Talk:Monty Hall problem/Archive 5

Please note: The conclusions of this article have been confirmed by experiment

There is no need to argue the factual accuracy of the conclusions in this article. The fact that switching improves your probability of winning is mathematically sound and has been confirmed numerous times by experiment. If you find the article's arguments unconvincing, then please feel free to use the space below to discuss improvements.

Archives
Archive 1: July 2004 Archive 2: July 2004 – July 2005 Archive 3: August 2005 – February 2006 Archive 4: March 2006 – November 2006

I've moved the existing talk page to Talk:Monty Hall problem/Archive2, so the edit history is now with the archive page. I've copied back a few recent threads. Older discussions are in Talk:Monty Hall problem/Archive1. Hope this helps, Wile E. Heresiarch 15:28, 28 July 2005 (UTC)

I've done similarly to produce Talk:Monty Hall problem/Archive3. In keeping with Wile E. Heresiarch I moved the page so the edit history is with the archive page, and copied back the current (March 2006) discussions. _└ ^UkPaolo/_talk^┐ 13:04, 10 March 2006 (UTC)

Likewise for Talk:Monty Hall problem/Archive4. Gzkn 06:15, 27 December 2006 (UTC)

Classic Riddle Misdirection

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:

• The player originally picked the door hiding goat number 1. The game host has shown the goat number 2.
• The player originally picked the door hiding goat number 2. The game host has shown the goat number 1.
• The player originally picked the door hiding the car. The game host has shown goat 1.
• The player originally picked the door hiding the car. The game host has shown goat 2.

OR

• The player originally picked a door hiding a goat. The game host has shown the other goat.
• The player originally picked the door hiding the car. The game host has shown either goat.

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)

Quote The original question is a classic case of riddle misdirection, nothing more - true, what is interesting is not the question, but the insistence by intelligent people on the wrong answer. Again and again, people with higher educations within the sciences (if not exactly mathematics) have published their incorrect "solutions", claiming that everyone defending the correct solution are missing something, are stupid, are a disgrace, or whatever. That is an interesting story, similar in character to what is dealt with in another featured article, 0.999....--Niels Ø 12:29, 8 November 2006 (UTC)

Actually, I may have to eat my words. I just created a simulator to prove my point, and seem to be wrong. I thought I had done so before, where I randomized the full scenario, at http://www.endlesspoetry.com/goatcar2.php (with the source at http://www.endlesspoetry.com/goatcar2.phps should anyone care to view it) with 20,000 iterations per run. However, a friend then pointed out that I had merely shown that if the player randomizes his choice to stay with the original door or switch to the remaining one, that the odds then change to 50/50. I modified the routine, so that the player always switched, at http://www.endlesspoetry.com/goatcar3.php (append an "s" for the source), and sure enough the player won 2/3rds of the time. Mvandemar 14:54, 8 November 2006 (UTC)

I have great respect for you for testing your own theory and listening to your friend. You might have already spotted where the error came in in the chain of reasoning you listed above, but in case you haven't, the answer is that you listed four possibilities above, and treated them as if all four were equally probable, when actually two of the possibilities are only half as probable, on the average, as the others.

The scenarios "The player originally picked the door hiding the car. The game host has shown goat 1" and "The player originally picked the door hiding the car. The game host has shown goat 2" can only happen when the precondition "The player originally picked the door hiding the car." is true. Since the player originally picks the door hiding the car only 1/3 of the time, those two possibilities combined must have a total probability of no more than 1/3.

Another way to look at it that might remove some of the misdirection is to realize that we are only interested in the car. It doesn't matter if what's behind the other doors is a goat, a boat, a baboon, a spitoon or nothing at all, so it doesn't matter if we change the problem so that behind every door there is either the car or just nothing at all. When we looked at the problem in terms of a car and two goats, it was easy to get misdirected into thinking that it matters which goat we are shown. But when we look at it in terms of just one car, and nothing else, it's fairly obvious that it doesn't matter which of two "nothings" Monty shows us -- all that matters is whether he is showing us a nothing from a set of two nothings (1/3 of the time) or a nothing from a set of a nothing and a car (2/3 of the time). -- Antaeus Feldspar 22:10, 9 November 2006 (UTC)

The solution in two completely intuitive sentences

1) Your first guess is most likely wrong.

