surely use the exact same logic for the original problem. lets forget probabilities here and return to the root of the question "are you more likely to win by switching or not?".
Response: it's not clear you can talk about "more likely" without discussing probabilities, so the answer is no, in the new game you are not more likely to win by switching because the odds are 50/50. Davkal 13:45, 29 November 2006 (UTC)
now i understand that the reason (in the original question) you will win is because you have essentially split the three doors into two sets, then after you pick between the two sets. thus you have a better chance of winning with the set of two (even tho one has been revealed, tho this is meaningless). surely if you were to repeat the experiment with the host not knowing and took the results of the games where he randomly chose the goat (discarding the games where he chose the car as the question states the chances if he picks the goat) you would still have a better chance of switching? mentioning such things as the odds of monty picking the car HAVE NOTHING TO DO WITH THE QUESTION. the question states he has chosen the door with a goat behind it. thus logically you still have a better chance of switching.
Response: if you were to repeat the experiment in the way described then all the cases that are discarded (the cases where Monty picks the car by accident) are cases where you would have won by switching (i.e, you didn't pick the car first). Overall, then, your score drops to 1 in 3 by switching, 1 in 3 by sticking and the "removed" 1 in 3 where Monty picks the car. So your choice to stick or switch is now between the two remaining 1 in 3 chances which equals 50/50 in those remaining cases. Davkal 13:45, 29 November 2006 (UTC)
imagine two of the "monty hall problems* playing side by side on a stage with a screen seperating them. on either side of the screen, both contestants pick their first box. on one side of the screen, the producer tells monty which of the remaining two boxes he must reveal (showing a goat), he does so. on the other side the host cannot here the producer very well and cannot be sure which box to pick. still he manages to get the one with the goat in (which he had a 2/3 chance of anyway). now according to what people have said on this site, even though both players games have *PHYSICALLY AND EMPIRICALLY* proceeded in exactly the same way, the first contestant should switch, while it does not make a difference if the second one does. now i know that *odds* can be made to show that it's equal chance, but odds aren't real life. to put it another way, in the first game, obviously the player has a better chance (2/3) if they switch. now imagine both games and the way they went, and if they have different odds, would the odds change if the producer of the second game confirmed to the host that his random guess was right?
Response: the difference is that in multiple trials one Monty would never pick the car where the other Monty (who can't hear the producer) would pick the car 1 in 3 times. The probabilities, then, for what happens in these cases are different, because what would actually happen in these cases IS different. The point being that your chances of picking the ace of spades out of a pack of cards at random are 1 in 52, where your chances of picking it out after looking through the deck to find it are 1 in 1. This is true even in cases where you have just picked the ace of spades out by random chance, and true even if you wanted to say (which I would not) the in both cases what has happened was physically and empirically identical. Davkal 13:45, 29 November 2006 (UTC)
ok i think i understand what you mean, the discarded ones (i.e the ones where monty picks the car) are scenarios where it would be beneficial to switch, so in a case of multiple games, the time when in would be beneficial to switch are halved. thanks for explaining this to me as it was really starting to get me annoyed. but if it was just one or two games, and this thing happened, it would still make sense to switch, right? i mean, in a real life situation, even if this was happening, it would still make sense to switch, as you still have a lesser chance of getting it right first time? 29/11/06
No, because all the multiple trials are intended to show here is what could be the case. And what they show is that in a one-off case where you pick a random door, and then Monty opens a random door from the other two and reveals a goat it's 50/50 between the remaining doors. This is because the one-off case is one of only two remaining possibilities - you picked the car and Monty picked a goat or you picked a goat and Monty picked a goat - in one you already have the car, in the other you don't. The third possibility, you picked a goat and Monty picks the car can be ruled out in the one-off case because we know it hasn't happened (i.e. Monty didn't pick the car).
Davkal
15:27, 29 November 2006 (UTC)
i'm sorry but now you've really confused me. consider this
"This is because the one-off case is one of only two remaining possibilities - you picked the car and Monty picked a goat or you picked a goat and Monty picked a goat "
how is this different from the original case!? monty is always going to pick a goat, so i fail to see how in a one-off one time only case, this changes the odds? say two games on the same stage, side by side one time only. both games proceed exactly the same however monty one is told by the producer to reveal a certain box at the second stage. monty two (who cannot see the other game) does not have a producer, but coincidentally picks the same box. the two contestants must have the same odds right? —Preceding unsigned comment added by 194.168.3.18 ( talk • contribs)
i'm sorry but i still don't agree with this. the question is not about options or possible scenarios, but if monty picks a goat, now consider this ;
":# The player picks the car. Monty opens the door containing the first goat. Winning strategy: STAY
in the strategy of the player, it doesn't matter which of the goats monty picks. these two scenarios are basically one. in the original scenario, imagine that the player picks the car, even if monty knows that the two remaining boxes are goats he still essentially "decides" which one to reveal, yet this obviously isn't counted in the analysis. the argument of the original solution was that because you were picking only one out of the three, the choice of switching was essentially a choice between staying with your one box out of three or switching to what was essentially two choices out of the original three. it has also been noted that the "reveal" by monty is a diversionary tactic and "red herring" of the puzzle designed to make it more counter-intuitive. the fact that in the "monty doesn't know" scenario, he has still picked the goat, means the choice to switch is still based on the "one choice versus two". the player is still more likely to have chosen the goat at first (how can you dispute this) thus switching will always be a better option (at least in a one off situation, the question is supposed to put us in the situation of the player and ask us what we would/should do. multiple situations, i.e. ones where monty would discard the games where he accidently picked the car would undoubtedly lower the benefits of switching). the reason (in the original game) why switching is beneficial is more based on the primary "split" than it is on monty's reveal. he will always reveal a goat, and in the "monty doesn't know" scenario, regardless of whether it was random or not, he has still picked a goat, thus the benefits of switching remain the same. your mathematics are undoubtedly right, yet i think the reasoning is wrong. ec 30/11 —Preceding unsigned comment added by 194.168.3.18 ( talk • contribs)
I think it's time for another experiment. Take three cards including the ace of spades. The ace of spades counts as the car. Step 1: shuffle the cards and then deal them face down so you don't know where the ace of spades is. Step 2: pretend you're the contestant and move one card to the side. Step 3: Pretend you're Monty and turn one of the remaining cards over. What happens is this: in about one-third of the trials the card you turn over as Monty will be the ace and you will have to abandon those trials; in the other two-thirds of the cases (where "Monty" reveals a "goat"), you will see that that the original "contestant's choice" card is the ace half the time and the third card picked by neither the "contestant" nor "Monty" is the ace half the time. That is, in the trials that reflect the new puzzle (Monty picks a goat/non-ace) you win half the time if you always switch and you win half the time if you always stick. Do it, try it, and then once you are satisfied that there is no advantage to be had in the new game from either switching or sticking, try to work out what is wrong with your arguments to the contrary above. Davkal 12:34, 30 November 2006 (UTC)
right gonna do it now at work to pass the time... —Preceding unsigned comment added by 194.168.3.18 ( talk • contribs)
Maybe this will help. Consider what happens if Monty has a very specific kind of forgetfulness, so that he attempts to execute the classical problem, but he always gets confused and opens the wrong door. Then if you pick a goat, Monty always shows you the car. So if Monty shows you a goat, that means you picked the car, and you shouldn't switch.
The forgetful problem (in which it doesn't matter if you switch) lies somewhere between the above problem, in which you shouldn't switch, and the classical problem, in which you should.
To be precise, if Monty's memory is such that he opens the wrong door with probability p, the probability that you picked the car, given that he reveals a goat, is given by
You can have fun plugging in values of 0, 1/2, and 1 for p. Melchoir 03:24, 30 November 2006 (UTC)
The long discussion above is similar to a decision between two possible rules for the "clueless host" game. Possible Rule 1: if Monty accidentally reveals a car, the contestant is allowed to choose it. Possible Rule 2: If Monty accidentally reveals a car, the game is canceled and the contestant gets to play again, presumably after the car and goats have been rearranged behind the doors.
Under Possible Rule 1, the original 1/3 vs 2/3 analysis applies ... except the bonus for switching is obvious because sometimes the contestant is looking at the car and can decide to take it. So under this rule, switching is still advisable -- when you see a car take it! When you see a goat, you lose nothing by switching to the non-revealed door, but you also gain nothing. The whole advantage of switching involves the car having been revealed.
Under Possible Rule 2, that advantage has been declared illegal. Consequently the advantage of switching has been removed, and there is no effective difference between switching and staying.
If a careless Monty Hall reveals a goat by accident, that's the same as operating under Possible Rule 2. If the real Monty Hall reveals a goat, that's the same as operating under Possible Rule 1, because in the cases in which he might have accidentally revealed the car he would open the other door instead. -- Lorenzo Traldi 04:21, 20 December 2006 (UTC)
I agree. The "possible rules" simply give reasonable whole-game contexts. I guess I wouldn't say Possible Rule 1 can never play a role in the problem under discussion, because the never seems to forbid Monty Hall's carelessness, which is after all the point in this variant. I'd rather simply say that once the contestant sees a goat we realize a car has not been revealed this time. (Technically, with the careless host the conditional probability of selecting the door with the car when a goat has been revealed is .50, either by switching to the still-closed door or staying with the door chosen first. The conditional probability of selecting the door with the car when that's the one Monty opened should certainly be 1.)-- Lorenzo Traldi 10:34, 20 December 2006 (UTC)
I think the use of "never" is what causes the confusion, as I said.-- Lorenzo Traldi 19:48, 21 December 2006 (UTC)
If Monty does not know, and picks a goat, it is random which goat he picks, and he will always pick a goat. If monty does know, it is random which goat he picks, and he will always pick a goat. Since nothing happens when he picks a car, well, he can't, because then there would be no strategy. If he knows, then he will not pick the car regardless. The results are completely the same. Saying they are different is like saying that if you have a two thirds chance of picking a pair of socks if you put on gloves, and you forget what you are supposed to do to get the two thirds chance, yet still put on gloves, you will have a fifty percent chance of picking socks. Again, He will pick the car then goat one or two (one third chance), goat number one, and then monty picks goat number two (one third chance), and goat number two then goat number one (one third chance). In fact, whether or not he knows, you should always switch if you get the chance, because if you got the chance, Monty most certainly picked a goat. Again, another analogy- if you pick up a tissue because you know your supposed to, and blow your nose with it, and if you pick up a tissue when you are not sure if your a supposed to, there is no chance of blowing your nose with it.
Your logic resembles the logic involved in the original problem. Your math and reasoning are completely off, because your judgement is impaired by incorrect reason. No offense, but it is kind of funny that you got the problem right, then got the problem wrong.
Given this logic, I think to avoid confusion it would be necessary to adjust the article to say that there is a ⅔s chance, because there is. Squarethecircle 00:07, 6 January 2007 (UTC)
It may be instructive to contrast the situation in the Monty Hall Problem with the situation that occurs in the game show Deal or No Deal when the player in Deal or No Deal is down to two cases and facing the decision to switch the case he or she originally selected with the only other remaining case. Suppose that in Deal or No Deal there are 26 cases containing different prize amounts and the highest prize possible is $1,000,000 as in the original version of the show. If there are only two prize amounts left on the board and one of them is $1,000,000 while the other is $0.01, the probability of winning $1,000,000 does not increase if the player switches the case he or she originally selected with the only other case remaining. However, if the host Howie Mandel eliminated the other 24 cases because he knew they didn't contain $1,000,000 then the probability that the player would win $1,000,000 by switching cases would be 25/26. Now what if Howie helped out in eliminating some of the 24 cases and the player got lucky in eliminating the others? :-) —Preceding unsigned comment added by 69.141.232.16 ( talk • contribs)
Vos Savant's answer makes it seem that the host's cluelessness makes a difference. In fact it does not. The contestant now knows that the open door hid a goat. As the contestant originally had only a 1/3 chance of choosing the door with the car, there is a 2/3 chance of getting the car by switching, just as in The solution above. -- Lorenzo Traldi 20:05, 19 December 2006 (UTC)
The host's cluelessness does make a difference because it introduces a chance that the host may ruin the game by accidentally opening a door to show a car and forcing a do-over.* This ruination cannot happen when the player has chosen the car (obviously) but it happens one out of every two times when the player chooses a goat. This means that out of six combinations of doors picked by the player and the host, there are two that result in the winning strategy being to switch, there are two that result in the winning strategy being to stay, and there are two that force do-overs. Assuming that we'll never hit an infinite series of do-overs, this means we really have four combinations: two that result in the winning strategy being to switch, and two that result in the winning strategy being to stay.
(*This is, anyways, the cleanest way to model the problem conditions which state that the forgetful host's random door opens to reveal a goat, even though clearly sometimes random choices would lead to the car being revealed instead.) -- Antaeus Feldspar 22:49, 19 December 2006 (UTC)
It looks like there are now two threads on this -- but anyway under the clueless host assumption, the advantage of switching is only that Monty might sometimes show the contestant the car. (That is, my comment above was wrong -- apologies.) If the contestant is allowed to keep the car under those circumstances, then "always switch" is still good advice -- 1/3 of the time you'll switch to the car because you're looking at it, 1/3 of the time you'll switch to the car because Monty has revealed a goat and your door has a goat, and 1/3 of the time you'll switch to a goat when you have the car and Monty has revealed a goat. In the four cases in which Monty has revealed a goat, switching is 50-50. In the two cases in which Monty has revealed the car switching is 100-0. :-) -- Lorenzo Traldi 05:05, 20 December 2006 (UTC)
Well it does explain where the 2/3 went.-- Lorenzo Traldi 23:51, 20 December 2006 (UTC)
It doesn't matter if there is a do over, because if there is, then there is no choice made. A choice can only be made if he picks the goat, eliminating the chance that he picks a car. Also, the page says he still picks a goat and it is one half. This is impossible. If you think it is not, then you probably should spend the rest of the doing what my user name implies, to knock some sense into you (no offense, it's probably just you confusing yourself. The average person would end up with a similar result if under the same condition). Feldspar, you are completely correct. I will not go into further depth on this situation because it has been explained above. Squarethecircle 00:13, 6 January 2007 (UTC)
"It doesn't matter if there is a do over" is simply not correct. For instance, suppose we toss a coin and call a do over every time we get heads. Then the only toss that counts is tails! -- Lorenzo Traldi 19:28, 14 January 2007 (UTC)
Personally I think "just happens to show a goat" is clearer, and it is the do over that is introducing a new rule that seems twisted ... of course this is merely an expository distinction, but I like the idea (discussed elsewhere) of first considering the general situation including the possibility that Monty Hall might show the car by accident, and then restricting one's attention to those cases in which he happens not to. Lorenzo Traldi 01:07, 16 January 2007 (UTC)
...If the host shows one of the goats the game provability is no more 1/3 or 2/3 or whatever: it becomes 1/2, right?
In fact the game was never 1/3 because one of the doors with a goat will always be eliminated, so the goat door 1 and the goat door 2 are, in a kind of way, the same door (aka the same result).It doesn't matter if you choose a door with a goat or not, when in the end you have to choose there are just two doors and one goat. In my point of view the game starts when there are only two doors and 50% of changes for each door.
PS: Feel free to correct me
PS2: I'm sure you will ;)
Thank you for the reply, now I can see it. I would have read the other disussion threads, but in this article there are too much. Great mike 10:27, 26 December 2006 (UTC)
This problem has probably confused so many people because the typical explanation for it is wrong even though it still gives the right answer. People poke holes in the argument only to find through some other means that the conclusion is still true.
It is fallacious to hold that you originally had a 1/3 chance of choosing the very goat that monty reveals that you did not choose when he opens a door. It is true that the knowledge that you did not choose the goat monty reveals would make it 50/50 you chose what was left, but the additional information that monty revealed that goat (which he is less likely to do because you chose the car than because you chose the other goat) readjusts the odds such that you are more likely to win by switching.
Using an alternate version of the problem where the odds of choosing each are the same but you know monty has a strategy for showing one of the 2 goats (which you can distinguish) shows how this reasoning fails in the general case. Here everything that is said in the explanaition of the original problem remains true but the conclusion is now false - because once again it is fallacious to hold that you had a 1/3 chance of choosing a goat to begin with even after it has been proven to you that you did not choose that very goat.
Here is the math, first a real explanation and then a alternate version of the problem where the wiki explanation fails.
