PARADE!
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The wording of the problem as given is indeed ambiguous as it doesn't state any of the 3 necessary conditions for the solution to work
1) Monty must in all cases offer the contestant an opportunity to switch doors
2) Monty must deliberately open a goat door (though this is implied by Monty's knowledge of where the prize is)
3) Given a choice of (goat) doors to open Monty picks one at random.
Without these conditions the contestant's chance of winning by switching can range fro 0% to 100%
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@PalmerEldritch said:
3) Given a choice of (goat) doors to open Monty picks one at random.
This condition is superfluous.@PalmerEldritch said:1) Monty must in all cases offer the contestant an opportunity to switch doors
Given that the question as posed concerns the contestant's best option in this case, this condition is guaranteed to apply.@PalmerEldritch said:2) Monty must deliberately open a goat door (though this is implied by Monty's knowledge of where the prize is)
Given only the assumption that a unique solution exists to the problem as stated, this condition is implied by the fact that the problem description includes information about Monty's knowledge.
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Condition 3) certainly isn't superfluous. The probability the car is behind Door2 given the contestant picked Door1 and the host opened Door3 can be expressed as:
1/(1+p), where p = probablity host opens Door 3 if the car is behind Door1. Only if p = 1/2 (the host picks a goat door at random) is the answer to the problem 2/3. If the host has say, a preference for opening Door 3 whenever possible (so p = 1) then there is no advantage in this case for the contestant to switch from Door 1 to Door 2.
Condition 1) has to be stated in the problem defintiton, otherwise the contestant in this case (and every other case) has no idea what his best option is. Is the host only offering the switch option because he's already picked the car? Impossible to say without this condition being specified.
Condition 2) is as I said implied , though I don't see why you assume that a unique solution exists to the problem as stated
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@PalmerEldritch said:
Is the host only offering the switch option because he's already picked the car?
No. The host is offering the switch option because conditions (1) and (2) apply. Given those, Monty gets an opportunity to choose one of two possible goat doors only when the contestant picks the car door at first; in this case and this case only, the contestant's winning move is to stay, not switch. This remains true regardless of which of the two possible goat doors Monty chooses, so the method of making any such choice is irrelevant, making condition (3) superfluous.
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@ceasar said:
If he knows which door has the car but instead chooses to flip a coin, that's the same as if he didn't know.
The possibilities if he flips a coin are:
1 You choose A, car is in A, Montey opens B. Stay wins.
2 You choose A, car is in A, Montey opens C. Stay wins.
3 You choose A, car is in B, Montey opens B. This did not happen since we know a goat was shown.
4 You choose A, car is in B, Montey opens C. Switch wins.
5 You choose A, car is in C, Montey opens B. Switch wins,
6 You choose A, car is in C, Montey opens C. This did not happen since we know a goat was shown.
(You could repeat for choosing B and C, but they'll be essentially the same.)
The 2/3 trick only works if Montey is forbidden from choosing 3 or 6. and must instead choose 4 or 5. If he was free to choose 3 or 6 and they merely didn't happen, that does nothing to differentiate 1 and 5 or 2 and 4 from the player's perspective.
By the way, this is wrong too. Eliminating the "did not happen" cases because they did not happen means that the probabilty that we have found ourselves in any such case becomes zero and we need to calculate the effect of that on the probabilities of the remaining cases; these don't automatically remain equally likely simply by virtue of having been listed.
The probabilities for cases 1 and 2 must sum to 1/3; those for cases 3 and 4 must sum to 1/3; those for cases 5 and 6 must sum to 1/3. This is because the only thing that distinguishes between the cases within each pair is Monty's reveal; the probability of which pair we're in depends solely on the relationship between the contestant's initial pick and the actual position of the car.
All Monty's reveal can do is distinguish case 1 from case 2, or case 3 from case 4, or case 5 from case 6. It can't change the fact that the probabilities for each pair of cases must sum to 1/3. Therefore, given that we know the probability of cases 3 and 6 are zero (because they demonstrably cannot have happened), we can deduce that the probabilities of cases 4 and 5 are now 1/3 each. We don't actually care about the individual probabilities of cases 1 and 2, because regardless of which of those we're dealing with, stay wins; and since those are the only cases in which stay does win, we only need the sum of their probabilities, which is 1/3. So there is one pair of cases in which stay wins, with an aggregate probability of 1/3; and two degenerate pairs in which switch wins, with probabilities of 1/3 each. The probability that switch wins therefore remains 2/3.