It has a 2 in 3 chance of being wrong.

2) If your first guess is wrong, and you switch, you will always win.

--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.

Well, not really. The fact is that under the given conditions, a strategy of always switching will succeed 2/3 of the time, whether or not the contestant:

• knows the 'hidden' facts of the game, such as that the host must always offer a choice,
• knows the 'open' facts of the game, such as that there is only one car and all the other doors are goats,
• understands the facts he does know about the game.

The contestant may think a different strategy is optimal, or may even be something like a robot that is capable of performing a simple strategy but incapable of 'thinking' about the strategy it performs. But the success of the strategy is still the same. -- Antaeus Feldspar 19:25, 23 November 2006 (UTC)

Wasn't this explaination originally in the article? I thought it was since I think it was the moment when it clicked in my mind. Or perhaps it just occured to me by myself. It is IMHO, one of the easiest ways to think about the problem. Nil Einne 18:08, 9 December 2006 (UTC)

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)

can someone please explain the whole "what if monty doesn't know"

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

Please see #Aids to understanding, above. -- Rick Block (talk) 15:01, 28 November 2006 (UTC)

I think it is severely confusing, actually, because for such an apparently small change, it has a profound effect on the entire basis of the problem.

The change may become clearer if we make it clearer which parts of the problem are random and which are not. Let's replace all random choices with die rolls. Since we're randomizing the method of choosing the doors, we can safely de-randomize the doors themselves; we'll call them A, B and C and specify that the car is always behind door A.

With these changes in place, let's look at the original problem, the one where the host knows what's behind each door. A die is rolled to choose a door for the player: a result of 1 or 2 gives him door A; 3 or 4 gives him door B; 5 or 6 gives him door C. If the player gets a 1 or 2, the host opens door B or door C (and which one it is doesn't matter, of course); if the player gets a 3, 4, 5, or 6, the host opens whichever one of doors B and C the player didn't get, and offers a chance to switch to door A. Out of six faces the die could have come up, four result in a situation where the player would win by switching.

Now let's look at the "forgotten goat" variant, the one where Nonty has forgotten or never knew where the goat was. Now Monty's choice is random, too, so we'll determine it by the same die roll: on an odd number, he picks the door which comes earlier in the alphabet, and on an even number, he picks the later door.

Now, our problem states that Monty, having forgotten which door contains the goat, guesses at the door and luckily guesses correctly because he opens it to reveal a goat. However, what if the die comes up a 3? According to the rules we've set out, a result of 3 means the player chooses door B and Monty chooses the earlier of the two doors left, A. But we are told this didn't happen; we are told that Monty made his random choice after forgetting which door contained the car and the door he chose contained a goat. If the die came up a 3, then the door he chose would have contained the car; by similar logic, the die coming up a 5 would also result in Monty showing the car.

The answer is that even though these are possible results of the die, they are results that we know did not happen this time around (just as if Monty were to open his door and show the car, we would know with 100% certainty that the player did not pick the car this time around.) There are only four die rolls which could have resulted in Monty opening a door at random and getting a goat: 1, 2, 4, and 6. Out of those die rolls, 1 and 2 represent the player's door having the car, while 4 and 6 represent the car being behind the remaining unpicked door. Thus, the chances in this variation, unlike the original problem, are 1 in 2. -- Antaeus Feldspar 17:41, 28 November 2006 (UTC)

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)

Both games DO NOT proceed in exactly the same way because in one Monty's behaviour is guided in a non-car revealing way, whereas in the other it is only by chance that this happens. This means that the probabilities are different. It might be clearer why this affects the odds if you think: what are the chances of Monty revealing a car in the original version (none); and what are the chances he will reveal a car in the new version (1 in 3) - this is what is different and this is what results in different odds after Monty's actions.Davkal 16:09, 29 November 2006 (UTC)

The fact that the second Monty would have to "coincidentally" pick the same box to make the situations come out the same (outwardly) should indicate that in fact the situations aren't the same at all. The first situation involves one random choice out of three (the player picking one of Monty's three doors) and one choice that isn't random (Monty selecting a door that he knows contains a goat), but the second situation involves two random choices (the player picking one of Monty's three doors, and Monty picking one of the two doors the player left behind, which might have the car behind them.)