A REAL sample space might look like
1/3 You chose Goat 1 Monty reveals Goat 2 Switching Wins
1/3 You chose Goat 2 Monty reveals Goat 1 Switching Wins
1/6 You chose car Monty reveals Goat 1 Switching Loses
1/6 You chose car Monty reveals Goat 2 Switching Loses
You can see how this is accurate because if Monty reveals Goat 2, then you can cross out thatyou chose goat 2 and that you chose car and monty reveals 1 i e
1/3 You chose Goat 1 Monty reveals Goat 2 Switching Wins
X1/3 You chose Goat 2 Monty reveals Goat 1 Switching Wins
X1/6 You chose car Monty reveals Goat 1 Switching Loses
1/6 You chose car Monty reveals Goat 2 Switching Loses
and then you get
2/3 You chose Goat 1 Monty reveals Goat 2 Switching Wins
0 You chose Goat 2 Monty reveals Goat 1 Switching Wins
0 You chose car Monty reveals Goat 1 Switching Loses
1/3 You chose car Monty reveals Goat 2 Switching Loses
Laying out the sample space in this correct manner also allows you to solve any other similar problem including the one where Monty chooses goat 1 with 2/3 probability when you choose the car:
1/3 You chose Goat 1, Monty reveals 2
1/3 You chose Goat 2, Monty reveals 1
2/9 You chose the car, Monty reveals 1
1/9 You chose the car, Monty reveals 2
Note in this case that although you had originally a 1/3 chance of choosing each goat or the car, collapsing the 1/3 chance of the revealed door into the unrevealed unchosen door to get 2/3 chance of car, or following any of the other rediculous suggestions in the wikipedia article does not give you the right answer. The odds can even be 50/50 for switching if monty always chooses one goat over the other.
-The author of this post abstains from signing his posts such as not to promote appeal to authority and ad hominem fallacy. —Preceding unsigned comment added by 69.252.158.32 ( talk • contribs)
1/3 You chose Goat 1, Monty reveals 2
1/3 You chose Goat 2, Monty reveals 1
1/3 You chose the car, Monty reveals 1
0 You chose the car, Monty reveals 2
It is stated in my examples that the goats are distinguishable. It is irrelevant whether or not you can distinguish the goats as far as the validity of the given solutions to the problem are concerned. Why should the fact that the goats are distinguishable suddenly invalidate a claim like "collapsing the 1/3 chance of the revealed door into the unrevealed unchosen door to get 2/3 chance of car"? It wouldn't - this claim was already invalid to begin with. It is no different than saying if you dance on your head on the night of a fool moon you will have a 2/3 chance of winning by switching. See look - it has been verified by experiment.
All you have to do to disprove the dancing on your head claim is do the problem without dancing on your head and see if you still get 2/3 chance to win by switching. And all you have to do to disprove the collapsing 2/3 between 2 doors claim into one door is to do a version of the problem where everything this explanation is based on is still true but a different answer is arrived at. In my version of the problem each door still has a 1/3 chance of holding the car so therefore if the statement regarding collapsing probabilities was ever true it should still hold. It does not hold so therefore it was never true. -The author abstains from signing posts so as not to endorse ad hominem and appeal to authority fallacies. Let everyone be no more or less than the strength of their arguments. —Preceding unsigned comment added by 69.252.158.32 ( talk • contribs)
I hereby propose that this section of the talk page be deleted as any attempts to discuss actual mathematics with 69.252.158.32 will only result in more people getting death threats. -- Antaeus Feldspar 01:15, 29 December 2006 (UTC)
I've copied the section just added to the article below. Before adding this analysis to the article I request a source be provided. Without a source it looks like original research to me (which is not allowed). Furthermore, it looks incorrect (per my comments above). The deleted section follows. -- Rick Block ( talk) 04:23, 30 December 2006 (UTC)
I've re-removed the section, per Rick Block's comment. A source needs to be provided for this; if we do include it, it needs some cleanup—I just don't understand the point that it's trying to get across. TenOfAllTrades( talk) 04:57, 30 December 2006 (UTC)
My comments were not clearly addressed. This whole thing is not worth arguing about. Please provide a reference substantiating this analysis. If it's not original research then there is a reference you should be able to provide. With 40 or more academic articles about this problem, it shouldn't be difficult to find one supporting your view if it has any validity whatsoever. Thanks. -- Rick Block ( talk) 05:34, 30 December 2006 (UTC)
I have read your response and fail to see how it addresses my comments. The odds in question are not conditional odds given a known exposed goat. In the problem as stated, the goats are not distinguishable. Any analysis involving distinguishable goats is simply irrelevant and confusing to the point. If you claim this analysis is already supported by the existing citations, please cite a specific one (paragraph or page number would be nice as well). Please sign your comments (or if you refuse to, at least don't delete the "unsigned" tag others add - who makes what comments can be easily determined by looking at the history of the article so you're not accomplishing anything by not signing). -- Rick Block ( talk) 05:58, 30 December 2006 (UTC)
At the point when the player is asked whether to switch, there are three possible situations corresponding to the player's initial choice, each with equal probability (1/3):
The player originally picked the door hiding goat number 1. The game host has shown the other goat.
The player originally picked the door hiding goat number 2. The game host has shown the other goat.
The player originally picked the door hiding the car. The game host has shown either of the two goats.
This proposed sample space is wrong in that it attributes a 1/3 probability of something happening that is already known to have not happened. Note at the top it says "At the point when the player is asked whether to switch" That means Monty has already eliminated one of the above possibilities - yet at the same time it is attributed a 1/3 probability. -K99 —The preceding
unsigned comment was added by
Kriminal99 (
talk •
contribs).
No further explanation is required to anyone that is not selectively blind. Your statements regarding the three choices are irrelevant because you are failing to realize an important fact. At the time the decision to switch is made, it is not an issue of those three choices and nothing else. It is an issue of 2 choices, and the probability of those choices must be determined using all available information including everything we know AT THAT POINT about the original 3 choices. At that point we know that one of the three that we did not choose did not hold the car, and we also know that Monty revealed this fact according to a certain algorithm. This information is totally different than the oversimplified 3 initial choice explanation which by shear coincidence happens to result in the same answer. —Preceding unsigned comment added by 69.252.158.32 ( talk • contribs)
The above explanation may be fallacious in that it requires the reader at a certain point to hold that the contestent had a 1 in 3 chance of choosing the goat that Monty has already revealed that the contestent did not choose. This is in violation of the principles of probability which would have someone change the probabilities of their initial choice with the new information given when the goat is revealed.
In this specific version of the Monty hall problem, the additional information given by the fact that the goat Monty reveals could not have been chosen is exactly offset by the information that Monty reveals that goat (which he is less likely to do when the contestant has chosen the car) Therefore the right answer can be arrived at incorrectly.
When Monty has a known strategy for revealing one of two distinguishable goats it is still true that there is initially a 1/3 chance of choosing each object and only in one of those 3 situations would switching allow you to win. However that has nothing to do with the probability of winning when one switches as this case shows, since the odds are no longer 2 to 3 in favor of switching. And the reason it has nothing to do with it, as mentioned before, is because it is fallacious to simultaneously hold that you had a one third chance of choosing a given goat and that you did not choose said goat because Monty has revealed that you did not.
Fallacious explanations for the outcome in the Monty Hall problem may have been the source of earlier widespread disagreement regarding the problem in which many members of the academic community expressed disagreement. Many people realize there is something wrong with common explanations of it only to find that there is still a 2/3 chance of winning by switching doors using other means of investigation.
This is a question that has already been asked and answered but it seems it needs to be addressed again -- if Monty has some particular pattern or method by which he picks a goat to show, and the player knows this and can distinguish between the goats, does it ever make some strategy other than "always switch" an optimal strategy?
The answer is "no". To prove that this is so, we will assume the opposite. We will assume that one of the goats is black and one is white, and that Monty has a bag with B black marbles and W white marbles; when he picks a black marble, he shows the black goat and when he picks a white marble he shows the white goat. B and W may be 0 but they cannot, of course, be negative numbers.
B+W represents the total number of times Monty chooses between the two goats, which he only does when the player initially picks the car. The player's chance of picking the black goat is equal to his chance of picking the white goat is equal to his chance of picking the car; therefore B+W is also the total number of times that the player sees a particular goat because he chose the other goat.
Therefore:
Now, let us assume that we can find values for B and W such that seeing a particular goat means the player should stay, and not switch. If there is such a goat, we can devise a strategy in the form "if you see this goat, stay, and otherwise switch" that beats the strategy of "always switch". If however the optimal strategy for each individual goat under all values of B and W is "switch" it means that the optimal strategy for the game is "always switch". So, we assume that for some goat (say the black goat) the player is actually more likely to have picked the car and therefore maximizes his chances by staying, rather than switching.
But here we run into a problem. Out of the B+W+B times the player sees the black goat, B is the number of times the player picked the car and B+W is the number of times the player picked the white goat. For staying to be an optimal strategy, B must be larger than B+W. W would have to be negative to fulfill this condition but we noted at the beginning, W represents a number of marbles and clearly cannot be negative. Even choosing a value of 0 for W only makes it possible to encounter a situation where switching provides no particular advantage, but: 1) no strategy in that situation can make the player's chances greater or less than 50%; 2) for every two times the player faces that situation, he gets a situation where switching pays off 100% of the time.
Can we even pick any values for B and W such that the overall success of a strategy of always switching even becomes something other than 2/3? No, we cannot. What is the proof? The proof is that staying always pays off when the player originally picked the car; switching always pays off when the player originally picked a goat. The ratio of times the player picks the car to times the player picks one of the two goats is constant, and is not affected by how Monty chooses which goat Monty chooses to show in those situations where he has a choice of goats. -- Antaeus Feldspar 02:27, 31 December 2006 (UTC)
This discussion goes around and around without a definitive end, like an argument over whether you should make clam chowder with milk or tomatoes. I think a graphic of the possible sequences of events puts an end to it, provided that people can agree that the graphic is an accurate model of the system. Here's my proposal:
File:ModelOfEvents-Probabilities.jpg
If this model is right, then its behavior is what you should program if you want to simulate the system. You can also infer the probabilities of each possible event in the Monty Hall sequence, like this:
The probability of each option (joint event) shown in this model is calculated by multiplying the probability of the single event times the probability of the prior event. The probability of a hidden prize is 1. [Consider some examples: The probability that I will initially choose Goat 1 is 1/3. If I choose Goat 1, then Monty is forced (probability = 1) to reveal Goat 2, so the probability of the sequence down to that point is (1/3)*(1)=1/3. Because Monty has his choice of two goats after I first choose the Car, the probability that Monty will choose Goat 2 in the joint event is (1/3)*(1/2)=1/6.]
The process is determined by three events:
1: I choose among the three doors.
2: Monty reveals a goat.
3: I choose one of the remaining two doors (by switching or not switching).
If this model is valid (modelling what it's supposed to model), then it disproves the original, peer-reviewed article. This is a mathematical proof, which is a pretty strong form of proof for this type of problem. Someone has already pointed out that disproof would be an interesting outcome, due to what it reveals about the limits of peer review. It's rare that a peer-reviewed decision can be tested in this way.
On this model, I have a .5 probability of coming out of the game with a car. This probability is determined by summing the probabilities of the sequences in which I get the car (1/6+1/6+1/12+1/12).
If this model is not valid, then someone should propose a different decomposition of the possible events in Monty's system.
I take it as axiomatic that:
1: The system can be analyzed as a series of three events.
2: The probabilities of the possible options for each event must sum to 1.
If someone does propose another decomposition of the possible sequences, the second requirement is a good logical test of the model.
I think that the model is correct and exhaustive, in which case we should be done with the question of what is right. That would leave us with a charming article that is fundamentally wrong. I can't suggest an equally charming article based on my analysis. I do have some suggestions, but they are inappropriate until commentators reach a consensus. I have been known to make mistakes from time to time. Etarking 03:38, 2 January 2007 (UTC)
I've restored the section about the variant recently discussed in Vos Savant's column. There's discussion about this in two separate sections (above). If the host doesn't know or forgets where the car is and opens a door anyway, it makes no difference whether you switch or not and you have a 50/50 chance either way. Please see the sections above for more discussion about this. -- Rick Block ( talk) 01:39, 3 January 2007 (UTC)
I apologize if I confused things by introducing the concept of the "do-over" to discussion of this variant. It seemed to be the easiest way to explain a difficult concept (or at least a concept that someone was indeed expressing difficulty with.)
The concept in question is that even though there are certain possibilities which could clearly come about by random chance, these possibilities are shown not to have happened by the evidence given as part of the problem statement. In this, the "forgotten goat" variant Monty Hall problem, there are six combinations of doors that can be chosen between the player and Monty; two of these are combinations where Monty does mistakenly show the car. However, the problem statement says that Monty does not show the car; thus, we are dealing only with the four possibilities that are not contradicted by the problem statement.
This seems to puzzle many people, who are not sure whether it is "legal" to simply 'prune' the probability tree and say that a thing doesn't happen when clearly it could. Many of us learned to work through probability puzzles through thought experiments employing an appropriate randomizer ("OK, if I assign two faces of the die to each of the three doors, and if I get every face of the die once...") However, the forgotten goat variant poses difficulty for those choosing this approach: how do you simulate a randomizer not coming up with results that, quite obviously, it could generate at any time?
The answer is to incorporate the "do-over" into the simulation: if you get a result that is contradicted by the evidence of the problem statement, simply ignore it and keep going until you get a result that isn't contradicted. However, it was only ever intended as an aid to help people run the simulation for themselves and see in a convincing way why the answer is what it is; it was never meant to be implied that the "do-over" was part of the actual problem. -- Antaeus Feldspar 22:15, 3 January 2007 (UTC)
But once the host has opened one of the doors, the choice has become out of TWO doors. The opened door can be forgotten. It's as if a new choice has been given: "A door has a car and a door has a goat" CHOOSE! :/ it 's a 50% choice!!! please help —The preceding unsigned comment was added by 194.204.127.214 ( talk • contribs).
Oooh thanks a lot! :D - markbri
Monty Hall problem has been nominated for a featured article review. Articles are typically reviewed for two weeks. Please leave your comments and help us to return the article to featured quality. If concerns are not addressed during the review period, articles are moved onto the Featured Article Removal Candidates list for a further period, where editors may declare "Keep" or "Remove" the article from featured status. The instructions for the review process are here. Reviewers' concerns are here. Gzkn 10:57, 7 January 2007 (UTC)
I wrote a flash simulation of the Monty Hall problem. Is there a way for me to upload it to wikipedia? —The preceding unsigned comment was added by Lax4mike ( talk • contribs) 03:22, 9 January 2007 (UTC).
These aren't Venn diagrams in the strict modern sense, but I'm not sure what else to call them. Venn might have disagreed. Septentrionalis PMAnderson 04:18, 9 January 2007 (UTC)
The following text was deleted:
on the grounds that it is not related, except in being a cognitive illusion. I disagree; I must not have made it clear, but I think (and so do several sources, including the paper cited, that it's the same cognitive illusion. The fallacious argument in the Paradox is
Discuss? Septentrionalis PMAnderson 02:56, 10 January 2007 (UTC)
Abscissa brings up a point worth discussing. Much of this article consists of different explanations of the same result, and they really are themselves quite similar.
Septentrionalis PMAnderson 03:36, 10 January 2007 (UTC)
the MH game a few times (handling the cards themselves: watching someone else is not very effective), that physical sense of the situation easily switches the intuition. After a few plays where you handle the cards yourself, it will become obvious that it pays to switch. But mere verbal instructions or even merely watching someone else handle the cards will easily fail. So a person in doubt about whether it really pays to switch ought to take a couple of minutes and deal the game a few times." -- Antaeus Feldspar 05:55, 14 January 2007 (UTC)
The reason the article is so repetitious is because the explanations are all wrong. Thus instead of a single valid explanation that would put any objection to rest a thousand invalid arguments are used instead. —Preceding unsigned comment added by 69.252.158.32 ( talk • contribs)
Is this picture helpful to anyone? When I first started to understand the problem this hindered more than helped. I understand that some people might have different ways of understanding the solution, but I believe this image is peddling a bad explaination on how to arrive at the solution. This problem has absolutly nothing to do with what is suggested in the picture. I move for its deletion, or at least a recreation which explains the solution better. This is not a case of just having 2 seperate guesses at the correct door (and thus 2/3 probability) 202.10.86.59 20:50, 13 January 2007 (UTC)
Well, then, since the people who have already spoken above saying "Presenting this analysis in the form of a JPEG adds nothing to presenting it as text" are apparently just not enough for the cartoon's provider, let me add yet another voice to that contingent. Perhaps then the cartoon's provider will be able to grant that yes, other editors have considered the merits of the cartoon and still don't think it's appropriate and -- here's the kicker -- should therefore not be re-added. The cartoon does not in any way shape or form illustrate the problem and therefore should not be in the article in the form of an illustration. This leaves the question of whether the text should be in the article in the form of text. Frankly, I do not think so; it is not nearly as clear as its author seems to think. It is hardly intuitive that "to open both the other doors instead ... [is] exactly the choice Monty is offering"; in fact, I am rather skeptical that it could even be described as true since "is exactly" and "is equivalent to" are two very different things. Now of course we could explain this equivalence, but I think the section "Combining doors" is already doing exactly that, very well. For these reasons, my recommendation is leave the cartoon out.