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@flabdablet said:
@ceasar said:
If he knows which door has the car but instead chooses to flip a coin, that's the same as if he didn't know.
The possibilities if he flips a coin are:
By the way, this is wrong too.
Therefore, given that we know the probability of cases 3 and 6 are zero (because they demonstrably cannot have happened), we can deduce that the probabilities of cases 4 and 5 are now 1/3 each.
Not in this case, where I specified that Monty could have picked the door with the car and decided with a coin flip. If we picked A and Monty is about to open B, that could be choice 1, 3, or 5, all equally likely. OK, so Monty opens B and B was a goat. Choice 3 is now eliminated. This does not double the odds of choice 5 and leave the odds of choice 1 the same.
Imagine that instead of 2 goats and a car, there is a car, a goat, and a kitten. Via coin flip, Monty opens a door you didn't pick and shows the goat. What exactly makes the car special vs the kitten? You want it more? But if the kitten was a goat, how could that possibly change the odds of winning the car?
Remember the game show "Deal or no deal"? There are 26 cases, opened one by one, and one case contains a million dollars. If the contestant gets all the way to the end where there are 2 cases left, they are given the opportunity to keep the case they originally picked, or switch with the remaining case. Say a contestant gets all the way to the final case and the million is still in play. Are you claiming that there is a 25/26 chance that switching cases will result in the million being won?
Or imagine there are 3 contestants, and each one picks a different door. After a goat is revealed behind door B, should the contestants who picked A and C BOTH switch with each other? Obviously both of their odds can't be 2/3.
(To differentiate from the problem where the answer IS higher if you switch - in THAT case, the winning item IS special because it is specifically forbidden to prematurely open the door it's behind, causing the host to alter his behavior.)
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@ceasar said:
@flabdablet said:
@ceasar said:
If he knows which door has the car but instead chooses to flip a coin, that's the same as if he didn't know.
By the way, this is wrong too.
Not in this case
Yeah, my bad.
P(switch-wins given goat-revealed) = P(switch-wins and goat-revealed) ÷ P(goat-revealed). I had managed to talk myself out of believing in the relevance of the fact that P(goat-revealed) ≠ 1 for the coin-flip case, via an unexamined assumption that P(switch-wins given goat-revealed) and P(switch-wins and goat-revealed) mean the same thing, which of course they absolutely don't.
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@ceasar said:
The possibilities if he flips a coin are:
But keys are that the pairs of possibilities — (1,2), (3,4) and (5,6) — are equally likely, and yet the individual possibilities are not equally likely. The contestant cannot distinguish between 1 and 2; they appear to be the same (as the contestant doesn't know our labelling). What's more, we know (from the setup) that Monty knows where the car is; it's not just that 3 and 6 don't happen, they cannot happen.- You choose A, car is in A, Montey opens B. Stay wins.
- You choose A, car is in A, Montey opens C. Stay wins.
- You choose A, car is in B, Montey opens B. This did not happen since we know a goat was shown.
- You choose A, car is in B, Montey opens C. Switch wins.
- You choose A, car is in C, Montey opens B. Switch wins,
- You choose A, car is in C, Montey opens C. This did not happen since we know a goat was shown.
But don't take my word for it. Code it up. Set the situation up with agents and virtual goats and so on. It's not really that much code. Run a bunch of tests, and you'll see the probabilities for what they really are.
Had I known it would encourage so many people to go so thoroughly wrong, I'd have mentioned it before!
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@dkf said:
What's more, we know (from the setup) that Monty knows where the car is; it's not just that 3 and 6 don't happen, they cannot happen.
No, this is the "Monty flipped a coin to pick his door and it happened to come up goat" side thread. If you wanted to stay on the 2/3 main line, you shouldn't have switched (it's all goats over here).
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@flabdablet said:
@dkf said:
But with conditional probability you get the same answers. Flipping a coin doesn't change that. (The original 1/3 comes from the probability that the initial guess was correct; Monty's actions might as well be deterministic.)What's more, we know (from the setup) that Monty knows where the car is; it's not just that 3 and 6 don't happen, they cannot happen.
No, this is the "Monty flipped a coin to pick his door and it happened to come up goat" side thread. If you wanted to stay on the 2/3 main line, you shouldn't have switched (it's all goats over here).If people are going to change the rules, you might as well assume that they can move the car around after the choice of whether to switch as well.