Here's a breakout of possibilities for the first situation:

1. The player picks the car. Monty opens either of the remaining two goat doors. Winning strategy: STAY
2. The player picks the first goat. Monty opens the door containing the second goat. Winning strategy: SWITCH
3. The player picks the second goat. Monty opens the door containing the first goat. Winning strategy: SWITCH

Now here's the breakout for the second situation:

1. The player picks the car. Monty opens the door containing the first goat. Winning strategy: STAY
2. The player picks the car. Monty opens the door containing the second goat. Winning strategy: STAY
3. The player picks the first goat. Monty opens the door containing the car.
4. The player picks the first goat. Monty opens the door containing the second goat. Winning strategy: SWITCH
5. The player picks the second goat. Monty opens the door containing the car.
6. The player picks the second goat. Monty opens the door containing the second goat. Winning strategy: SWITCH

The player makes a choice between three doors; Monty then makes a choice between two. This means there are six possibilities. However, we are told that when Monty opens the door, it has a goat behind it. This means that any "possibility" that would result in Monty opening a door with the car behind it didn't happen. Out of those which are still possible, half of them have the car behind the player's door. -- Antaeus Feldspar 01:00, 30 November 2006 (UTC)

Hang on, you've just counted your options here and said "there are four valid options and it's half-and-half." You've completely forgotten that your first two options, wherein "the player picks the car," comprise only 1/3 probability out of the entire set of options. Remember that the player chooses first and there is only a 1/3 chance that the player picked the car.

There is no quantitative difference between "Monty knew which door had a goat" and "Monty correctly guessed which door had a goat" as long as he didn't open the player's door. Maelin (Talk | Contribs) 02:23, 30 November 2006 (UTC)

No, because two out of the six intial options is 1/3. The four options that remain to make it 50/50 excludes the two cases where Monty picks the car because were told that hasn't happpened. Davkal 02:37, 30 November 2006 (UTC)

"You've completely forgotten that your first two options, wherein "the player picks the car," comprise only 1/3 probability out of the entire set of options." No, I haven't forgotten that. They do comprise only 1/3 probability out of the entire set of options, but only if by "entire set of options" you mean "all the possible outcomes from the player picking one of three doors and Monty picking one of the two doors left." When you restrict it to "all the possible outcomes from the player picking one of three doors and Monty picking one of the two doors left that result in Monty's door containing a goat," you've eliminated two of the six possible outcomes.

Suppose you roll a die without looking at the result. The chances of a 1 coming up are 1/6, and so are the chances of a 2, so are the chances of a 3, et cetera; you have six possible outcomes and each of them has a 1/6 chance of being what actually happened when you rolled the die. But suppose you ask a friend to look at the die and tell you if it's a 6. "No," he says, "it's not a 6." What does this do to the chance that you rolled a 6? Obviously, that's gone down to nothing, since you know through your friend that it didn't come up as a 6. What does this do to the chances for all the other faces? They can't still each be 1/6, because then they only add up to 5/6 and not to 1. What actually happens is that the chances for the five remaining faces each go up to 1/5. In just the same way, in the variant problem, each possible outcome has a probability of 1/6 until the door Monty picked by chance is shown to contain a goat; this eliminates two of the six "possible" outcomes by showing them not to be possible and leaves each of the remaining outcomes with a probability of 1/4.