If I sound frustrated it's because I am. "I can put it back against consensus" does not mean "I should put it back against consensus". -- Antaeus Feldspar 01:20, 21 February 2007 (UTC)
I think the cartoon is excellent. It detracts nothing from the article and instead adds a little bit of humour as well as clever explanation in pictorial form. I can't imagine why anyone would be dead set against it. The arguments offered above to that effect seem to me to be peculiar. To start arguing about identity versus equality when talking about a cartoon is bizarre. And I'm not even from the Kingdom of Fife. Davkal 01:47, 21 February 2007 (UTC)
Infact, I should have been more general. I do not believe the Venn diagram is an accurate way to represent the problem. In the Venn diagram, doors 2 and 3 are grouped, seemingly because they are physically next to each other. What if door one were next to door three? If there is indeed a sound reason why, I think it should be presented in the article, in the Venn diagram section. There is a 2/3 chance it could be behind doors 2&3, then it must follow that there is a 2/3 chance of it being behind doors 1&3. So then, doesn't the Venn diagram present a choice that is 50/50? Clearly, this does not agree with the actual solution. In conclusion, the Venn diagram seems like an over-simplification, if it is even correct. WIth regards to the picture, any tool that seems to aid understanding is fine with me, however, I don't think the Venn solution is accurate, and therefore the same is said for the picture, in my opinion. 202.10.86.59 05:58, 14 January 2007 (UTC)
Because of the very nature of the paradox, the usual 50:50 response and the amount of debate that it generates, I would like to see more text in the intro on the background leading into the analysis. The analysis is so long and uses so many methods, without fully explicating the controversy that has been generated in the public sphere, that many readers will not bother with the analysis, and will write if off -- some people were prompting to take the page down as incorrect and a near-hoax, for instance. I would suggest using text based on the following:
"When Marilyn vos Savant quoted this puzzle in the US a few years ago [citation, date], she received over 10,000 letters mostly telling her she was wrong.
One was from Robert Sachs, a professor of mathematics at George Mason University in Fairfax, Va. who said "As a professional mathematician, I'm very concerned with the general public's lack of mathematical skills. Please help by confessing your error and, in the future, being more careful."
However a week later and Dr. Sachs wrote her another letter telling her that "after removing my foot from my mouth I'm now eating humble pie. I vowed as penance to answer all the people who wrote to castigate me. It's been an intense professional embarrassment."
Well, we said it was counter-intuitive. Even professional mathematicians get it wrong." ( http://www.grand-illusions.com/monty.htm) That is just a framework suggestion, not suggesting direct plagiarism of the text. Or similar words around what has driven the controversy most recently, being the 'Marilyn vos Savant' article. The so-called Marilyn vos Savant character is presumably a pen name and may represent a number of contributors [verify].
Also, it would be good to include links to online simulators in a separate section, or make reference in the current heading to them. (examples also found at the same site: http://www.grand-illusions.com/monty2.htm) -- Sean01 07:37, 15 January 2007 (UTC)
I think this is an excellent article, I like the formal discussion and and also the artwork, even the goofy one at the top. You might want to consider moving the history section up a bit and add to it the above posting (on this talk page) about why this has caused confusion. Maybe a good place would be just before the "understanding aids", especially if the above information is added then the reader will be even more interested to read the excellent analysis. But I mean, even as it currently stands, it's awesome, simply awesome! -- Merzul 02:28, 17 January 2007 (UTC)
Is the extended Bayesian analysis recently added from some particular source? If so, can someone please add a reference? The article is currently undergong a featured article review, which includes making sure that everything that should be, is referenced. I fear some folks might view this extended analysis as approaching original research, which is prohibited (see WP:NOR). -- Rick Block ( talk) 14:24, 17 January 2007 (UTC)
I have noted that there has been some unproductive edit skirmishing over how to render simple fractions. I'd like to see us discuss the relevant advantages and disadvantages of our options, rather than just reverting each other without such discussion. There are three main approaches, and I'm just going to list the advantages and disadvantages that I see to each. -- Antaeus Feldspar
<math> \frac{1}{3} </math>
(which becomes ); <math> \tfrac{2}{3} </math>
(which becomes )
As you might be able to guess, the last option is my least favorite, but I think my reasoning (readers should be able to read the math we are presenting) is rather unassailable. Anyone else? -- Antaeus Feldspar 01:42, 20 January 2007 (UTC)
Since this article primarily includes only very simple fractions (mostly 1/3 and 2/3), I think the advantages of using plain text which renders everywhere outweigh any disadvantages with this approach. Does anyone violently object to this solution? -- Rick Block ( talk) 02:40, 23 January 2007 (UTC)
I believe all current specific suggestions from Wikipedia:Featured article review/Monty Hall problem have been addressed, except this comment from user:SandyGeorgia:
There's a thread above, #About the Introduction, and online simulators, with a suggestion for the lead as well. I'm not quite sure what to with these comments. The problem statement in the lead is deliberately a quote to deter wordsmithing (which has been a problem). The spoiler is there since the solution is specified. I've recently added a sentence about the response to the 1990 Parade column. I'm considering shortening the lead by deleting everything after the spoiler warning (and the warning), which would include deleting the cartoon about the solution and revising the text following the quoted problem statement, sort of like:
Thoughts on this? -- Rick Block ( talk) 17:21, 20 January 2007 (UTC)
I frankly would rather see the solution deleted from the lead. There are of course many different standards for what should go in the lead and what should be saved for after the lead, but one that I favor is that the lead should be what a programmer calls a " black box" -- one sees its inputs and outputs but not its inner functioning. Describing how it has confused people because of its counter-intuitive solution -- that's input/output. Describing what its counter-intuitive solution is and why -- that's the inner functioning.
That being said, the solution should be at most the second thing after the lead, since if someone is interested in anything else besides the "inputs and outputs" of the problem, it will be the solution. The only reason for making it the second thing after the lead, rather than the first thing, is that the solution will often appear nonsensical or wrong to people if they've misunderstood the exact problem constraints, so there is a strong argument to be made for presenting the unambiguous problem statement first and then spelling out the solution within those constraints. -- Antaeus Feldspar 02:46, 22 January 2007 (UTC)
In response to further comments from user:SandyGeorgia at Wikipedia:Featured article review/Monty Hall problem, I think she's suggesting condensing the lead down to a paragraph or two, perhaps something like:
Immediately followed by the section describing the problem (moving the quoted version from Parade and the associated image there), then followed by the section describing the solution (with the cartoon graphic). Anyone have any objections to this approach or alternate suggestions for how to address SandyGeorgia's comments? -- Rick Block ( talk) 14:37, 25 January 2007 (UTC)
I removed this addition by an IP from aids to understanding - not sure if it's needed or a repeat:
Another way of thinking of this is: If I initially choose the car, he will show either one of the goats, in this case it is best to stay. If I initially choose the first goat, he will show the second goat, it is best to switch. If I initially choose the second goat, he will show the first goat, it is best to switch. Notice that even after he exposes a goat you don't know if your door has a goat or car. Since ⅔ of the time it is best to switch, switch! Thus, simply put, your first choice was likely a goat (a ⅔ chance), so when he shows you where the other goat is, it is in your best interest to choose the other door.
SandyGeorgia ( Talk) 04:03, 26 January 2007 (UTC)
I removed this addition showing a connection between two different aids to understanding sections:
Note that the both the prior probabilities , all equal to 1/3, as well as those conditional probabilities which are nonzero are displayed in the above decision tree. The Bayesian analysis which follows at this point is essentially equivalent to the decision tree analysis given previously.
This, and similarly motivated paragraphs are redundant. All proper proofs of the winning strategy are equivalent, but this does not imply that the article should spell out at length the N*(N-1)/2 ways in which N "aids to understanding" relate to each other: it would only add unneeded clutter. Rather, the "aids to understanding" should, IMHO, be left as alternatives to each other - different ways for different minds to arrive to a common conclusion. The Glopk 16:50, 26 January 2007 (UTC)
I've moved the following text here from the article:
The aquarium analogy
An excellent analogy helping to understand is suggested by Andrea Gennari of Rome, Italy. This is the so called "Aquarium analogy". An aquarium contains 100 mugs of water (the example can be made using different measurement units without affecting the result, in the example a "mug" of water can be a 1 liter mug or a 1000 liter mug). Now assume that the player cannot see what is inside the aquarium i.e. the aquarium is hidden to sight. The aquarium contains a fish. For the sake of simplicity, assume the fish can be contained in a large enough mug of water. As for the 100 doors example, the player is given the possibility of finding the fish by filling a mug with water so that he has to pick up a mug and put it in the aquarium. What is the chance of the fish being in the mug the player has just filled in with water from the aquarium? Clearly the probability is 1 out of 100 or 1%. Now we place the hidden mug aside without showing the content and we progressively fill 98 mugs with water from the same aquarium, this time showing that in each mug there is no fish. Progressively the aquarium is emptied but we take care of never picking the fish when we fill in the 98 mugs. The fish remains in the aquarium. So at the end of the process we have the original first mug filled with water, the content of which is hidden to the player, the 98 mugs filled with water and no fish, the content of which is shown to the player, and the aquarium filled with the equivalent of only one mug of water and always hidden to the player. The player is then given the possibility of switching between the original mug chosen and the aquarium (containing a volume of water equivalent to a mug). What is the most logical action to maximise the probability of finding the fish? Or, in other terms, is it more likely that the fish is in the original one mug chosen at the beginning or in the one mug volume of water contained in the aquarium? From the example it is clear that the fish can only be in one of the two, but it is now intuitive that, while the probability of the fish being in the first original mug picked by the player is 1%, the probability of the fish being in the last remaining hidden volume of one mug (contained in the aquarium) should be 99%, because when the first mug was filled at the beginning, the probability of the fish being in the other 99% of the volume of the aquarium (equivalent to 99 mugs) should have been clearly 99% and none of the 98 mugs "extracted" contained the fish. The example mirrors exactly the 100 doors example. This example helps understanding in that it makes use of volumes and "probability density". The probability density of the fish being in any given volume is the same therefore if we split the volume in two smaller volumes (a "1 mug" volume and a "99" mugs volume) the probability of the fish being in the "99" mugs volume is clearly 99%. When we show the content of 98 out of the 99 mugs volume, what we do is to concentrate the probability of 99% in the remaining one mug volume of the aquarium. The same example can be done with a "three mug" aquarium analogy thus reflecting the original Monty Hall problem.
Despite the fact that the analogy is attributed to "Andrea Gennari of Rome, Italy", it is not cited. We have no idea where this appeared or who Andrea Gennari is. And frankly, even though I understand the Monty Hall problem pretty well, I can't follow this analogy which is supposedly clearer than the existing explanations. -- Antaeus Feldspar 01:13, 1 February 2007 (UTC)
I undid change 104755257 by user David Eppstein, who had changed some of the "inline" fractions in the equations of the Bayes Theorem sections into LaTeX \frac{}{} forms. The result of this change was a marked inconsistency in the equations that looked visually unpleasant. In particular, the multi-line formula evaluating the normalizing constant mixed fractions of the two types, probably because having the final result in a \frac form made for a confusing read next to the previous line.. Also, the line spacing on my monitor became quite "jumpy". Something needs to be done to improve the rendering of math in Wikipoedia, but until it is I think we should edit conservatively.... The Glopk 15:24, 1 February 2007 (UTC)
The lead paragraph should state the problem in a simple and low-drama way. In particular, I think that the mention to any elements of "Let's Make a Deal" should be minimized because this is not about a game show, it is about a logic problem. We should stick to the correct problem statement, the answer, and then mention that some find it non-intuitive. Unless somebody is going to have a heart attack over this, I will attempt to do so.-- 199.33.32.40 00:17, 2 February 2007 (UTC)
Sorry Anteaus. It is not my intent to present the "game show version" of the problem in the lead section except at a historical curiousity. It is my intent to present the adult "more than three doors" version of the problem in the lead paragraph, and without the doors, the klieg lights, pretty girls, prizes etc. This FA is under the category of Math. Cold, unforgiving, brutal Math. I know it is less entertaining that way, but that is the subject of the article. Wikipedia is not a game show. I present the clear, bone-crushingly obvious and boring version of the problem first. The subject is not Monty Hall. The subject is the Math (more specifically Logic) Problem. -- 199.33.32.40 01:15, 2 February 2007 (UTC)
Anteaus, much of the fighting and confusion of this article is due to the treacle the project puts out via the mentioning in the lead section of females like Marilyn vos Savant and her ludicrous errors in attempting to even address this problem. Here you guys are coming to the consensus that some pretty face is more important than the subject of the article. The project is supposed to be educational, not just a bunch of pot stirring. Consensus is all fine and lovely but Math is unforgiving enough as it is without that female muddying the waters deliberately, mistakenly or simply for the publicity. She barely rates making it into the Trivia section of this article.-- 199.33.32.40 01:25, 2 February 2007 (UTC)
This article is a disappointment. Marilyn got it wrong and within this article she should be relegated to some obscure footnote. Based on the dramatic header you have at the top of this talk page, it is no wonder that you fail to educate some readers about the true nature of the subject. -- 199.33.32.40 01:49, 2 February 2007 (UTC)
That some female who did little but cause more confusion gets into the lead section while the problem statement itself does not might be entertaining, but it is not educational. Does this really belong in the "Media" group rather than the "Math" group of FA articles? Well, if it does not belong in the "Math" group, then here is what was tried as the problem statement in the lead section:
At least this is about Math rather than about some entertaining "savant". BTW: Once a decent, subject-oriented version of this article is created, I strongly suspect that it will become more obvious that it deserves the "Low" importance rating that it gets from the Math WikiProject and that it is an unworthy subject of a "Math" FA. I actually want this to remain in Math and get permanantly demoted. -- 64.9.233.132 03:15, 2 February 2007 (UTC)
The References section contains entries that are not cited anywhere in the article body. One was added today, presumably by its author. What is the policy or custom about such references? Leave them alone? In academic papers (at least in my field), having such "dangling" references is highly frowned upon. The Glopk 17:19, 3 February 2007 (UTC)
I've been away for a couple of weeks. In the intervening period, this article seems to have morphed into a dusty old text book. That's very sad, given that Wikipedia has the opportunity to be so accessible and so different. I say: delete things that are wrong, not things that are simply not to your liking. De gustibus non est disputandum, after all! StuFifeScotland 13:18, 12 February 2007 (UTC)
See also: #MontyHallProblemMadeEasy.jpg
Can we reach a consensus for whether the cartoon image should be in the article or not? The reasons I prefer it not to be in the article are listed below. Please, let's not vote about this, but discuss (and respond to) reasons for or against. -- Rick Block ( talk) 02:07, 21 February 2007 (UTC)
We're not reaching a consensus. Let's have a vote. -- Doradus 07:08, 23 February 2007 (UTC)
Glopk, tis one of the mysteries of life that, for some reason, some people find it easier to relate to something that has a face (e.g. a cartoon) than they do to a dry dusty textbook. You may not like this, and you may feel that it is wrong, but that doesn't change the fact. Another fact that is not changed by mere repetition of the contrary viewpoint, is that the picture is actually a picture (even with text). It does not merely repeat the text as you suggest, but does other things as well - and those other things are primarily what makes it a picture. Your failure to see this point is one of the main reasons why your thoughts on why it should be removed don't really count for very much. That is, the thing you want removed (a mere repetition of the text) isn't actually in the article and nobody is suggesting it should be. Davkal 03:21, 22 February 2007 (UTC)
1. We already have testimony from uninvolved individuals to the effect that they think the picture is good. That is of value to them, and that it will be used by them in teaching etc. It is therefore not me who is trying to foist something onto everyone, but you who is intent on denying everyone something that has already been shown to be useful because you don't like it.
2. I did read your comments tiwce and did not embarrass myself - I have read them again and still stand by what I said. I was pointing out that you still don't seem to understand the value of pictures, even though you say you do, because, for example, the example you give involving Eureka might work quite well in a variety of ways yet you dismiss it out of hand and think it supports your point rather than actually supporting the contrary. The point being that pictures work in a variety of ways, and your example may well act as, say, an excellent aid to memory. Similarly, the cartoon here may well help someone grasp a main point in a way that pages and pages of plain text does not. That, I think, is one of things that many who support the cartoon have in mind. Facilitating understanding though the use of various media. And it is one of the reasons why the repetition of text argument is a non-starter.