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@dkf said:
But with conditional probability you get the same answers.
No, you really don't unless P(monty-reveals-goat) = 1; what you're after is P(switch-wins given monty-reveals-goat), which is P(monty-reveals-goat and switch-wins) ÷ P(monty-reveals-goat). If you interpret the problem statement such that Monty is not obliged to reveal a goat, but merely happens to have done so in this instance, then P(monty-reveals-goat) < 1 and the resulting conditional probability changes accordingly.
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@flabdablet said:
@dkf said:
But with conditional probability you get the same answers.
No, you really don't unless P(monty-reveals-goat) = 1; what you're after is P(switch-wins given monty-reveals-goat), which is P(monty-reveals-goat and switch-wins) ÷ P(monty-reveals-goat). If you interpret the problem statement such that Monty is not obliged to reveal a goat, but merely happens to have done so in this instance, then P(monty-reveals-goat) < 1 and the resulting conditional probability changes accordingly.However, we can guarantee that switch will not win if the game is unwinnable. Therefore, P(!monty-reveals-goat and switch-wins) is 0, making P(monty-reveals-goat and switch-wins) = P(switch-wins).
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Not to be confused with the 'Monty Haul Problem'.
[url=http://castle.chirpingmustard.com][img]http://imgur.com/n4X6yS9.png[/img][/url]
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@dkf said:
@flabdablet said:
@dkf said:
But with conditional probability you get the same answers. Flipping a coin doesn't change that. (The original 1/3 comes from the probability that the initial guess was correct; Monty's actions might as well be deterministic.)What's more, we know (from the setup) that Monty knows where the car is; it's not just that 3 and 6 don't happen, they cannot happen.
No, this is the "Monty flipped a coin to pick his door and it happened to come up goat" side thread. If you wanted to stay on the 2/3 main line, you shouldn't have switched (it's all goats over here).If people are going to change the rules, you might as well assume that they can move the car around after the choice of whether to switch as well.
I didn't change the rules. I pointed out that the rules were ambiguous.
So do you think switching provides a 25/26 chance of winning on Deal Or No Deal? If not, how does that differ?
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@ceasar said:
So do you think switching provides a 25/26 chance of winning on Deal Or No Deal? If not, how does that differ?
Well, there's much more variation in the prizes, the host never picks what to reject, and you can only pick one, not 25. These all change the nature of the game so much that you don't use the same analysis at all (where it is much more about working out how good the alternate offer is with respect to the expected winnings given the remaining prizes).
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@flabdablet said:
Monty gets an opportunity to choose one of two possible goat doors only when the contestant picks the car door at first; in this case and this case only, the contestant's winning move is to stay, not switch. This remains true regardless of which of the two possible goat doors Monty chooses, so the method of making any such choice is irrelevant, making condition (3) superfluous.
Mathematics disagrees with you. The method by which Monty chooses between two goat doors determines the probability of the car being behind Door2, as I explained in my previous post.
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At risk of being painted a hypocrit given my part in the first three pages of this thread:
lol @ u guys
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@PalmerEldritch said:
The method by which Monty chooses between two goat doors determines the probability of the car being behind Door2, as I explained in my previous post.
No it doesn't, since if Monty has two goat doors to choose from, the probability of the car being behind either one of them is exactly zero.
Conditions (1) and (2) are sufficient to establish that Monty must reveal a goat and then offer the contestant a chance to switch doors. Condition (3) specifies Monty's method for choosing between candidate goat doors; but in the only case where Monty actually gets an opportunity to exercise any such method, both candidate doors are completely equivalent losing propositions from the contestant's point of view. Therefore, whether Monty will pick one of them vs. the other is completely irrelevant to the contestant's chance of winning by switching, from which it follows that the method by which Monty makes any such choice is equally irrelevant.
Which is why, if you posit conditions (1) and (2), condition (3) is superfluous. Clear?
If you're going to pick holes in my argument you'd be far better off concentrating on my claim about the necessary truth of condition (1) which was, as I have already said, mistaken.
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@flabdablet said:
@PalmerEldritch said:
The method by which Monty chooses between two goat doors determines the probability of the car being behind Door2, as I explained in my previous post.
No it doesn't, since if Monty has two goat doors to choose from, the probability of the car being behind either one of them is exactly zero.