Why doesn't the same thing happen in the original problem? Because nowhere in the original problem do we get evidence that eliminates possible outcomes. No matter which door the player chooses, Monty always has a door chosen from the other two which he can open to show a goat. If the player chooses door A and Monty opens door C, it could be because door B contains the car... or door A holds the car and Monty just happened to choose C though he could have as easily chosen B. The fact that he opens a door to reveal a goat doesn't eliminate any outcomes and so it doesn't change any probabilities. -- Antaeus Feldspar 05:44, 30 November 2006 (UTC)

I think the problem here is where we're trying to analyse a probability situation in which we have arbitrarily decided that a random event turned out "just so". I'm not sure the same mathematical analysis is valid here and I'm not convinced about this assumption that when you chop off a branch of a probability tree by saying "it happens to not turn out like that" that you can just redistribute the "lost" probability evenly among the remaining branches. However I'm not so sure of my own reasoning here so I'll wait until someone can clarify this. Maelin (Talk | Contribs) 11:57, 30 November 2006 (UTC)

Well, maybe a visual demonstration will make it clearer why redistributing the probability of a "lost" (disproven) outcome evenly among the remaining possible outcomes is the correct thing to do.

Suppose you want to get a random color from the set of Black and White, but the only randomizer you have produces a random color each time you use it from the set of Black, Red and White. (Let's say it's a special die from a board game, and for some reason you don't have a coin to flip.) You decide that if the die comes up Black or White, you'll accept that as the outcome, and if it comes up Red, you'll simply roll another result.

Now, let's visually depict this. Draw a long, thin box lengthwise across a sheet of paper; this represents the entire probability space. Now divide the box into equal thirds; mark the section on the left as "Black" and the section on the right as "White". These represent the portions of the probability space where "Black" or "White" came up on the first roll.

The section in the middle represents the portion of the probability space where the first roll came up "Red", but it doesn't get marked "Red" because Red is our signal to roll again. So we divide this middle section into three equal portions again, marking the left one "Black" and the right "White". Now we've marked all portions of the probability space where we got an answer on the first or the second roll.

If you keep doing this, I think you'll see the following:

• Each time we color in Black and White sections to represent those colors coming up on the nth roll, the portion of the probability space that is unmarked decreases. Continuing to infinity (to represent rolling, and rolling, and rolling, until we get a non-Red result) would result in the entire probability space being colored in.
• Each time we color in Black and White sections, we add them in precisely equal amounts, so the proportion of Black to White always remains the same: if we continue to infinity, the entire probability space will be split evenly between Black and White.
• This simulates our variant problem: Black represents the cases where the player gets the car on the first try; White represents the cases where the player gets a goat, and Monty opens the door that shows the other goat; Red represents the cases where the player gets a goat and Monty opens the door to show the car. Since our problem specifies that Monty does not open the door to show the car (just as we specified that "Red" is not a valid result) we keep re-randomizing until we get a result that is valid -- which, as we just saw, is equivalent to dividing the probability space among only those results which are valid. -- Antaeus Feldspar 22:13, 30 November 2006 (UTC)

I think I get it now. It's because the contestant who knows that Monty forgot but was lucky can no longer think, "What choice pays off most frequently in this situation?" but instead needs to think. "What are the possible ituations I might be in, and for each one, what is the probability of it being the case in this instance?" Maelin (Talk | Contribs) 08:21, 4 December 2006 (UTC)

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

1. The player picks the car. Monty opens the door containing the second 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)

In the original problem, it doesn't matter which of the goats Monty picks because Monty's knowledge always guarantees it to be a goat. In the variant problem, Monty is picking at random: when the player has picked the car, Monty's random guessing will always pick a goat (since goats are all that are left); when the player has picked a goat, it's only half the time that Monty's random choice will pick a goat. Yes, the chance that the player didn't initially pick the car is twice the chance that he did -- but only half the outcomes where the player didn't pick the car match what we know, that Monty opened a door to reveal a goat. The fact that only half the player-picked-a-goat outcomes are Monty-shows-a-goat outcomes cancels out the fact that they are twice as frequent as player-picked-a-car outcomes. -- Antaeus Feldspar 17:59, 30 November 2006 (UTC)

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)

The extreme case

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

${\frac {1\cdot (1/3)}{1\cdot (1/3)+(1-p)\cdot (2/3)}}={\frac {1}{3-2p}}.$

You can have fun plugging in values of 0, 1/2, and 1 for p. Melchoir 03:24, 30 November 2006 (UTC)