3. If you want to measure illustrations against Michelangelo's work then you'd probably need to remove every picture in Wiki. And while I'm on that point, why do we have the picture of the goat at the top of the article which has to be accompanied by five lines of caption merely repeating the text in order to explain it. That picture would better if we had a little cartoon Monty saying "Pick a door" and a cartoon contestant saing "I pick door 1" then monty saying "door 3 has a goat, would you like to change your mind". That picture, the goat, actually stands up a whole lot less well to the arguments but nobody is suggesting we remove it. Davkal 14:40, 22 February 2007 (UTC)
The above are arguments as to why it augments the article. It adds variety, it is a nice pictorial represenation of a sound point along with a humourous way of showing the puzzlement that can be caused by the MHP. Davkal 04:37, 21 February 2007 (UTC)
Glopke's argument against me is basically to insult me gratuitously. Isn't there a rule about this? Isn't this contrary to Wikipedia policy? Davkai's response to his insult is on point. Bill Jefferys 14:51, 22 February 2007 (UTC)
I apologize for misspelling your name; this was unintended. But I am puzzled about your claim that I violated "no-soapbox twice on this very article before you reverted my edits." This because the only thing I did was to revert a previous edit to the main article, with a short comment saying that I thought the picture useful. I had not commented before, and after my reversion my reversion was reverted. I have not made any changes to the main article since, and have only presented my views as a person who has used the Monty Hall example often in my classes. Please point to where I was violating "no soapbox" at any time before you reverted my edit (singular) or even later. If I have violated policy, I will apologize of course, but first I have to know what I have done wrong, specifically, not based on vague accusations but on specific example. Bill Jefferys 22:22, 22 February 2007 (UTC)
Everyone here should re-read WP:NPA, specifically "Comment on content, not on the contributor." In addition, I suspect some WP:OWN at work here as well. In any case, it's becoming evident consensus is difficult to reach on this issue and there're at least a few editors on each side, maybe it's time to invite an admin over for arbitration. Tendancer 17:55, 22 February 2007 (UTC)
So far we have a number of people who feel that the cartoon adds to the article and a number who feel it does not. What that means though is that for some people who read the article the cartoon will be a useful addition, and for others it will make no difference. The point being that it's all very well to say I don't find it funny and that humour is subjective, but this is completely irrelevent since is such cases nothing is lost by including it. What is missing from the against side are the negative things that the cartoon is supposed to do - someone finding it offensive or something, that would be a negative. But so far we simply have plus points and zeros. I don't think anybody supporting the cartoon thinks its the best, or funniest, or most aesthetically pleasing thing they've ever seen, but so what, to simply point these things out is not to give a reason for removal. For that it is necessary to show how its inclusion damages the article. Davkal 19:22, 22 February 2007 (UTC)
The lot of you need to grow up. I don't have an opinion on the image, but it's clear to me that the right thing to do is keep the controversial material off the page until either a consensus can be reached, or a vote can be taken to resolve the issue. -- Doradus 20:04, 22 February 2007 (UTC)
So the comment on content, not on the contributor, rule doesn't apply in special Feldspar cases. How surprising that you are the only one allowed leeway in this respect. I am commenting on your base motives. 1. you think you own the article. 2. you think you have the authority to treat everyone who disagrees with anything you say in a condascending manner. 3. You think you know what a picture is. I wouldn't normally draw attention to this claim because 99% of people do, but when you're one of the few who actually doesn't I think it needs to be mentioned. Davkal 23:50, 23 February 2007 (UTC)
You're obviously just having a lugh. When you're ready to discuss it sensibly I'll discuss it with you. Davkal 05:10, 21 February 2007 (UTC)
Both of you. Unless you really don't know what a picture is, then the only assumption left is that you're having a laugh. What else have we had, oooh the poor Germans, oooh the poor server, oooh what about the poor people with poor eyesight who don't know how to click on an image. What next? If these are supposed to be serious good faith edits then God help you. Davkal 05:47, 21 February 2007 (UTC)
Yes, inasmuch as how else do you explain to someone who insists that something isn't a picture but is merely text when that thing is obviously a picture. I did it by pointing out the little face on it and text bubbles etc. It's perhaps a bit sarcastic but it's clearly in good faith. Taking up too much room on the server and people who don't speak English not being able to understand it and people with poor eyesight and no IT skills not being able to read it are, on the other hand, simple examples of opening your mouth and letting any old thing come out.
Davkal
16:14, 21 February 2007 (UTC)
Put me in the camp I'm for the image. I think there may be a bit of intellectual elitism at work here, and that's making some editors against it because the image is a bit too cartoonish/dumbed-down unconventional/un-mathematical interpretation for their understanding of the problem. However, the layman reads this, and I'm sure this article was not once chosen as a featured article because it was mathematically sound (though that would be one of the reasons)--all the cartoons, simplified explanations for the non-mathematical layperson certainly must've been a giant contributing factor.
Personally, I felt the cartoon helped and presented a new--even if not very mathematical--way of looking at the problem, and that's a plus. This article is starting to read and more like a textbook and that's a terrible thing. While a mathematician whose life around numbers may prefer to have this article read like a doctoral thesis, nevertheless this is still an encyclopedia that's generally accessible--and at some point an article is no longer informative if it's too boring/dreary/sets too high a bar to be read. Of course the goal is not to get it so dumbed down so any doofus who couldn't pass algebra would read this article and magically get a better understanding the problem just because "oooh there are pictures" (and that particular doofus probably wouldn't want to read this article anyway)...but this article is straying down a very grey path. Do we really want this to become like e.g. the article for the Poisson Bracket. Nothing against Poisoon Brackets but, there's an example of an article that won't be featured or get marked as even a "good article" any year soon. Tendancer 17:38, 21 February 2007 (UTC)
I agree entirely with all of the above. As I said earlier, I think it brings variety to the article and illustrates one point quite nicely in a slightly humourous and novel way. I can't for the life of me see any genuine reason for not including it.
Davkal
18:42, 21 February 2007 (UTC)
I've chosen one door, my chances are obviously 1 in 3. |
I would rather be able to open 2 doors... etc... |
I'm all for having a quick little synopsis of the solution offset from the main text. However, many good points are made above against the image presentation. Personally, I don't like the image's text itself. The explanation poor, the logic unsound, and in its current format it can't be iteratively revised wiki-style. How about using a simple colored box? See demo to the right. (My colors are hideous, but you get the point...) ~ Booya Bazooka 05:18, 21 February 2007 (UTC)
Just curious what do both camps (keep picture/remove picture) think of this suggestion? It appears to be the closest thing we have to something that may reach a consensus and does satisfy arguments from both camps. Suppose Flatscan/Booyabazooka/me adds this to the article, will it placate both sides or will start another revert war (as happened with the last 10 edits on this article)? Tendancer 19:15, 22 February 2007 (UTC)
How many mathematicians does it take to change a lightbulb? None, they don't do jokes. See above! Davkal 20:54, 23 February 2007 (UTC)
I think we should remove the goat picture at the top of the article.
Reasons for removal.
1. It's not funny.
2. It needs five lines of accompanying text, which merely repeats what is said in the article, in order to explain what it is supposed to be about.
3. Chinese people might not recognise the markings on the doors.
4. The wiki server must be creaking under the weight of the storage requirements for such images.
5. The numbers on the doors are too small for some people to read.
6. You can't edit it if, for example, you don't like the colour of the goat or the doors.
7. The goat's facing the wrong way, goats always face to their right (especially on gameshows).
8. It's not as good a picture as the Mona Lisa.
Davkal 15:41, 22 February 2007 (UTC)
I think we should replace it with this
Davkal 16:48, 22 February 2007 (UTC)
The cartoon we're discussing is not my picture. I did the one above as, wait for it, a joke. No jokes in Wiki obviously, not even in the talk pages - all serious grown up stuff here - I do apologise. And, nobody is saying that the cartoon would work without the text, it doesn't need to, it's simply a novel way of explaining a point that many find useful. What's wrong with putting the text into coloured boxes? Nothing really I suppose, but it's not as good as the cartoon. Davkal 20:41, 22 February 2007 (UTC)
Right then, since nobody seems to have any objection I'll remove the goat. Davkal 00:45, 23 February 2007 (UTC)
-- Niels Ø (noe) 07:20, 23 February 2007 (UTC)
The reasons in the list above are almost identical to the reasons given for removing the cartoon. What I fail to see, is how the first goat picture differs in any substantial way from the cartoon. That is, all the reasons that have been suggested for removing the cartoon apply equally (and in some cases more so) to the first goat picture. The question, then, is: what's so different about the first goat picture that means it shouldn't be removed from the article? Davkal 13:03, 23 February 2007 (UTC)
Both are pictorial illustrations of the article (the goat is an illustration of the goat behind door three being revealed) and the cartoon is an illustratioon oif the puzzlement nd thought processes of a contestant who solves the puzzle). Both need copious amounts of text to explain the full point made. The cartoon - as cartoons do - includes that text as part and parcel of the illustration, whereas the goat needs a five line caption to explain which part it is supposed to represent.
Davkal
14:43, 23 February 2007 (UTC)
The point, the main point, against the cartoon was that it merely repeated text from the article and added nothing to the explanation of the point. Exactly the same can be said for the goat+caption. Another complaint was that the cartoon wasn't funny, well the goat didn't exactly have me splitting my sides. It seems now, then, that the only original argument left for removing the cartoon is that it can't be used in foreign versions. As well as being quite proposterous, this can be seen to be no more than a red herring by reference to your final point about standards of writing and presentation. The real reason, then, for exclusion, is that the cartoon is not worthy of inclusion in such a beautifully crafted piece (partly your work I take it). In other words, intellectual elitism and ownership of the article are all that were ever really behind this. Davkal 17:40, 23 February 2007 (UTC)
A picture doesn't need a reason to be a picture. The fact that it is a picture is enough to make it a picture. And the fact that you don't appear to know what a picture is, is sad, startling and frightening in equal measure.
Davkal
23:41, 23 February 2007 (UTC)
It's not my image ya clown Davkal 23:29, 23 February 2007 (UTC)
And weakest has an "a" in it. Thus clown above
Davkal
23:30, 23 February 2007 (UTC)
In case anyone failed to notice, there is an issue with the copyright of Image:MontyHallProblemMadeEasy.jpg (which I have asked the originator to address, see User talk:StuFifeScotland). Pending resolution of this issue, this image must not be put on this or any other page. I suspect this will be seen by some as a bad faith preemption of the above discussion. This is not my intent. Discuss away. But don't put the image back. I also strongly urge everyone commenting about this issue to refrain from further incivilities. -- Rick Block ( talk) 06:02, 24 February 2007 (UTC)
Honestly, f*ck this. If I knew it was going to set off such a drama war I would never have put the image back in the first place. And now that I think more carefully about it, it isn't a persuasive solution at all. The only real solution is to map out the probability tree. Hope you folks had fun. Gazpacho 11:58, 24 February 2007 (UTC)
The reasoning in the text box is fallacious. It has been oversimplified to the point that it is not accurate. As it is currently worded, it makes it seem as if switching is the winning move EVERY time, which it isn't. NipokNek 12:42, 24 February 2007 (UTC)
The reason this problem causes so much confusion is because the current explanation of the problem is incorrect given the need to consider the sample space at the time the decision is made and not at the beginning of the problem. At the time the choice is made it is not the case that there is a 1/3 chance of having chosen each item. There are two options that have not been revealed, and the only reason there is not a 1/2 chance of each one containing the car is because Monty reveals the goat you see more often when you have chose the other goat than he does because you have chosen the car. The explanations given on the article page do not logically necessitate the outcome, and in other scenarios where no signfigant changes are made those approaches would result in a wrong answer. <comments removed by Xiner> —Preceding unsigned comment added by 69.252.158.32 ( talk • contribs)
Yes I agree regarding the Venn Diagram. Using one you can make this argument very simple. I also think that when considering how likely this argument is to convince people who disagree, they are failing to consider that it is an issue of logical fallacy that makes the current explanations confusing and not simplicity. The common probability space used clearly does not represent the state of things at the time the decision is made and their reasoning is self contradicting. That is why they need 10 different explanations and people still have trouble with it.
I have moved here this paragraph, added as an introduction to the "Aids to understanding" section. While I think it may be useful to have some mnemonicals, they should probably come after more thorough explanations. Otherwise, for example, it is not clear at all that the "Bet on it!" of mantra (1) is anything but an arbitrary choice. The Glopk 23:30, 17 March 2007 (UTC)
There are more problems than simply the placement. Addressing the reader is discouraged by the Manual of Style, and the tone is very informal. Perhaps these issues can be fixed, but I'm not sure how. I've reverted another addition of this same text. -- Rick Block ( talk) 03:24, 20 March 2007 (UTC)
The assumes, without explicitly stating, that the car is preferable to the goat. This might be so obvious that it goes without saying, but it may help clarify.-- Blackmagicfish 05:37, 20 March 2007 (UTC)
To the person who reverted my addition to “Other Host Behaviors”: Why was my addition removed for being unreferenced? I do not see any references for the list of other host behaviors, so why is one unreferenced comment preferable to the others. I was simply pointing out that, in all the host behaviors where the host is not forcing a 100% result, switching either gives you a 50/50 chance, or a 2/3 chance. Therefore, at worst switching gives you the same odds as staying, and at best it gives you better odds. How is it original research to point that out? 75.31.42.44 05:38, 23 March 2007 (UTC)
It needs some cleanup, perhaps some rearrangement, and maybe conversion into an .svg, but I think it's a good start.
-- Father Goose 10:14, 23 March 2007 (UTC)
I've cleaned it up and tweaked the layout. It's got various artifacts when you zoom in but I feel it's presentable and I'm happy with the composition. Maybe someone will take it upon themselves to clean it further or convert it to .svg. In it goes. And now, for my next act...-- Father Goose 00:37, 24 March 2007 (UTC)
Player's initial choice | Probability | Host reveals | Outcome if switching | ||||
---|---|---|---|---|---|---|---|
Car | Goat 1 | Goat 2 | 1/3 | Car | Goat 1 | Other goat | |
Car | Goat 1 | Goat 2 | 1/3 | Car | Goat 1 | Car | |
Car | Goat 1 | Goat 2 | 1/3 | Car | Goat 2 | Car |
Player's initial choice | Probability | Host reveals | Outcome if switching | ||||
---|---|---|---|---|---|---|---|
Car | Goat 1 | Goat 2 | 1/3 | Either goat | Car | Goat 1 | Goat 2 |
Car | Goat 1 | Goat 2 | 1/3 | Goat 2 | Car | Goat 1 | Goat 2 |
Car | Goat 1 | Goat 2 | 1/3 | Goat 1 | Car | Goat 1 | Goat 2 |
Well, I've reworked the entire thing. Artwork is up to professional standards now, and the text is removed from the images themselves: the whole thing is now a set of 6 images in a table with editable text captions (except for the "A" and "B" labels next to the goats). As to whether readers are "lazy" or if an illustration with accompanying captions is of use (particularly to laypeople), I do beg to differ.-- Father Goose 04:33, 29 March 2007 (UTC)
Player's initial choice | Probability | Host reveals | Outcome if switching |
---|---|---|---|
1/3 | Either goat | ||
1/3 | Goat 2 | ||
1/3 | Goat 1 |
I've added a "Why the odds are 2/3" section, drawing from material already in the article in scattered places (with some rewriting). I've moved the "forgetful host" variant towards the top of the article in this restructure, as comparing it to the original puzzle brings to light (IMO) exactly where the unexpected odds arise. I'll pen a more detailed justification for this change if objections are raised to it.
Moving some material into it left other material orphaned, namely the somewhat arbitrarily-grouped material found at the top of the "Aids to understanding" section. I've put the two paragraphs which deal with the host's anticipated behavior if the puzzle were a real-world situation into a "Sources of confusion" section, and moved the remaining free-radical paragraph into the "Combining doors" section.
At this point, it's my feeling that the "combining doors" section and "venn diagram" section overlap heavily and should be, er, combined. But that would involve removing some of the redundant material, and in the spirit of no deletion without justification, I'm going to hold off on doing it for now.-- Father Goose 09:44, 29 March 2007 (UTC)
The page includes this sentence. "In the table below, the host's picks from the table above are circled." Am I missing the circles? Lorenzo Traldi 19:10, 1 April 2007 (UTC)
In the aids to understanding, it might be helpful to point out that if the host picks a door at random, you still have a 2/3 chance of winning just not in the same way. If the host picks a door at random, his choice has a 1/3 chance of being the car (if it is you choose to switch obviously) and if it's not you then go with either one of the remaining doors (which one doesn't matter). All in all, you have a 2/3 chance of winning Nil Einne 10:40, 15 April 2007 (UTC)
This is an archive of past discussions. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page. |
Archive 1 | ← | Archive 3 | Archive 4 | Archive 5 | Archive 6 | Archive 7 | → | Archive 10 |
Ok, the main problem I see is with the invalid grouping of outcomes, which causes the number of possibilities to be reduced. The following 2 are syntactically identical ways of stating the same thing, both of which illustrate the number of possible outcomes:
OR
In the first example, if the player chooses to switch, the car is won in the first two cases and lost in the second two. A player choosing to stay with the initial choice loses in the first two and wins in the third and fourth cases. In the second example, if the player stays they lose in one and win in the other.
The original question is a classic case of riddle misdirection, nothing more. The fact that it is as old as it is and people still do not see it astounds me. If one of the incorrect doors is eliminated, and the player is then forced to make a decision, it changes to a straight 50/50 chance of getting the correct door. The initial choice can then be discarded. This article seriously needs to be revised.
Mvandemar 05:30, 8 November 2006 (UTC)
How can the probability magically change??? What if we assume their are only two doors to choose from at the start of the game, the third door being an attempt to persuade participants into falsely believing three choices exist. The question of the zero probability door being revealed in the middle of the game should be irrelevant because we know at the onset of the game that a door with zero probability will be revealed before the end. -- Nbritton 17:05, 9 March 2007 (UTC)
It has a 2 in 3 chance of being wrong.