Conditions (1) and (2) are sufficient to establish that Monty must reveal a goat and then offer the contestant a chance to switch doors. Condition (3) specifies Monty's method for choosing between candidate goat doors; but in the only case where Monty actually gets an opportunity to exercise any such method, both candidate doors are completely equivalent losing propositions from the contestant's point of view. Therefore, whether Monty will pick one of them vs. the other is completely irrelevant to the contestant's chance of winning by switching, from which it follows that the method by which Monty makes any such choice is equally irrelevant.
Which is why, if you posit conditions (1) and (2), condition (3) is superfluous. Clear?
I understand what you're saying but your analysis isn't complete. In the problem as stated the probability switching to Door2 wins can be calculated using Bayes Theory as = 1/(1+p) where p = probability Monty opens Door3 if the car is behind Door1, and unless p = 1/2 (i.e. Monty picks a goat door at random) switching is NOT = 2/3.
Now you might well infer that p=1/2 by symmetry, just as you might infer conditions (1) and (2), but you can't simply say condition (3) is irrelevant.
I'm not trying to pick holes in your argument, I'm just saying the problem as posed is ambiguous, and there are 3 conditions that need to be assumed (if they're not stated) to make the probability of winning by switching doors = 2/3.
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@PalmerEldritch said:
I'm just saying the problem as posed is ambiguous, and there are 3 conditions that need to be assumed (if they're not stated) to make the probability of winning by switching doors = 2/3.
And I'm saying you're wrong, because as soon as you assume your first two conditions, which between them specify unambiguously that Monty must open a door to reveal a goat before giving the contestant the option of switching, then the probability of winning by switching becomes 2/3 without needing your third condition; that is, your third condition is superfluous.It's not that the probability of Monty's picking one goat door over the other is assumed to be 1/2; the probability of Monty's picking one goat door over the other is irrelevant. The contestant has a 1/3 chance of picking the door with the car initially. Monty therefore has a 2/3 chance of getting a forced choice of door (because one of the remaining doors hides the car) in which case switching wins, and a 1/3 chance of getting an irrelevant choice between the two goat doors, in which case switching loses. Bayes is not applicable.
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I'm surprised how many people can still get this problem wrong. And after 5 pages of discussion and with the internet available to do research.
Just to reiterate the problem (this is what wikipedia claims to be the original question):
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
There doesn't seem to be any discussion that your original choice has a 1/3 chance of being correct.
There is one condition that matters and that is, is the host allowed to show a car? The problem states that the host knows where the car is, and yet he chose not to. Seeing as it is a game show, and they're formulaic, I think it's to be expected that the host will never reveal the car before giving the contestant a chance to switch. But it's likely that in retelling the conditions vary. This bit, that the host never shows the car, is important, and probably where a lot of confusion stems from. Still, I think it's something to be expected given everything stated in the original problem (that the host knows where the car is and didn't show it).
If your original choice had a 1/3 chance, then the probability of the car being in either of the other two doors was 2/3. Since I established before (for the purpose of this analysis) that the host always shows a goat and never a car, when he reveals the goat you still have a one in three chance of there being a car behind your pick.
Consider a similar situation in which you are NOT allowed to switch, and Monty just opens one of the other doors to create a bit of tension. You'd still expect to win the car in one of three times. That he revealed a goat in another door doesn't magically make you win the car half the times. In this setup, the show would go: you choose, host shows a goat, and one in three times you win a car. Two in three times, then, the car must be in the door you didn't pick and he didn't open.
If, then, you were allowed to change your original pick to the door he didn't open, you'd get to win the car two out of three times. Which means that switching would actually be better for you (if you prefer cars to goats).
What happens if the host is allowed to show the car, though (and end the game before asking you if you want to switch)? Well, let's look at the example where you can't switch again.
Without the implicit condition, the show would go: you pick a door, host shows a car or a goat, you win that car in one of three shows. You obviously don't win the car in any of the shows when he reveals the car, which work out to one third of the times (never when you pick the car, meaning 0*1/3, and half the times when you didn't pick it, 0.5*2/3 = 1/3). Of the remaining 2/3 of the shows, half the times you win the car and the other half you don't. So in this scenario, that he shows a goat doesn't help you learn anything. It just means you are still in the game. Even if you could, switching wouldn't help you.
So, to conclude, if we take as a given that the host will never show a car, then it is to your benefit to switch, as it changes your chances of winning from 1/3 to 2/3. If the host is allowed to show the car, however, it doesn't matter if you switch or not. Finally, if you don't know if the host is allowed to show the car, you should switch. It can't hurt your chances, and it might improve them.