I was initially confused by this, too, but Marylin is right. The way I've always envisaged the Monty Hall problem is about information: does Monty's choice tell you (as the player) anything new? The Venn diagram representation demonstrates that in the original problem the answer is, NO. Monty might as well ask you to take either your first choice or switch to take whatever is behind both the other doors, for all the good his revealing the goat does. Now in this situation where Monty picks from the remaining doors at random, YES, his picking the goat does tell you something new. He is more likely to pick the goat if you picked the car. The fact that he did pick the goat means the conditional probability that you picked the car is higher. As it turns out, for the case where he picks at random it comes out that your chances of winning are 50:50 whether you switch or not. In the 'extreme' case, if Monty is guaranteed to mess up and reveal the car if you picked a goat, then again, YES, you do learn something new, and in this extreme you can be sure that if Monty picked a goat, you have the car and should not switch, just as Melchoir's formula reveals. Ironically, the reasoning that people incorrectly apply to the original problem produces the correct answer to the 'forgetful Monty' version, and vice versa! BJC 16:45, 14 December 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 believe I follow what you're saying, Lorenzo, but in actuality, your "Possible Rule 1" can never play a role in the problem under discussion. Even in the variant problem, where it is clear that Monty could pick the door that has the car behind it, the problem statement clearly states that he doesn't. Which, I freely concede, is a damn confusing thing to grasp: on the one hand he can because the random choices make that possible but on the other hand he can't because the problem statement says he didn't? But that's actually the case. It's not any different from saying "You flip two coins and at least one of them comes up heads; what is the chance that the other one is tails?" Obviously the randomizer (the two coins) offers the possibility of "both coins come up tails" but our problem statement rules out that possibility by telling us that "at least one of them comes up heads".

Now, "Possible Rule 2" is perhaps an easier way to model the probabilities involved in situations like this of "all these outcomes are possible according to our randomizer but only some of them are possible according to the problem statement". It is not necessary for "do-overs" to actually, literally occur (which makes it somewhat unfortunate that someone added "the game ends prematurely one-third of the time" to discussion of the variant problem, as that seems to miss the point that the problem statement states what the host's door reveals and it is never the car.) However, calling a "do-over" every time you try to simulate the problem and get preconditions contrary to the problem statement makes it possible to do some of the traditional forms of analysis such as, well, simulating the problem with rolling dice and flipping coins. -- Antaeus Feldspar 05:49, 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 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." You're confusing two different things. You're confusing what is possible according to the random events and what is possible according to the evidence. The random events do indeed hold the possibility that Monty will open the door and show the car but that does not matter because the evidence says that he did not. If a problem says "You flipped a coin and it came up heads" then anything you say starting with "but if it comes up tails instead" is irrelevant. -- Antaeus Feldspar 17:39, 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)

Deal or No Deal vs. Let's Make a Deal

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)

Well from my knowledge of Deal or No Deal from the Australian version, it's more complicated/different. In Deal or No Deal, if you end up with two cases, one with $0.01 and one with $1,000,000, your choices are to take the bank offer which would be about $500k or assume your cases is the $1 million. If your case has the $0.01, you end up with $0.01. You can't choose the other case as such, you only choose between whatever is in your case and the bank offer. If the host knowingly opens 24/26 cases without the $1 million, it will be a very, very bad idea for you to think your case has 1$ million since there's a 1/26 chance it does whereas as 25/26 chance it only has $0.01. So the chances are very likely you'd end up with $0.01. Far better to accept the bank deal. Of course, if the host does do that, then it doesn't really make sense for the bank to offer you about $500k either Nil Einne 18:17, 9 December 2006 (UTC)

Yeah, I'd take the $500,000 bank offer. In that situation, a 'fair' bank offer would be $1,000,000 * (1/26) + $0.01 (25/26) ~= $38,461.55. That reminds me of something I've wondered about Deal or No Deal: Suppose, in the normal version of the show where the host doesn't help out, that there are three cases left, $0.01, $500,000 and $1,000,000 and the player turns down a reasonable offer, say $450,000. If the player then opens the case containing $500,000, the banker could force the player to get lucky by offering a really low amount, like $0.02. I wonder if there's anything stated in the rules of the show about the offers or if the player is supposed to just trust that the banker wouldn't do such a thing.  :-) 69.141.232.16 21:30, 9 December 2006 (UTC)