-- 220.237.67.125 15:58, 8 November 2006 (UTC)
The explanation above is only true for external obser who knows the host choice is not random. However in the perspective of the contestant who assumes the choice of the host is random the chance is still 50/50. I suggest you emphasise this fact in the article. Nice and clean explanation though from user above.
Actually this explanation seems perfectly correct, whether or not the host choice is random. If the contestant's first choice is incorrect then opening the second door must indicate the correct choice. Either Monty Hall opens the second door and reveals a goat, in which case the contestant should switch to the last remaining door, or else Monty Hall opens a door and reveals the car, in which case the contestant should obviously switch to that door.-- Lorenzo Traldi 10:41, 20 December 2006 (UTC)
ok so a lot of people have been saying that if Monty forgets where the car is but luckily picks a door with a goat in to "eliminate" this makes the remaining choice 50/50. i am sorry but i thought i was understanding this thing up until that. how does it? how can odds change based on the host who is surely peripheral to the experiment? maybe i've missed it but i consider this to be unexplained and SEVERLY confusing 28/11
A simple solution to the "Monty doesn't know" problem is that if Monty doesn't know, then his choice is between two doors that each hav a 1 in 3 chance (the other 1 in 3 chance door being already picked by the contestant). Assuming Monty picks the goat door (as we must) this simply removes one of the 1 in 3 chance doors and we are left with two doors that each had a 1 in 3 chance (the same chance)when there were three and so now as equal probabilities have a 1 in 2 chance since there are only two. Davkal 18:09, 28 November 2006 (UTC)
yes but how is the immediate above different from the original problem?? you could use surely use the exact same logic for the original problem. lets forget probabilities here and return to the root of the question "are you more likely to win by switching or not?".
Response: it's not clear you can talk about "more likely" without discussing probabilities, so the answer is no, in the new game you are not more likely to win by switching because the odds are 50/50. Davkal 13:45, 29 November 2006 (UTC)
now i understand that the reason (in the original question) you will win is because you have essentially split the three doors into two sets, then after you pick between the two sets. thus you have a better chance of winning with the set of two (even tho one has been revealed, tho this is meaningless). surely if you were to repeat the experiment with the host not knowing and took the results of the games where he randomly chose the goat (discarding the games where he chose the car as the question states the chances if he picks the goat) you would still have a better chance of switching? mentioning such things as the odds of monty picking the car HAVE NOTHING TO DO WITH THE QUESTION. the question states he has chosen the door with a goat behind it. thus logically you still have a better chance of switching.
Response: if you were to repeat the experiment in the way described then all the cases that are discarded (the cases where Monty picks the car by accident) are cases where you would have won by switching (i.e, you didn't pick the car first). Overall, then, your score drops to 1 in 3 by switching, 1 in 3 by sticking and the "removed" 1 in 3 where Monty picks the car. So your choice to stick or switch is now between the two remaining 1 in 3 chances which equals 50/50 in those remaining cases. Davkal 13:45, 29 November 2006 (UTC)
imagine two of the "monty hall problems* playing side by side on a stage with a screen seperating them. on either side of the screen, both contestants pick their first box. on one side of the screen, the producer tells monty which of the remaining two boxes he must reveal (showing a goat), he does so. on the other side the host cannot here the producer very well and cannot be sure which box to pick. still he manages to get the one with the goat in (which he had a 2/3 chance of anyway). now according to what people have said on this site, even though both players games have *PHYSICALLY AND EMPIRICALLY* proceeded in exactly the same way, the first contestant should switch, while it does not make a difference if the second one does. now i know that *odds* can be made to show that it's equal chance, but odds aren't real life. to put it another way, in the first game, obviously the player has a better chance (2/3) if they switch. now imagine both games and the way they went, and if they have different odds, would the odds change if the producer of the second game confirmed to the host that his random guess was right?
Response: the difference is that in multiple trials one Monty would never pick the car where the other Monty (who can't hear the producer) would pick the car 1 in 3 times. The probabilities, then, for what happens in these cases are different, because what would actually happen in these cases IS different. The point being that your chances of picking the ace of spades out of a pack of cards at random are 1 in 52, where your chances of picking it out after looking through the deck to find it are 1 in 1. This is true even in cases where you have just picked the ace of spades out by random chance, and true even if you wanted to say (which I would not) the in both cases what has happened was physically and empirically identical. Davkal 13:45, 29 November 2006 (UTC)
ok i think i understand what you mean, the discarded ones (i.e the ones where monty picks the car) are scenarios where it would be beneficial to switch, so in a case of multiple games, the time when in would be beneficial to switch are halved. thanks for explaining this to me as it was really starting to get me annoyed. but if it was just one or two games, and this thing happened, it would still make sense to switch, right? i mean, in a real life situation, even if this was happening, it would still make sense to switch, as you still have a lesser chance of getting it right first time? 29/11/06
No, because all the multiple trials are intended to show here is what could be the case. And what they show is that in a one-off case where you pick a random door, and then Monty opens a random door from the other two and reveals a goat it's 50/50 between the remaining doors. This is because the one-off case is one of only two remaining possibilities - you picked the car and Monty picked a goat or you picked a goat and Monty picked a goat - in one you already have the car, in the other you don't. The third possibility, you picked a goat and Monty picks the car can be ruled out in the one-off case because we know it hasn't happened (i.e. Monty didn't pick the car).
Davkal
15:27, 29 November 2006 (UTC)
i'm sorry but now you've really confused me. consider this
"This is because the one-off case is one of only two remaining possibilities - you picked the car and Monty picked a goat or you picked a goat and Monty picked a goat "
how is this different from the original case!? monty is always going to pick a goat, so i fail to see how in a one-off one time only case, this changes the odds? say two games on the same stage, side by side one time only. both games proceed exactly the same however monty one is told by the producer to reveal a certain box at the second stage. monty two (who cannot see the other game) does not have a producer, but coincidentally picks the same box. the two contestants must have the same odds right? —Preceding unsigned comment added by 194.168.3.18 ( talk • contribs)
i'm sorry but i still don't agree with this. the question is not about options or possible scenarios, but if monty picks a goat, now consider this ;
":# The player picks the car. Monty opens the door containing the first goat. Winning strategy: STAY
in the strategy of the player, it doesn't matter which of the goats monty picks. these two scenarios are basically one. in the original scenario, imagine that the player picks the car, even if monty knows that the two remaining boxes are goats he still essentially "decides" which one to reveal, yet this obviously isn't counted in the analysis. the argument of the original solution was that because you were picking only one out of the three, the choice of switching was essentially a choice between staying with your one box out of three or switching to what was essentially two choices out of the original three. it has also been noted that the "reveal" by monty is a diversionary tactic and "red herring" of the puzzle designed to make it more counter-intuitive. the fact that in the "monty doesn't know" scenario, he has still picked the goat, means the choice to switch is still based on the "one choice versus two". the player is still more likely to have chosen the goat at first (how can you dispute this) thus switching will always be a better option (at least in a one off situation, the question is supposed to put us in the situation of the player and ask us what we would/should do. multiple situations, i.e. ones where monty would discard the games where he accidently picked the car would undoubtedly lower the benefits of switching). the reason (in the original game) why switching is beneficial is more based on the primary "split" than it is on monty's reveal. he will always reveal a goat, and in the "monty doesn't know" scenario, regardless of whether it was random or not, he has still picked a goat, thus the benefits of switching remain the same. your mathematics are undoubtedly right, yet i think the reasoning is wrong. ec 30/11 —Preceding unsigned comment added by 194.168.3.18 ( talk • contribs)
I think it's time for another experiment. Take three cards including the ace of spades. The ace of spades counts as the car. Step 1: shuffle the cards and then deal them face down so you don't know where the ace of spades is. Step 2: pretend you're the contestant and move one card to the side. Step 3: Pretend you're Monty and turn one of the remaining cards over. What happens is this: in about one-third of the trials the card you turn over as Monty will be the ace and you will have to abandon those trials; in the other two-thirds of the cases (where "Monty" reveals a "goat"), you will see that that the original "contestant's choice" card is the ace half the time and the third card picked by neither the "contestant" nor "Monty" is the ace half the time. That is, in the trials that reflect the new puzzle (Monty picks a goat/non-ace) you win half the time if you always switch and you win half the time if you always stick. Do it, try it, and then once you are satisfied that there is no advantage to be had in the new game from either switching or sticking, try to work out what is wrong with your arguments to the contrary above. Davkal 12:34, 30 November 2006 (UTC)
right gonna do it now at work to pass the time... —Preceding unsigned comment added by 194.168.3.18 ( talk • contribs)
Maybe this will help. Consider what happens if Monty has a very specific kind of forgetfulness, so that he attempts to execute the classical problem, but he always gets confused and opens the wrong door. Then if you pick a goat, Monty always shows you the car. So if Monty shows you a goat, that means you picked the car, and you shouldn't switch.
The forgetful problem (in which it doesn't matter if you switch) lies somewhere between the above problem, in which you shouldn't switch, and the classical problem, in which you should.
To be precise, if Monty's memory is such that he opens the wrong door with probability p, the probability that you picked the car, given that he reveals a goat, is given by
You can have fun plugging in values of 0, 1/2, and 1 for p. Melchoir 03:24, 30 November 2006 (UTC)
The long discussion above is similar to a decision between two possible rules for the "clueless host" game. Possible Rule 1: if Monty accidentally reveals a car, the contestant is allowed to choose it. Possible Rule 2: If Monty accidentally reveals a car, the game is canceled and the contestant gets to play again, presumably after the car and goats have been rearranged behind the doors.
Under Possible Rule 1, the original 1/3 vs 2/3 analysis applies ... except the bonus for switching is obvious because sometimes the contestant is looking at the car and can decide to take it. So under this rule, switching is still advisable -- when you see a car take it! When you see a goat, you lose nothing by switching to the non-revealed door, but you also gain nothing. The whole advantage of switching involves the car having been revealed.
Under Possible Rule 2, that advantage has been declared illegal. Consequently the advantage of switching has been removed, and there is no effective difference between switching and staying.
If a careless Monty Hall reveals a goat by accident, that's the same as operating under Possible Rule 2. If the real Monty Hall reveals a goat, that's the same as operating under Possible Rule 1, because in the cases in which he might have accidentally revealed the car he would open the other door instead. -- Lorenzo Traldi 04:21, 20 December 2006 (UTC)
I agree. The "possible rules" simply give reasonable whole-game contexts. I guess I wouldn't say Possible Rule 1 can never play a role in the problem under discussion, because the never seems to forbid Monty Hall's carelessness, which is after all the point in this variant. I'd rather simply say that once the contestant sees a goat we realize a car has not been revealed this time. (Technically, with the careless host the conditional probability of selecting the door with the car when a goat has been revealed is .50, either by switching to the still-closed door or staying with the door chosen first. The conditional probability of selecting the door with the car when that's the one Monty opened should certainly be 1.)-- Lorenzo Traldi 10:34, 20 December 2006 (UTC)
I think the use of "never" is what causes the confusion, as I said.-- Lorenzo Traldi 19:48, 21 December 2006 (UTC)
If Monty does not know, and picks a goat, it is random which goat he picks, and he will always pick a goat. If monty does know, it is random which goat he picks, and he will always pick a goat. Since nothing happens when he picks a car, well, he can't, because then there would be no strategy. If he knows, then he will not pick the car regardless. The results are completely the same. Saying they are different is like saying that if you have a two thirds chance of picking a pair of socks if you put on gloves, and you forget what you are supposed to do to get the two thirds chance, yet still put on gloves, you will have a fifty percent chance of picking socks. Again, He will pick the car then goat one or two (one third chance), goat number one, and then monty picks goat number two (one third chance), and goat number two then goat number one (one third chance). In fact, whether or not he knows, you should always switch if you get the chance, because if you got the chance, Monty most certainly picked a goat. Again, another analogy- if you pick up a tissue because you know your supposed to, and blow your nose with it, and if you pick up a tissue when you are not sure if your a supposed to, there is no chance of blowing your nose with it.
Your logic resembles the logic involved in the original problem. Your math and reasoning are completely off, because your judgement is impaired by incorrect reason. No offense, but it is kind of funny that you got the problem right, then got the problem wrong.
Given this logic, I think to avoid confusion it would be necessary to adjust the article to say that there is a ⅔s chance, because there is. Squarethecircle 00:07, 6 January 2007 (UTC)
It may be instructive to contrast the situation in the Monty Hall Problem with the situation that occurs in the game show Deal or No Deal when the player in Deal or No Deal is down to two cases and facing the decision to switch the case he or she originally selected with the only other remaining case. Suppose that in Deal or No Deal there are 26 cases containing different prize amounts and the highest prize possible is $1,000,000 as in the original version of the show. If there are only two prize amounts left on the board and one of them is $1,000,000 while the other is $0.01, the probability of winning $1,000,000 does not increase if the player switches the case he or she originally selected with the only other case remaining. However, if the host Howie Mandel eliminated the other 24 cases because he knew they didn't contain $1,000,000 then the probability that the player would win $1,000,000 by switching cases would be 25/26. Now what if Howie helped out in eliminating some of the 24 cases and the player got lucky in eliminating the others? :-) —Preceding unsigned comment added by 69.141.232.16 ( talk • contribs)
Vos Savant's answer makes it seem that the host's cluelessness makes a difference. In fact it does not. The contestant now knows that the open door hid a goat. As the contestant originally had only a 1/3 chance of choosing the door with the car, there is a 2/3 chance of getting the car by switching, just as in The solution above. -- Lorenzo Traldi 20:05, 19 December 2006 (UTC)
The host's cluelessness does make a difference because it introduces a chance that the host may ruin the game by accidentally opening a door to show a car and forcing a do-over.* This ruination cannot happen when the player has chosen the car (obviously) but it happens one out of every two times when the player chooses a goat. This means that out of six combinations of doors picked by the player and the host, there are two that result in the winning strategy being to switch, there are two that result in the winning strategy being to stay, and there are two that force do-overs. Assuming that we'll never hit an infinite series of do-overs, this means we really have four combinations: two that result in the winning strategy being to switch, and two that result in the winning strategy being to stay.
(*This is, anyways, the cleanest way to model the problem conditions which state that the forgetful host's random door opens to reveal a goat, even though clearly sometimes random choices would lead to the car being revealed instead.) -- Antaeus Feldspar 22:49, 19 December 2006 (UTC)
It looks like there are now two threads on this -- but anyway under the clueless host assumption, the advantage of switching is only that Monty might sometimes show the contestant the car. (That is, my comment above was wrong -- apologies.) If the contestant is allowed to keep the car under those circumstances, then "always switch" is still good advice -- 1/3 of the time you'll switch to the car because you're looking at it, 1/3 of the time you'll switch to the car because Monty has revealed a goat and your door has a goat, and 1/3 of the time you'll switch to a goat when you have the car and Monty has revealed a goat. In the four cases in which Monty has revealed a goat, switching is 50-50. In the two cases in which Monty has revealed the car switching is 100-0. :-) -- Lorenzo Traldi 05:05, 20 December 2006 (UTC)
Well it does explain where the 2/3 went.-- Lorenzo Traldi 23:51, 20 December 2006 (UTC)
It doesn't matter if there is a do over, because if there is, then there is no choice made. A choice can only be made if he picks the goat, eliminating the chance that he picks a car. Also, the page says he still picks a goat and it is one half. This is impossible. If you think it is not, then you probably should spend the rest of the doing what my user name implies, to knock some sense into you (no offense, it's probably just you confusing yourself. The average person would end up with a similar result if under the same condition). Feldspar, you are completely correct. I will not go into further depth on this situation because it has been explained above. Squarethecircle 00:13, 6 January 2007 (UTC)
"It doesn't matter if there is a do over" is simply not correct. For instance, suppose we toss a coin and call a do over every time we get heads. Then the only toss that counts is tails! -- Lorenzo Traldi 19:28, 14 January 2007 (UTC)
Personally I think "just happens to show a goat" is clearer, and it is the do over that is introducing a new rule that seems twisted ... of course this is merely an expository distinction, but I like the idea (discussed elsewhere) of first considering the general situation including the possibility that Monty Hall might show the car by accident, and then restricting one's attention to those cases in which he happens not to. Lorenzo Traldi 01:07, 16 January 2007 (UTC)
...If the host shows one of the goats the game provability is no more 1/3 or 2/3 or whatever: it becomes 1/2, right?
In fact the game was never 1/3 because one of the doors with a goat will always be eliminated, so the goat door 1 and the goat door 2 are, in a kind of way, the same door (aka the same result).It doesn't matter if you choose a door with a goat or not, when in the end you have to choose there are just two doors and one goat. In my point of view the game starts when there are only two doors and 50% of changes for each door.
PS: Feel free to correct me
PS2: I'm sure you will ;)
Thank you for the reply, now I can see it. I would have read the other disussion threads, but in this article there are too much. Great mike 10:27, 26 December 2006 (UTC)
This problem has probably confused so many people because the typical explanation for it is wrong even though it still gives the right answer. People poke holes in the argument only to find through some other means that the conclusion is still true.