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@Kian said:
If the host is allowed to show the car, however, it doesn't matter if you switch or not.
That depends on the method the host uses in order to decide whether or not to show the car. In the extreme case where the host only chooses to open a goat door if you've already picked the door with the car, switching is guaranteed to make you lose.@Kian said:Finally, if you don't know if the host is allowed to show the car, you should switch. It can't hurt your chances, and it might improve them.
Should be pretty clear from the above that there are indeed variants of the non-mandatory goat-reveal versions of this game where switching does hurt your chances. So perhaps your best bet is to punch Monty in the head, steal the keys and piss off in the car.
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@ceasar said:
@flabdablet said:
@ceasar said:
If he knows which door has the car but instead chooses to flip a coin, that's the same as if he didn't know.
The possibilities if he flips a coin are:
By the way, this is wrong too.
Therefore, given that we know the probability of cases 3 and 6 are zero (because they demonstrably cannot have happened), we can deduce that the probabilities of cases 4 and 5 are now 1/3 each.
Not in this case, where I specified that Monty could have picked the door with the car and decided with a coin flip. If we picked A and Monty is about to open B, that could be choice 1, 3, or 5, all equally likely. OK, so Monty opens B and B was a goat. Choice 3 is now eliminated. This does not double the odds of choice 5 and leave the odds of choice 1 the same.
Imagine that instead of 2 goats and a car, there is a car, a goat, and a kitten. Via coin flip, Monty opens a door you didn't pick and shows the goat. What exactly makes the car special vs the kitten? You want it more? But if the kitten was a goat, how could that possibly change the odds of winning the car?
Remember the game show "Deal or no deal"? There are 26 cases, opened one by one, and one case contains a million dollars. If the contestant gets all the way to the end where there are 2 cases left, they are given the opportunity to keep the case they originally picked, or switch with the remaining case. Say a contestant gets all the way to the final case and the million is still in play. Are you claiming that there is a 25/26 chance that switching cases will result in the million being won?
Or imagine there are 3 contestants, and each one picks a different door. After a goat is revealed behind door B, should the contestants who picked A and C BOTH switch with each other? Obviously both of their odds can't be 2/3.
(To differentiate from the problem where the answer IS higher if you switch - in THAT case, the winning item IS special because it is specifically forbidden to prematurely open the door it's behind, causing the host to alter his behavior.)
Yes, if you change the premise, then the answers change too. But that's not how the Lets Make a Deal's [url=http://en.wikipedia.org/wiki/Let%27s_Make_a_Deal#Big_Deal]Big Deal[/url] works and the Monty Hall Problem describes the single-trader version of this.
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@flabdablet said:
In the extreme case where the host only chooses to open a goat door if you've already picked the door with the car, switching is guaranteed to make you lose.
Game shows aren't structured like that. It would put the host at risk of physical assault from the contestant.
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@dkf said:
@flabdablet said:
In the extreme case where the host only chooses to open a goat door if you've already picked the door with the car, switching is guaranteed to make you lose.
Game shows aren't structured like that. It would put the host at risk of physical assault from the contestant."You can leave by the stairs, or the window."
"The stairs."
"Would you like to switch?"
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You know what's the real WTF about this problem? The number of apparent computer science professionals who never took basic logic, probability or set theory courses. Yes, it seems counter-intuitive that switches increases the probability of finding the car. That happens a lot, and it's why you were taught, in your very first algebra class, to show your work. Look at it this way:
Chance of the car being in the first pick - 1 out of 3.
Chance of the car being in one of the two doors left over - 2 out of 3.
Chance of Monty removing the goat door if you didn't pick the car - 100% (Here's the only thing that you can even try to argue about, whether Monty will always show the remaining goat door, or will sometimes show the car door. You'd be wrong to argue it, since the whole problem relies on Monty always showing a goat, but you can try)
Number of doors left after Monty shows the goat - 1 out of 1.
Chance of the car being in the one door left - (2 out of 3) * (1 out of 1)The important thing to remember here is that Monty ALWAYS waits until you've already picked a door to reveal the goat and ALWAYS reveals a goat. It's this act that changes the probabilities involved and throws off your intuition, because you're still thinking about the car being in one of 3 random doors, but it's no longer purely random after Monty shows the goat.The difference here with the "deal or no deal" situation is that in deal or no deal, there's a chance of having the "money" suit-case revealed before you get to the last case.