Comment about clueless host

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)

In actual fact it does make a difference, please see #can someone please explain the whole "what if monty doesn't know" (above). -- Rick Block (talk) 20:19, 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)

The difference in the clueless host stems from the fact that in the original problem, it's a "clean" probability situation - you can consider all the options and draw out a standard probability tree. In the clueless situation, however, you need to arbitrarily lop branches off. Instead of "Any of these options could have happened", it's "any of these options could have happened, except for these two which, just by luck, didn't. You're restricting yourself to analysing a certain subset of all possible states, a subtree of the total probability tree, in which a particular random event went a specific way. Maelin (Talk | Contribs) 01:01, 20 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)

I'm sorry, Lorenzo, but talking about 'what happens in those situations where Monty opens the door and it shows the car' is only going to confuse people (including yourself) to no benefit, because the problem states that this does not happen. Introducing a case that does not happen and saying "when this happens, the player is offered a chance to switch to the car that has been revealed" has nothing to do with the problem under discussion because it is part of the problem statement that the door opened reveals a goat. -- Antaeus Feldspar 15:33, 20 December 2006 (UTC)

Well it does explain where the 2/3 went.--Lorenzo Traldi 23:51, 20 December 2006 (UTC)

Clarifying that instead of having three results, all of which are possible from our randomizer and all of which are possible according to the evidence, we have six results, all of which are possible from our randomizer but only four of which are possible according to the evidence, is indeed a good thing. But to speculate on what happens "when" those two results which our problem statement makes clear don't happen, happen, is not really clarifying anything. -- Antaeus Feldspar 14:18, 21 December 2006 (UTC)

I may sound stupid, but...

...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 ;)

What you're saying isn't stupid; however, it is incorrect. You are saying that when there are only two doors, there are only two possibilities: either the car is behind the player's door, or it is behind the other unopened door. If the probabilities of these two possibilities are equal, then they must each be 1/2.

However, it is not true that the probability of these two possibilities are equal, because there are twice as many ways to get to "the car is behind the other unopened door" as there are to get to "the car is behind the player's door". If the player picks the first goat (1/3 probability), Monty shows the second goat, and the car is behind the other unopened door. If the player picks the second goat (1/3 probability), Monty shows the first goat, and the car is behind the other unopened door. Only if the player picks the car (1/3 probability) does the other unopened door contain a goat. -- Antaeus Feldspar 20:27, 25 December 2006 (UTC)

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)

The explanation is wrong and contains a logical fallacy and arrives at the right answer by coincidence

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)

I'm not sure I completely understand your objection, but let's try an example case where Monty chooses goat 1 with 100% probability when you choose the car. I think you're saying this means the starting probabilities are:

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

So, if Monty reveals Goat 1 then you have a 50% chance of winning the car by switching and (similarly) if Monty reveals Goat 2 then you have a 0% chance of winning the car by switching. Although I agree these are the correct conditional probabilities, I'm not sure this is a helpful analysis since the goats are indistinguishable. Since you don't know which goat Monty has revealed, I think you need to combine these two probabilities and when you do this you end up needing to look at all of the possibilities (not just those involving the goat that Monty has revealed). -- Rick Block (talk) 19:14, 28 December 2006 (UTC)

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)

It seems to me that a user violating terms of service wouldn't be a reason to eliminate past work done by someone, even when it was the case. But having followed your link it appears that you have made a false accusation - a straw man. A death threat is when someone threatens to kill you. Not when someone argues that you deserve to be killed. It appears it was this writer's intention to cause you to face the reality that repeated use of passive-aggressive, directly aggressive, or any other tactics designed to overcome reason with force only motivates acts in kind and that it is rediculous to encourage behavior like yours yet prevent other acts of force in retaliation. Although death may seem a bit extreme, it is no less moral than acts of deception and other means of forcing people to act according to your views instead of their own. The only difference is that no amends can be made after it has been carried out, and often the prospect of it is enough to force someone to realize that trying to force their ignorant views on others is not worth the associated costs. -K99