It is fallacious to hold that you originally had a 1/3 chance of choosing the very goat that monty reveals that you did not choose when he opens a door. It is true that the knowledge that you did not choose the goat monty reveals would make it 50/50 you chose what was left, but the additional information that monty revealed that goat (which he is less likely to do because you chose the car than because you chose the other goat) readjusts the odds such that you are more likely to win by switching.
Using an alternate version of the problem where the odds of choosing each are the same but you know monty has a strategy for showing one of the 2 goats (which you can distinguish) shows how this reasoning fails in the general case. Here everything that is said in the explanaition of the original problem remains true but the conclusion is now false - because once again it is fallacious to hold that you had a 1/3 chance of choosing a goat to begin with even after it has been proven to you that you did not choose that very goat.
Here is the math, first a real explanation and then a alternate version of the problem where the wiki explanation fails.
A REAL sample space might look like
1/3 You chose Goat 1 Monty reveals Goat 2 Switching Wins
1/3 You chose Goat 2 Monty reveals Goat 1 Switching Wins
1/6 You chose car Monty reveals Goat 1 Switching Loses
1/6 You chose car Monty reveals Goat 2 Switching Loses
You can see how this is accurate because if Monty reveals Goat 2, then you can cross out thatyou chose goat 2 and that you chose car and monty reveals 1 i e
1/3 You chose Goat 1 Monty reveals Goat 2 Switching Wins
X1/3 You chose Goat 2 Monty reveals Goat 1 Switching Wins
X1/6 You chose car Monty reveals Goat 1 Switching Loses
1/6 You chose car Monty reveals Goat 2 Switching Loses
and then you get
2/3 You chose Goat 1 Monty reveals Goat 2 Switching Wins
0 You chose Goat 2 Monty reveals Goat 1 Switching Wins
0 You chose car Monty reveals Goat 1 Switching Loses
1/3 You chose car Monty reveals Goat 2 Switching Loses
Laying out the sample space in this correct manner also allows you to solve any other similar problem including the one where Monty chooses goat 1 with 2/3 probability when you choose the car:
1/3 You chose Goat 1, Monty reveals 2
1/3 You chose Goat 2, Monty reveals 1
2/9 You chose the car, Monty reveals 1
1/9 You chose the car, Monty reveals 2
Note in this case that although you had originally a 1/3 chance of choosing each goat or the car, collapsing the 1/3 chance of the revealed door into the unrevealed unchosen door to get 2/3 chance of car, or following any of the other rediculous suggestions in the wikipedia article does not give you the right answer. The odds can even be 50/50 for switching if monty always chooses one goat over the other.
-The author of this post abstains from signing his posts such as not to promote appeal to authority and ad hominem fallacy. —Preceding unsigned comment added by 69.252.158.32 ( talk • contribs)
1/3 You chose Goat 1, Monty reveals 2
1/3 You chose Goat 2, Monty reveals 1
1/3 You chose the car, Monty reveals 1
0 You chose the car, Monty reveals 2
It is stated in my examples that the goats are distinguishable. It is irrelevant whether or not you can distinguish the goats as far as the validity of the given solutions to the problem are concerned. Why should the fact that the goats are distinguishable suddenly invalidate a claim like "collapsing the 1/3 chance of the revealed door into the unrevealed unchosen door to get 2/3 chance of car"? It wouldn't - this claim was already invalid to begin with. It is no different than saying if you dance on your head on the night of a fool moon you will have a 2/3 chance of winning by switching. See look - it has been verified by experiment.
All you have to do to disprove the dancing on your head claim is do the problem without dancing on your head and see if you still get 2/3 chance to win by switching. And all you have to do to disprove the collapsing 2/3 between 2 doors claim into one door is to do a version of the problem where everything this explanation is based on is still true but a different answer is arrived at. In my version of the problem each door still has a 1/3 chance of holding the car so therefore if the statement regarding collapsing probabilities was ever true it should still hold. It does not hold so therefore it was never true. -The author abstains from signing posts so as not to endorse ad hominem and appeal to authority fallacies. Let everyone be no more or less than the strength of their arguments. —Preceding unsigned comment added by 69.252.158.32 ( talk • contribs)
I hereby propose that this section of the talk page be deleted as any attempts to discuss actual mathematics with 69.252.158.32 will only result in more people getting death threats. -- Antaeus Feldspar 01:15, 29 December 2006 (UTC)
I've copied the section just added to the article below. Before adding this analysis to the article I request a source be provided. Without a source it looks like original research to me (which is not allowed). Furthermore, it looks incorrect (per my comments above). The deleted section follows. -- Rick Block ( talk) 04:23, 30 December 2006 (UTC)
I've re-removed the section, per Rick Block's comment. A source needs to be provided for this; if we do include it, it needs some cleanup—I just don't understand the point that it's trying to get across. TenOfAllTrades( talk) 04:57, 30 December 2006 (UTC)
My comments were not clearly addressed. This whole thing is not worth arguing about. Please provide a reference substantiating this analysis. If it's not original research then there is a reference you should be able to provide. With 40 or more academic articles about this problem, it shouldn't be difficult to find one supporting your view if it has any validity whatsoever. Thanks. -- Rick Block ( talk) 05:34, 30 December 2006 (UTC)
I have read your response and fail to see how it addresses my comments. The odds in question are not conditional odds given a known exposed goat. In the problem as stated, the goats are not distinguishable. Any analysis involving distinguishable goats is simply irrelevant and confusing to the point. If you claim this analysis is already supported by the existing citations, please cite a specific one (paragraph or page number would be nice as well). Please sign your comments (or if you refuse to, at least don't delete the "unsigned" tag others add - who makes what comments can be easily determined by looking at the history of the article so you're not accomplishing anything by not signing). -- Rick Block ( talk) 05:58, 30 December 2006 (UTC)
At the point when the player is asked whether to switch, there are three possible situations corresponding to the player's initial choice, each with equal probability (1/3):
The player originally picked the door hiding goat number 1. The game host has shown the other goat.
The player originally picked the door hiding goat number 2. The game host has shown the other goat.
The player originally picked the door hiding the car. The game host has shown either of the two goats.
This proposed sample space is wrong in that it attributes a 1/3 probability of something happening that is already known to have not happened. Note at the top it says "At the point when the player is asked whether to switch" That means Monty has already eliminated one of the above possibilities - yet at the same time it is attributed a 1/3 probability. -K99 —The preceding
unsigned comment was added by
Kriminal99 (
talk •
contribs).
No further explanation is required to anyone that is not selectively blind. Your statements regarding the three choices are irrelevant because you are failing to realize an important fact. At the time the decision to switch is made, it is not an issue of those three choices and nothing else. It is an issue of 2 choices, and the probability of those choices must be determined using all available information including everything we know AT THAT POINT about the original 3 choices. At that point we know that one of the three that we did not choose did not hold the car, and we also know that Monty revealed this fact according to a certain algorithm. This information is totally different than the oversimplified 3 initial choice explanation which by shear coincidence happens to result in the same answer. —Preceding unsigned comment added by 69.252.158.32 ( talk • contribs)
The above explanation may be fallacious in that it requires the reader at a certain point to hold that the contestent had a 1 in 3 chance of choosing the goat that Monty has already revealed that the contestent did not choose. This is in violation of the principles of probability which would have someone change the probabilities of their initial choice with the new information given when the goat is revealed.
In this specific version of the Monty hall problem, the additional information given by the fact that the goat Monty reveals could not have been chosen is exactly offset by the information that Monty reveals that goat (which he is less likely to do when the contestant has chosen the car) Therefore the right answer can be arrived at incorrectly.
When Monty has a known strategy for revealing one of two distinguishable goats it is still true that there is initially a 1/3 chance of choosing each object and only in one of those 3 situations would switching allow you to win. However that has nothing to do with the probability of winning when one switches as this case shows, since the odds are no longer 2 to 3 in favor of switching. And the reason it has nothing to do with it, as mentioned before, is because it is fallacious to simultaneously hold that you had a one third chance of choosing a given goat and that you did not choose said goat because Monty has revealed that you did not.
Fallacious explanations for the outcome in the Monty Hall problem may have been the source of earlier widespread disagreement regarding the problem in which many members of the academic community expressed disagreement. Many people realize there is something wrong with common explanations of it only to find that there is still a 2/3 chance of winning by switching doors using other means of investigation.
This is a question that has already been asked and answered but it seems it needs to be addressed again -- if Monty has some particular pattern or method by which he picks a goat to show, and the player knows this and can distinguish between the goats, does it ever make some strategy other than "always switch" an optimal strategy?
The answer is "no". To prove that this is so, we will assume the opposite. We will assume that one of the goats is black and one is white, and that Monty has a bag with B black marbles and W white marbles; when he picks a black marble, he shows the black goat and when he picks a white marble he shows the white goat. B and W may be 0 but they cannot, of course, be negative numbers.
B+W represents the total number of times Monty chooses between the two goats, which he only does when the player initially picks the car. The player's chance of picking the black goat is equal to his chance of picking the white goat is equal to his chance of picking the car; therefore B+W is also the total number of times that the player sees a particular goat because he chose the other goat.
Therefore:
Now, let us assume that we can find values for B and W such that seeing a particular goat means the player should stay, and not switch. If there is such a goat, we can devise a strategy in the form "if you see this goat, stay, and otherwise switch" that beats the strategy of "always switch". If however the optimal strategy for each individual goat under all values of B and W is "switch" it means that the optimal strategy for the game is "always switch". So, we assume that for some goat (say the black goat) the player is actually more likely to have picked the car and therefore maximizes his chances by staying, rather than switching.
But here we run into a problem. Out of the B+W+B times the player sees the black goat, B is the number of times the player picked the car and B+W is the number of times the player picked the white goat. For staying to be an optimal strategy, B must be larger than B+W. W would have to be negative to fulfill this condition but we noted at the beginning, W represents a number of marbles and clearly cannot be negative. Even choosing a value of 0 for W only makes it possible to encounter a situation where switching provides no particular advantage, but: 1) no strategy in that situation can make the player's chances greater or less than 50%; 2) for every two times the player faces that situation, he gets a situation where switching pays off 100% of the time.
Can we even pick any values for B and W such that the overall success of a strategy of always switching even becomes something other than 2/3? No, we cannot. What is the proof? The proof is that staying always pays off when the player originally picked the car; switching always pays off when the player originally picked a goat. The ratio of times the player picks the car to times the player picks one of the two goats is constant, and is not affected by how Monty chooses which goat Monty chooses to show in those situations where he has a choice of goats. -- Antaeus Feldspar 02:27, 31 December 2006 (UTC)
This discussion goes around and around without a definitive end, like an argument over whether you should make clam chowder with milk or tomatoes. I think a graphic of the possible sequences of events puts an end to it, provided that people can agree that the graphic is an accurate model of the system. Here's my proposal:
File:ModelOfEvents-Probabilities.jpg
If this model is right, then its behavior is what you should program if you want to simulate the system. You can also infer the probabilities of each possible event in the Monty Hall sequence, like this:
The probability of each option (joint event) shown in this model is calculated by multiplying the probability of the single event times the probability of the prior event. The probability of a hidden prize is 1. [Consider some examples: The probability that I will initially choose Goat 1 is 1/3. If I choose Goat 1, then Monty is forced (probability = 1) to reveal Goat 2, so the probability of the sequence down to that point is (1/3)*(1)=1/3. Because Monty has his choice of two goats after I first choose the Car, the probability that Monty will choose Goat 2 in the joint event is (1/3)*(1/2)=1/6.]
The process is determined by three events:
1: I choose among the three doors.
2: Monty reveals a goat.
3: I choose one of the remaining two doors (by switching or not switching).
If this model is valid (modelling what it's supposed to model), then it disproves the original, peer-reviewed article. This is a mathematical proof, which is a pretty strong form of proof for this type of problem. Someone has already pointed out that disproof would be an interesting outcome, due to what it reveals about the limits of peer review. It's rare that a peer-reviewed decision can be tested in this way.
On this model, I have a .5 probability of coming out of the game with a car. This probability is determined by summing the probabilities of the sequences in which I get the car (1/6+1/6+1/12+1/12).
If this model is not valid, then someone should propose a different decomposition of the possible events in Monty's system.
I take it as axiomatic that:
1: The system can be analyzed as a series of three events.
2: The probabilities of the possible options for each event must sum to 1.
If someone does propose another decomposition of the possible sequences, the second requirement is a good logical test of the model.
I think that the model is correct and exhaustive, in which case we should be done with the question of what is right. That would leave us with a charming article that is fundamentally wrong. I can't suggest an equally charming article based on my analysis. I do have some suggestions, but they are inappropriate until commentators reach a consensus. I have been known to make mistakes from time to time. Etarking 03:38, 2 January 2007 (UTC)
I've restored the section about the variant recently discussed in Vos Savant's column. There's discussion about this in two separate sections (above). If the host doesn't know or forgets where the car is and opens a door anyway, it makes no difference whether you switch or not and you have a 50/50 chance either way. Please see the sections above for more discussion about this. -- Rick Block ( talk) 01:39, 3 January 2007 (UTC)
I apologize if I confused things by introducing the concept of the "do-over" to discussion of this variant. It seemed to be the easiest way to explain a difficult concept (or at least a concept that someone was indeed expressing difficulty with.)
The concept in question is that even though there are certain possibilities which could clearly come about by random chance, these possibilities are shown not to have happened by the evidence given as part of the problem statement. In this, the "forgotten goat" variant Monty Hall problem, there are six combinations of doors that can be chosen between the player and Monty; two of these are combinations where Monty does mistakenly show the car. However, the problem statement says that Monty does not show the car; thus, we are dealing only with the four possibilities that are not contradicted by the problem statement.
This seems to puzzle many people, who are not sure whether it is "legal" to simply 'prune' the probability tree and say that a thing doesn't happen when clearly it could. Many of us learned to work through probability puzzles through thought experiments employing an appropriate randomizer ("OK, if I assign two faces of the die to each of the three doors, and if I get every face of the die once...") However, the forgotten goat variant poses difficulty for those choosing this approach: how do you simulate a randomizer not coming up with results that, quite obviously, it could generate at any time?
The answer is to incorporate the "do-over" into the simulation: if you get a result that is contradicted by the evidence of the problem statement, simply ignore it and keep going until you get a result that isn't contradicted. However, it was only ever intended as an aid to help people run the simulation for themselves and see in a convincing way why the answer is what it is; it was never meant to be implied that the "do-over" was part of the actual problem. -- Antaeus Feldspar 22:15, 3 January 2007 (UTC)
But once the host has opened one of the doors, the choice has become out of TWO doors. The opened door can be forgotten. It's as if a new choice has been given: "A door has a car and a door has a goat" CHOOSE! :/ it 's a 50% choice!!! please help —The preceding unsigned comment was added by 194.204.127.214 ( talk • contribs).
Oooh thanks a lot! :D - markbri
Monty Hall problem has been nominated for a featured article review. Articles are typically reviewed for two weeks. Please leave your comments and help us to return the article to featured quality. If concerns are not addressed during the review period, articles are moved onto the Featured Article Removal Candidates list for a further period, where editors may declare "Keep" or "Remove" the article from featured status. The instructions for the review process are here. Reviewers' concerns are here. Gzkn 10:57, 7 January 2007 (UTC)
I wrote a flash simulation of the Monty Hall problem. Is there a way for me to upload it to wikipedia? —The preceding unsigned comment was added by Lax4mike ( talk • contribs) 03:22, 9 January 2007 (UTC).
These aren't Venn diagrams in the strict modern sense, but I'm not sure what else to call them. Venn might have disagreed. Septentrionalis PMAnderson 04:18, 9 January 2007 (UTC)
The following text was deleted:
on the grounds that it is not related, except in being a cognitive illusion. I disagree; I must not have made it clear, but I think (and so do several sources, including the paper cited, that it's the same cognitive illusion. The fallacious argument in the Paradox is
Discuss? Septentrionalis PMAnderson 02:56, 10 January 2007 (UTC)
Abscissa brings up a point worth discussing. Much of this article consists of different explanations of the same result, and they really are themselves quite similar.
Septentrionalis PMAnderson 03:36, 10 January 2007 (UTC)
the MH game a few times (handling the cards themselves: watching someone else is not very effective), that physical sense of the situation easily switches the intuition. After a few plays where you handle the cards yourself, it will become obvious that it pays to switch. But mere verbal instructions or even merely watching someone else handle the cards will easily fail. So a person in doubt about whether it really pays to switch ought to take a couple of minutes and deal the game a few times." -- Antaeus Feldspar 05:55, 14 January 2007 (UTC)
The reason the article is so repetitious is because the explanations are all wrong. Thus instead of a single valid explanation that would put any objection to rest a thousand invalid arguments are used instead. —Preceding unsigned comment added by 69.252.158.32 ( talk • contribs)
Is this picture helpful to anyone? When I first started to understand the problem this hindered more than helped. I understand that some people might have different ways of understanding the solution, but I believe this image is peddling a bad explaination on how to arrive at the solution. This problem has absolutly nothing to do with what is suggested in the picture. I move for its deletion, or at least a recreation which explains the solution better. This is not a case of just having 2 seperate guesses at the correct door (and thus 2/3 probability) 202.10.86.59 20:50, 13 January 2007 (UTC)
Well, then, since the people who have already spoken above saying "Presenting this analysis in the form of a JPEG adds nothing to presenting it as text" are apparently just not enough for the cartoon's provider, let me add yet another voice to that contingent. Perhaps then the cartoon's provider will be able to grant that yes, other editors have considered the merits of the cartoon and still don't think it's appropriate and -- here's the kicker -- should therefore not be re-added. The cartoon does not in any way shape or form illustrate the problem and therefore should not be in the article in the form of an illustration. This leaves the question of whether the text should be in the article in the form of text. Frankly, I do not think so; it is not nearly as clear as its author seems to think. It is hardly intuitive that "to open both the other doors instead ... [is] exactly the choice Monty is offering"; in fact, I am rather skeptical that it could even be described as true since "is exactly" and "is equivalent to" are two very different things. Now of course we could explain this equivalence, but I think the section "Combining doors" is already doing exactly that, very well. For these reasons, my recommendation is leave the cartoon out.
If I sound frustrated it's because I am. "I can put it back against consensus" does not mean "I should put it back against consensus". -- Antaeus Feldspar 01:20, 21 February 2007 (UTC)
I think the cartoon is excellent. It detracts nothing from the article and instead adds a little bit of humour as well as clever explanation in pictorial form. I can't imagine why anyone would be dead set against it. The arguments offered above to that effect seem to me to be peculiar. To start arguing about identity versus equality when talking about a cartoon is bizarre. And I'm not even from the Kingdom of Fife. Davkal 01:47, 21 February 2007 (UTC)
Infact, I should have been more general. I do not believe the Venn diagram is an accurate way to represent the problem. In the Venn diagram, doors 2 and 3 are grouped, seemingly because they are physically next to each other. What if door one were next to door three? If there is indeed a sound reason why, I think it should be presented in the article, in the Venn diagram section. There is a 2/3 chance it could be behind doors 2&3, then it must follow that there is a 2/3 chance of it being behind doors 1&3. So then, doesn't the Venn diagram present a choice that is 50/50? Clearly, this does not agree with the actual solution. In conclusion, the Venn diagram seems like an over-simplification, if it is even correct. WIth regards to the picture, any tool that seems to aid understanding is fine with me, however, I don't think the Venn solution is accurate, and therefore the same is said for the picture, in my opinion. 202.10.86.59 05:58, 14 January 2007 (UTC)
Because of the very nature of the paradox, the usual 50:50 response and the amount of debate that it generates, I would like to see more text in the intro on the background leading into the analysis. The analysis is so long and uses so many methods, without fully explicating the controversy that has been generated in the public sphere, that many readers will not bother with the analysis, and will write if off -- some people were prompting to take the page down as incorrect and a near-hoax, for instance. I would suggest using text based on the following:
"When Marilyn vos Savant quoted this puzzle in the US a few years ago [citation, date], she received over 10,000 letters mostly telling her she was wrong.
One was from Robert Sachs, a professor of mathematics at George Mason University in Fairfax, Va. who said "As a professional mathematician, I'm very concerned with the general public's lack of mathematical skills. Please help by confessing your error and, in the future, being more careful."
However a week later and Dr. Sachs wrote her another letter telling her that "after removing my foot from my mouth I'm now eating humble pie. I vowed as penance to answer all the people who wrote to castigate me. It's been an intense professional embarrassment."
Well, we said it was counter-intuitive. Even professional mathematicians get it wrong." ( http://www.grand-illusions.com/monty.htm) That is just a framework suggestion, not suggesting direct plagiarism of the text. Or similar words around what has driven the controversy most recently, being the 'Marilyn vos Savant' article. The so-called Marilyn vos Savant character is presumably a pen name and may represent a number of contributors [verify].
Also, it would be good to include links to online simulators in a separate section, or make reference in the current heading to them. (examples also found at the same site: http://www.grand-illusions.com/monty2.htm) -- Sean01 07:37, 15 January 2007 (UTC)
I think this is an excellent article, I like the formal discussion and and also the artwork, even the goofy one at the top. You might want to consider moving the history section up a bit and add to it the above posting (on this talk page) about why this has caused confusion. Maybe a good place would be just before the "understanding aids", especially if the above information is added then the reader will be even more interested to read the excellent analysis. But I mean, even as it currently stands, it's awesome, simply awesome! -- Merzul 02:28, 17 January 2007 (UTC)
Is the extended Bayesian analysis recently added from some particular source? If so, can someone please add a reference? The article is currently undergong a featured article review, which includes making sure that everything that should be, is referenced. I fear some folks might view this extended analysis as approaching original research, which is prohibited (see WP:NOR). -- Rick Block ( talk) 14:24, 17 January 2007 (UTC)
I have noted that there has been some unproductive edit skirmishing over how to render simple fractions. I'd like to see us discuss the relevant advantages and disadvantages of our options, rather than just reverting each other without such discussion. There are three main approaches, and I'm just going to list the advantages and disadvantages that I see to each. -- Antaeus Feldspar
<math> \frac{1}{3} </math>
(which becomes ); <math> \tfrac{2}{3} </math>
(which becomes )
As you might be able to guess, the last option is my least favorite, but I think my reasoning (readers should be able to read the math we are presenting) is rather unassailable. Anyone else? -- Antaeus Feldspar 01:42, 20 January 2007 (UTC)
Since this article primarily includes only very simple fractions (mostly 1/3 and 2/3), I think the advantages of using plain text which renders everywhere outweigh any disadvantages with this approach. Does anyone violently object to this solution? -- Rick Block ( talk) 02:40, 23 January 2007 (UTC)
I believe all current specific suggestions from Wikipedia:Featured article review/Monty Hall problem have been addressed, except this comment from user:SandyGeorgia:
There's a thread above, #About the Introduction, and online simulators, with a suggestion for the lead as well. I'm not quite sure what to with these comments. The problem statement in the lead is deliberately a quote to deter wordsmithing (which has been a problem). The spoiler is there since the solution is specified. I've recently added a sentence about the response to the 1990 Parade column. I'm considering shortening the lead by deleting everything after the spoiler warning (and the warning), which would include deleting the cartoon about the solution and revising the text following the quoted problem statement, sort of like:
Thoughts on this? -- Rick Block ( talk) 17:21, 20 January 2007 (UTC)
I frankly would rather see the solution deleted from the lead. There are of course many different standards for what should go in the lead and what should be saved for after the lead, but one that I favor is that the lead should be what a programmer calls a " black box" -- one sees its inputs and outputs but not its inner functioning. Describing how it has confused people because of its counter-intuitive solution -- that's input/output. Describing what its counter-intuitive solution is and why -- that's the inner functioning.
That being said, the solution should be at most the second thing after the lead, since if someone is interested in anything else besides the "inputs and outputs" of the problem, it will be the solution. The only reason for making it the second thing after the lead, rather than the first thing, is that the solution will often appear nonsensical or wrong to people if they've misunderstood the exact problem constraints, so there is a strong argument to be made for presenting the unambiguous problem statement first and then spelling out the solution within those constraints. -- Antaeus Feldspar 02:46, 22 January 2007 (UTC)
In response to further comments from user:SandyGeorgia at Wikipedia:Featured article review/Monty Hall problem, I think she's suggesting condensing the lead down to a paragraph or two, perhaps something like:
Immediately followed by the section describing the problem (moving the quoted version from Parade and the associated image there), then followed by the section describing the solution (with the cartoon graphic). Anyone have any objections to this approach or alternate suggestions for how to address SandyGeorgia's comments? -- Rick Block ( talk) 14:37, 25 January 2007 (UTC)
I removed this addition by an IP from aids to understanding - not sure if it's needed or a repeat:
Another way of thinking of this is: If I initially choose the car, he will show either one of the goats, in this case it is best to stay. If I initially choose the first goat, he will show the second goat, it is best to switch. If I initially choose the second goat, he will show the first goat, it is best to switch. Notice that even after he exposes a goat you don't know if your door has a goat or car. Since ⅔ of the time it is best to switch, switch! Thus, simply put, your first choice was likely a goat (a ⅔ chance), so when he shows you where the other goat is, it is in your best interest to choose the other door.
SandyGeorgia ( Talk) 04:03, 26 January 2007 (UTC)
I removed this addition showing a connection between two different aids to understanding sections:
Note that the both the prior probabilities , all equal to 1/3, as well as those conditional probabilities which are nonzero are displayed in the above decision tree. The Bayesian analysis which follows at this point is essentially equivalent to the decision tree analysis given previously.
This, and similarly motivated paragraphs are redundant. All proper proofs of the winning strategy are equivalent, but this does not imply that the article should spell out at length the N*(N-1)/2 ways in which N "aids to understanding" relate to each other: it would only add unneeded clutter. Rather, the "aids to understanding" should, IMHO, be left as alternatives to each other - different ways for different minds to arrive to a common conclusion. The Glopk 16:50, 26 January 2007 (UTC)
I've moved the following text here from the article:
The aquarium analogy
An excellent analogy helping to understand is suggested by Andrea Gennari of Rome, Italy. This is the so called "Aquarium analogy". An aquarium contains 100 mugs of water (the example can be made using different measurement units without affecting the result, in the example a "mug" of water can be a 1 liter mug or a 1000 liter mug). Now assume that the player cannot see what is inside the aquarium i.e. the aquarium is hidden to sight. The aquarium contains a fish. For the sake of simplicity, assume the fish can be contained in a large enough mug of water. As for the 100 doors example, the player is given the possibility of finding the fish by filling a mug with water so that he has to pick up a mug and put it in the aquarium. What is the chance of the fish being in the mug the player has just filled in with water from the aquarium? Clearly the probability is 1 out of 100 or 1%. Now we place the hidden mug aside without showing the content and we progressively fill 98 mugs with water from the same aquarium, this time showing that in each mug there is no fish. Progressively the aquarium is emptied but we take care of never picking the fish when we fill in the 98 mugs. The fish remains in the aquarium. So at the end of the process we have the original first mug filled with water, the content of which is hidden to the player, the 98 mugs filled with water and no fish, the content of which is shown to the player, and the aquarium filled with the equivalent of only one mug of water and always hidden to the player. The player is then given the possibility of switching between the original mug chosen and the aquarium (containing a volume of water equivalent to a mug). What is the most logical action to maximise the probability of finding the fish? Or, in other terms, is it more likely that the fish is in the original one mug chosen at the beginning or in the one mug volume of water contained in the aquarium? From the example it is clear that the fish can only be in one of the two, but it is now intuitive that, while the probability of the fish being in the first original mug picked by the player is 1%, the probability of the fish being in the last remaining hidden volume of one mug (contained in the aquarium) should be 99%, because when the first mug was filled at the beginning, the probability of the fish being in the other 99% of the volume of the aquarium (equivalent to 99 mugs) should have been clearly 99% and none of the 98 mugs "extracted" contained the fish. The example mirrors exactly the 100 doors example. This example helps understanding in that it makes use of volumes and "probability density". The probability density of the fish being in any given volume is the same therefore if we split the volume in two smaller volumes (a "1 mug" volume and a "99" mugs volume) the probability of the fish being in the "99" mugs volume is clearly 99%. When we show the content of 98 out of the 99 mugs volume, what we do is to concentrate the probability of 99% in the remaining one mug volume of the aquarium. The same example can be done with a "three mug" aquarium analogy thus reflecting the original Monty Hall problem.
Despite the fact that the analogy is attributed to "Andrea Gennari of Rome, Italy", it is not cited. We have no idea where this appeared or who Andrea Gennari is. And frankly, even though I understand the Monty Hall problem pretty well, I can't follow this analogy which is supposedly clearer than the existing explanations. -- Antaeus Feldspar 01:13, 1 February 2007 (UTC)
I undid change 104755257 by user David Eppstein, who had changed some of the "inline" fractions in the equations of the Bayes Theorem sections into LaTeX \frac{}{} forms. The result of this change was a marked inconsistency in the equations that looked visually unpleasant. In particular, the multi-line formula evaluating the normalizing constant mixed fractions of the two types, probably because having the final result in a \frac form made for a confusing read next to the previous line.. Also, the line spacing on my monitor became quite "jumpy". Something needs to be done to improve the rendering of math in Wikipoedia, but until it is I think we should edit conservatively.... The Glopk 15:24, 1 February 2007 (UTC)
The lead paragraph should state the problem in a simple and low-drama way. In particular, I think that the mention to any elements of "Let's Make a Deal" should be minimized because this is not about a game show, it is about a logic problem. We should stick to the correct problem statement, the answer, and then mention that some find it non-intuitive. Unless somebody is going to have a heart attack over this, I will attempt to do so.-- 199.33.32.40 00:17, 2 February 2007 (UTC)
Sorry Anteaus. It is not my intent to present the "game show version" of the problem in the lead section except at a historical curiousity. It is my intent to present the adult "more than three doors" version of the problem in the lead paragraph, and without the doors, the klieg lights, pretty girls, prizes etc. This FA is under the category of Math. Cold, unforgiving, brutal Math. I know it is less entertaining that way, but that is the subject of the article. Wikipedia is not a game show. I present the clear, bone-crushingly obvious and boring version of the problem first. The subject is not Monty Hall. The subject is the Math (more specifically Logic) Problem. -- 199.33.32.40 01:15, 2 February 2007 (UTC)
Anteaus, much of the fighting and confusion of this article is due to the treacle the project puts out via the mentioning in the lead section of females like Marilyn vos Savant and her ludicrous errors in attempting to even address this problem. Here you guys are coming to the consensus that some pretty face is more important than the subject of the article. The project is supposed to be educational, not just a bunch of pot stirring. Consensus is all fine and lovely but Math is unforgiving enough as it is without that female muddying the waters deliberately, mistakenly or simply for the publicity. She barely rates making it into the Trivia section of this article.-- 199.33.32.40 01:25, 2 February 2007 (UTC)
This article is a disappointment. Marilyn got it wrong and within this article she should be relegated to some obscure footnote. Based on the dramatic header you have at the top of this talk page, it is no wonder that you fail to educate some readers about the true nature of the subject. -- 199.33.32.40 01:49, 2 February 2007 (UTC)
That some female who did little but cause more confusion gets into the lead section while the problem statement itself does not might be entertaining, but it is not educational. Does this really belong in the "Media" group rather than the "Math" group of FA articles? Well, if it does not belong in the "Math" group, then here is what was tried as the problem statement in the lead section:
At least this is about Math rather than about some entertaining "savant". BTW: Once a decent, subject-oriented version of this article is created, I strongly suspect that it will become more obvious that it deserves the "Low" importance rating that it gets from the Math WikiProject and that it is an unworthy subject of a "Math" FA. I actually want this to remain in Math and get permanantly demoted. -- 64.9.233.132 03:15, 2 February 2007 (UTC)
The References section contains entries that are not cited anywhere in the article body. One was added today, presumably by its author. What is the policy or custom about such references? Leave them alone? In academic papers (at least in my field), having such "dangling" references is highly frowned upon. The Glopk 17:19, 3 February 2007 (UTC)
I've been away for a couple of weeks. In the intervening period, this article seems to have morphed into a dusty old text book. That's very sad, given that Wikipedia has the opportunity to be so accessible and so different. I say: delete things that are wrong, not things that are simply not to your liking. De gustibus non est disputandum, after all! StuFifeScotland 13:18, 12 February 2007 (UTC)
See also: #MontyHallProblemMadeEasy.jpg
Can we reach a consensus for whether the cartoon image should be in the article or not? The reasons I prefer it not to be in the article are listed below. Please, let's not vote about this, but discuss (and respond to) reasons for or against. -- Rick Block ( talk) 02:07, 21 February 2007 (UTC)
We're not reaching a consensus. Let's have a vote. -- Doradus 07:08, 23 February 2007 (UTC)
Glopk, tis one of the mysteries of life that, for some reason, some people find it easier to relate to something that has a face (e.g. a cartoon) than they do to a dry dusty textbook. You may not like this, and you may feel that it is wrong, but that doesn't change the fact. Another fact that is not changed by mere repetition of the contrary viewpoint, is that the picture is actually a picture (even with text). It does not merely repeat the text as you suggest, but does other things as well - and those other things are primarily what makes it a picture. Your failure to see this point is one of the main reasons why your thoughts on why it should be removed don't really count for very much. That is, the thing you want removed (a mere repetition of the text) isn't actually in the article and nobody is suggesting it should be. Davkal 03:21, 22 February 2007 (UTC)
1. We already have testimony from uninvolved individuals to the effect that they think the picture is good. That is of value to them, and that it will be used by them in teaching etc. It is therefore not me who is trying to foist something onto everyone, but you who is intent on denying everyone something that has already been shown to be useful because you don't like it.
2. I did read your comments tiwce and did not embarrass myself - I have read them again and still stand by what I said. I was pointing out that you still don't seem to understand the value of pictures, even though you say you do, because, for example, the example you give involving Eureka might work quite well in a variety of ways yet you dismiss it out of hand and think it supports your point rather than actually supporting the contrary. The point being that pictures work in a variety of ways, and your example may well act as, say, an excellent aid to memory. Similarly, the cartoon here may well help someone grasp a main point in a way that pages and pages of plain text does not. That, I think, is one of things that many who support the cartoon have in mind. Facilitating understanding though the use of various media. And it is one of the reasons why the repetition of text argument is a non-starter.
3. If you want to measure illustrations against Michelangelo's work then you'd probably need to remove every picture in Wiki. And while I'm on that point, why do we have the picture of the goat at the top of the article which has to be accompanied by five lines of caption merely repeating the text in order to explain it. That picture would better if we had a little cartoon Monty saying "Pick a door" and a cartoon contestant saing "I pick door 1" then monty saying "door 3 has a goat, would you like to change your mind". That picture, the goat, actually stands up a whole lot less well to the arguments but nobody is suggesting we remove it. Davkal 14:40, 22 February 2007 (UTC)
The above are arguments as to why it augments the article. It adds variety, it is a nice pictorial represenation of a sound point along with a humourous way of showing the puzzlement that can be caused by the MHP. Davkal 04:37, 21 February 2007 (UTC)
Glopke's argument against me is basically to insult me gratuitously. Isn't there a rule about this? Isn't this contrary to Wikipedia policy? Davkai's response to his insult is on point. Bill Jefferys 14:51, 22 February 2007 (UTC)
I apologize for misspelling your name; this was unintended. But I am puzzled about your claim that I violated "no-soapbox twice on this very article before you reverted my edits." This because the only thing I did was to revert a previous edit to the main article, with a short comment saying that I thought the picture useful. I had not commented before, and after my reversion my reversion was reverted. I have not made any changes to the main article since, and have only presented my views as a person who has used the Monty Hall example often in my classes. Please point to where I was violating "no soapbox" at any time before you reverted my edit (singular) or even later. If I have violated policy, I will apologize of course, but first I have to know what I have done wrong, specifically, not based on vague accusations but on specific example. Bill Jefferys 22:22, 22 February 2007 (UTC)
Everyone here should re-read WP:NPA, specifically "Comment on content, not on the contributor." In addition, I suspect some WP:OWN at work here as well. In any case, it's becoming evident consensus is difficult to reach on this issue and there're at least a few editors on each side, maybe it's time to invite an admin over for arbitration. Tendancer 17:55, 22 February 2007 (UTC)
So far we have a number of people who feel that the cartoon adds to the article and a number who feel it does not. What that means though is that for some people who read the article the cartoon will be a useful addition, and for others it will make no difference. The point being that it's all very well to say I don't find it funny and that humour is subjective, but this is completely irrelevent since is such cases nothing is lost by including it. What is missing from the against side are the negative things that the cartoon is supposed to do - someone finding it offensive or something, that would be a negative. But so far we simply have plus points and zeros. I don't think anybody supporting the cartoon thinks its the best, or funniest, or most aesthetically pleasing thing they've ever seen, but so what, to simply point these things out is not to give a reason for removal. For that it is necessary to show how its inclusion damages the article. Davkal 19:22, 22 February 2007 (UTC)
The lot of you need to grow up. I don't have an opinion on the image, but it's clear to me that the right thing to do is keep the controversial material off the page until either a consensus can be reached, or a vote can be taken to resolve the issue. -- Doradus 20:04, 22 February 2007 (UTC)
So the comment on content, not on the contributor, rule doesn't apply in special Feldspar cases. How surprising that you are the only one allowed leeway in this respect. I am commenting on your base motives. 1. you think you own the article. 2. you think you have the authority to treat everyone who disagrees with anything you say in a condascending manner. 3. You think you know what a picture is. I wouldn't normally draw attention to this claim because 99% of people do, but when you're one of the few who actually doesn't I think it needs to be mentioned. Davkal 23:50, 23 February 2007 (UTC)
You're obviously just having a lugh. When you're ready to discuss it sensibly I'll discuss it with you. Davkal 05:10, 21 February 2007 (UTC)
Both of you. Unless you really don't know what a picture is, then the only assumption left is that you're having a laugh. What else have we had, oooh the poor Germans, oooh the poor server, oooh what about the poor people with poor eyesight who don't know how to click on an image. What next? If these are supposed to be serious good faith edits then God help you. Davkal 05:47, 21 February 2007 (UTC)
Yes, inasmuch as how else do you explain to someone who insists that something isn't a picture but is merely text when that thing is obviously a picture. I did it by pointing out the little face on it and text bubbles etc. It's perhaps a bit sarcastic but it's clearly in good faith. Taking up too much room on the server and people who don't speak English not being able to understand it and people with poor eyesight and no IT skills not being able to read it are, on the other hand, simple examples of opening your mouth and letting any old thing come out.
Davkal
16:14, 21 February 2007 (UTC)
Put me in the camp I'm for the image. I think there may be a bit of intellectual elitism at work here, and that's making some editors against it because the image is a bit too cartoonish/dumbed-down unconventional/un-mathematical interpretation for their understanding of the problem. However, the layman reads this, and I'm sure this article was not once chosen as a featured article because it was mathematically sound (though that would be one of the reasons)--all the cartoons, simplified explanations for the non-mathematical layperson certainly must've been a giant contributing factor.
Personally, I felt the cartoon helped and presented a new--even if not very mathematical--way of looking at the problem, and that's a plus. This article is starting to read and more like a textbook and that's a terrible thing. While a mathematician whose life around numbers may prefer to have this article read like a doctoral thesis, nevertheless this is still an encyclopedia that's generally accessible--and at some point an article is no longer informative if it's too boring/dreary/sets too high a bar to be read. Of course the goal is not to get it so dumbed down so any doofus who couldn't pass algebra would read this article and magically get a better understanding the problem just because "oooh there are pictures" (and that particular doofus probably wouldn't want to read this article anyway)...but this article is straying down a very grey path. Do we really want this to become like e.g. the article for the Poisson Bracket. Nothing against Poisoon Brackets but, there's an example of an article that won't be featured or get marked as even a "good article" any year soon. Tendancer 17:38, 21 February 2007 (UTC)
I agree entirely with all of the above. As I said earlier, I think it brings variety to the article and illustrates one point quite nicely in a slightly humourous and novel way. I can't for the life of me see any genuine reason for not including it.
Davkal
18:42, 21 February 2007 (UTC)
I've chosen one door, my chances are obviously 1 in 3. |
I would rather be able to open 2 doors... etc... |
I'm all for having a quick little synopsis of the solution offset from the main text. However, many good points are made above against the image presentation. Personally, I don't like the image's text itself. The explanation poor, the logic unsound, and in its current format it can't be iteratively revised wiki-style. How about using a simple colored box? See demo to the right. (My colors are hideous, but you get the point...) ~ Booya Bazooka 05:18, 21 February 2007 (UTC)
Just curious what do both camps (keep picture/remove picture) think of this suggestion? It appears to be the closest thing we have to something that may reach a consensus and does satisfy arguments from both camps. Suppose Flatscan/Booyabazooka/me adds this to the article, will it placate both sides or will start another revert war (as happened with the last 10 edits on this article)? Tendancer 19:15, 22 February 2007 (UTC)
How many mathematicians does it take to change a lightbulb? None, they don't do jokes. See above! Davkal 20:54, 23 February 2007 (UTC)
I think we should remove the goat picture at the top of the article.
Reasons for removal.
1. It's not funny.
2. It needs five lines of accompanying text, which merely repeats what is said in the article, in order to explain what it is supposed to be about.
3. Chinese people might not recognise the markings on the doors.
4. The wiki server must be creaking under the weight of the storage requirements for such images.
5. The numbers on the doors are too small for some people to read.
6. You can't edit it if, for example, you don't like the colour of the goat or the doors.
7. The goat's facing the wrong way, goats always face to their right (especially on gameshows).
8. It's not as good a picture as the Mona Lisa.
Davkal 15:41, 22 February 2007 (UTC)
I think we should replace it with this
Davkal 16:48, 22 February 2007 (UTC)
The cartoon we're discussing is not my picture. I did the one above as, wait for it, a joke. No jokes in Wiki obviously, not even in the talk pages - all serious grown up stuff here - I do apologise. And, nobody is saying that the cartoon would work without the text, it doesn't need to, it's simply a novel way of explaining a point that many find useful. What's wrong with putting the text into coloured boxes? Nothing really I suppose, but it's not as good as the cartoon. Davkal 20:41, 22 February 2007 (UTC)
Right then, since nobody seems to have any objection I'll remove the goat. Davkal 00:45, 23 February 2007 (UTC)
-- Niels Ø (noe) 07:20, 23 February 2007 (UTC)
The reasons in the list above are almost identical to the reasons given for removing the cartoon. What I fail to see, is how the first goat picture differs in any substantial way from the cartoon. That is, all the reasons that have been suggested for removing the cartoon apply equally (and in some cases more so) to the first goat picture. The question, then, is: what's so different about the first goat picture that means it shouldn't be removed from the article? Davkal 13:03, 23 February 2007 (UTC)
Both are pictorial illustrations of the article (the goat is an illustration of the goat behind door three being revealed) and the cartoon is an illustratioon oif the puzzlement nd thought processes of a contestant who solves the puzzle). Both need copious amounts of text to explain the full point made. The cartoon - as cartoons do - includes that text as part and parcel of the illustration, whereas the goat needs a five line caption to explain which part it is supposed to represent.
Davkal
14:43, 23 February 2007 (UTC)
The point, the main point, against the cartoon was that it merely repeated text from the article and added nothing to the explanation of the point. Exactly the same can be said for the goat+caption. Another complaint was that the cartoon wasn't funny, well the goat didn't exactly have me splitting my sides. It seems now, then, that the only original argument left for removing the cartoon is that it can't be used in foreign versions. As well as being quite proposterous, this can be seen to be no more than a red herring by reference to your final point about standards of writing and presentation. The real reason, then, for exclusion, is that the cartoon is not worthy of inclusion in such a beautifully crafted piece (partly your work I take it). In other words, intellectual elitism and ownership of the article are all that were ever really behind this. Davkal 17:40, 23 February 2007 (UTC)
A picture doesn't need a reason to be a picture. The fact that it is a picture is enough to make it a picture. And the fact that you don't appear to know what a picture is, is sad, startling and frightening in equal measure.
Davkal
23:41, 23 February 2007 (UTC)
It's not my image ya clown Davkal 23:29, 23 February 2007 (UTC)
And weakest has an "a" in it. Thus clown above
Davkal
23:30, 23 February 2007 (UTC)
In case anyone failed to notice, there is an issue with the copyright of Image:MontyHallProblemMadeEasy.jpg (which I have asked the originator to address, see User talk:StuFifeScotland). Pending resolution of this issue, this image must not be put on this or any other page. I suspect this will be seen by some as a bad faith preemption of the above discussion. This is not my intent. Discuss away. But don't put the image back. I also strongly urge everyone commenting about this issue to refrain from further incivilities. -- Rick Block ( talk) 06:02, 24 February 2007 (UTC)
Honestly, f*ck this. If I knew it was going to set off such a drama war I would never have put the image back in the first place. And now that I think more carefully about it, it isn't a persuasive solution at all. The only real solution is to map out the probability tree. Hope you folks had fun. Gazpacho 11:58, 24 February 2007 (UTC)
The reasoning in the text box is fallacious. It has been oversimplified to the point that it is not accurate. As it is currently worded, it makes it seem as if switching is the winning move EVERY time, which it isn't. NipokNek 12:42, 24 February 2007 (UTC)
The reason this problem causes so much confusion is because the current explanation of the problem is incorrect given the need to consider the sample space at the time the decision is made and not at the beginning of the problem. At the time the choice is made it is not the case that there is a 1/3 chance of having chosen each item. There are two options that have not been revealed, and the only reason there is not a 1/2 chance of each one containing the car is because Monty reveals the goat you see more often when you have chose the other goat than he does because you have chosen the car. The explanations given on the article page do not logically necessitate the outcome, and in other scenarios where no signfigant changes are made those approaches would result in a wrong answer. <comments removed by Xiner> —Preceding unsigned comment added by 69.252.158.32 ( talk • contribs)
Yes I agree regarding the Venn Diagram. Using one you can make this argument very simple. I also think that when considering how likely this argument is to convince people who disagree, they are failing to consider that it is an issue of logical fallacy that makes the current explanations confusing and not simplicity. The common probability space used clearly does not represent the state of things at the time the decision is made and their reasoning is self contradicting. That is why they need 10 different explanations and people still have trouble with it.
I have moved here this paragraph, added as an introduction to the "Aids to understanding" section. While I think it may be useful to have some mnemonicals, they should probably come after more thorough explanations. Otherwise, for example, it is not clear at all that the "Bet on it!" of mantra (1) is anything but an arbitrary choice. The Glopk 23:30, 17 March 2007 (UTC)
There are more problems than simply the placement. Addressing the reader is discouraged by the Manual of Style, and the tone is very informal. Perhaps these issues can be fixed, but I'm not sure how. I've reverted another addition of this same text. -- Rick Block ( talk) 03:24, 20 March 2007 (UTC)
The assumes, without explicitly stating, that the car is preferable to the goat. This might be so obvious that it goes without saying, but it may help clarify.-- Blackmagicfish 05:37, 20 March 2007 (UTC)
To the person who reverted my addition to “Other Host Behaviors”: Why was my addition removed for being unreferenced? I do not see any references for the list of other host behaviors, so why is one unreferenced comment preferable to the others. I was simply pointing out that, in all the host behaviors where the host is not forcing a 100% result, switching either gives you a 50/50 chance, or a 2/3 chance. Therefore, at worst switching gives you the same odds as staying, and at best it gives you better odds. How is it original research to point that out? 75.31.42.44 05:38, 23 March 2007 (UTC)
It needs some cleanup, perhaps some rearrangement, and maybe conversion into an .svg, but I think it's a good start.
-- Father Goose 10:14, 23 March 2007 (UTC)
I've cleaned it up and tweaked the layout. It's got various artifacts when you zoom in but I feel it's presentable and I'm happy with the composition. Maybe someone will take it upon themselves to clean it further or convert it to .svg. In it goes. And now, for my next act...-- Father Goose 00:37, 24 March 2007 (UTC)
Player's initial choice | Probability | Host reveals | Outcome if switching | ||||
---|---|---|---|---|---|---|---|
Car | Goat 1 | Goat 2 | 1/3 | Car | Goat 1 | Other goat | |
Car | Goat 1 | Goat 2 | 1/3 | Car | Goat 1 | Car | |
Car | Goat 1 | Goat 2 | 1/3 | Car | Goat 2 | Car |
Player's initial choice | Probability | Host reveals | Outcome if switching | ||||
---|---|---|---|---|---|---|---|
Car | Goat 1 | Goat 2 | 1/3 | Either goat | Car | Goat 1 | Goat 2 |
Car | Goat 1 | Goat 2 | 1/3 | Goat 2 | Car | Goat 1 | Goat 2 |
Car | Goat 1 | Goat 2 | 1/3 | Goat 1 | Car | Goat 1 | Goat 2 |
Well, I've reworked the entire thing. Artwork is up to professional standards now, and the text is removed from the images themselves: the whole thing is now a set of 6 images in a table with editable text captions (except for the "A" and "B" labels next to the goats). As to whether readers are "lazy" or if an illustration with accompanying captions is of use (particularly to laypeople), I do beg to differ.-- Father Goose 04:33, 29 March 2007 (UTC)
Player's initial choice | Probability | Host reveals | Outcome if switching |
---|---|---|---|
1/3 | Either goat | ||
1/3 | Goat 2 | ||
1/3 | Goat 1 |
I've added a "Why the odds are 2/3" section, drawing from material already in the article in scattered places (with some rewriting). I've moved the "forgetful host" variant towards the top of the article in this restructure, as comparing it to the original puzzle brings to light (IMO) exactly where the unexpected odds arise. I'll pen a more detailed justification for this change if objections are raised to it.
Moving some material into it left other material orphaned, namely the somewhat arbitrarily-grouped material found at the top of the "Aids to understanding" section. I've put the two paragraphs which deal with the host's anticipated behavior if the puzzle were a real-world situation into a "Sources of confusion" section, and moved the remaining free-radical paragraph into the "Combining doors" section.
At this point, it's my feeling that the "combining doors" section and "venn diagram" section overlap heavily and should be, er, combined. But that would involve removing some of the redundant material, and in the spirit of no deletion without justification, I'm going to hold off on doing it for now.-- Father Goose 09:44, 29 March 2007 (UTC)
The page includes this sentence. "In the table below, the host's picks from the table above are circled." Am I missing the circles? Lorenzo Traldi 19:10, 1 April 2007 (UTC)
In the aids to understanding, it might be helpful to point out that if the host picks a door at random, you still have a 2/3 chance of winning just not in the same way. If the host picks a door at random, his choice has a 1/3 chance of being the car (if it is you choose to switch obviously) and if it's not you then go with either one of the remaining doors (which one doesn't matter). All in all, you have a 2/3 chance of winning Nil Einne 10:40, 15 April 2007 (UTC)