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@Snooder said:
You know what's the real WTF about this problem? The number of apparent computer science professionals who never took basic logic, probability or set theory courses.
I disagree. There have been plenty of people with advanced degrees in those things who had trouble understanding this problem. That said, I agree that CS types not having a handle on those things is a WTF.
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@flabdablet said:
@PalmerEldritch said:
I'm just saying the problem as posed is ambiguous, and there are 3 conditions that need to be assumed (if they're not stated) to make the probability of winning by switching doors = 2/3.
And I'm saying you're wrong, because as soon as you assume your first two conditions, which between them specify unambiguously that Monty must open a door to reveal a goat before giving the contestant the option of switching, then the probability of winning by switching becomes 2/3 without needing your third condition; that is, your third condition is superfluous.It's not that the probability of Monty's picking one goat door over the other is assumed to be 1/2; the probability of Monty's picking one goat door over the other is irrelevant. The contestant has a 1/3 chance of picking the door with the car initially. Monty therefore has a 2/3 chance of getting a forced choice of door (because one of the remaining doors hides the car) in which case switching wins, and a 1/3 chance of getting an irrelevant choice between the two goat doors, in which case switching loses. Bayes is not applicable.
I 'm not wrong. It's a conditional probability problem: "What is the probability the car is behind Door2 given that you picked Door1 and Monty opened Door3?", and Bayes is most certainly applicable (why would there be hundreds of websites demonstating a Bayes solution to the MHP if it wasn't?). If you don't believe me look it up.
The prior probabilities are P(Car Door1) = P(car Door2) = P(car Door3) = 1/3.
Monty opens Door3 so P(car Door3) = 0. so the car is either behind Door1 or Door2 (with equal prior probabilities)
A) What is the probabilitiy the Car is behind Door1 given Monty opens Door3? B) What is the probabilitiy the Car is behind Door2 given Monty opens Door3?
The reason the answer to A) is 1/3 and the answer to B) is 2/3 is because if the car is behind Door2 the probability Monty opens Door3 = 1, whereas if the Car is behind Door1 the probability Monty opens Door3 = 1/2 (since he is just as likely to open Door2). So it is twice as likely he opened Door3 because he had to then because he chose chose to, therefore it's twice as likely the car is behind Door2 than Door1. However, if he opens Door3 with a probability of 1 when the car is behind Door1then the answer to A) and B) are the same and = 1/2.
Play the game 300 times (always picking Door1, and Monty always opens Door3 if possible).
In 100 trials the car is behind Door1 and Monty opens Door3.
In 100 trials the car is behind Door2 and Monty opens Door3.
In 100 trials the car is behind Door3 and Monty opens Door2,
In the 200 trials where Monty opens Door3 you win 100 by switching and 100 by staying (there is no benefit in switching).
Now change the scenario to where Monty opens a goat door at random
In 100 trials the car is behind Door1, Monty opens Door3 50 times and Door2 50 times
In 100 trials the car is behind Door2 and Monty opens Door3.
In 100 trials the car is behind Door3 and Monty opens Door2,
In the 150 trials where Monty opens Door3 you win 100 by switching and 50 by staying
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@PalmerEldritch said:
@flabdablet said:
@PalmerEldritch said:
I'm just saying the problem as posed is ambiguous, and there are 3 conditions that need to be assumed (if they're not stated) to make the probability of winning by switching doors = 2/3.
And I'm saying you're wrong, because as soon as you assume your first two conditions, which between them specify unambiguously that Monty must open a door to reveal a goat before giving the contestant the option of switching, then the probability of winning by switching becomes 2/3 without needing your third condition; that is, your third condition is superfluous.It's not that the probability of Monty's picking one goat door over the other is assumed to be 1/2; the probability of Monty's picking one goat door over the other is irrelevant. The contestant has a 1/3 chance of picking the door with the car initially. Monty therefore has a 2/3 chance of getting a forced choice of door (because one of the remaining doors hides the car) in which case switching wins, and a 1/3 chance of getting an irrelevant choice between the two goat doors, in which case switching loses. Bayes is not applicable.
I 'm not wrong. It's a conditional probability problem: "What is the probability the car is behind Door2 given that you picked Door1 and Monty opened Door3?", and Bayes is most certainly applicable (why would there be hundreds of websites demonstating a Bayes solution to the MHP if it wasn't?). If you don't believe me look it up.
The prior probabilities are P(Car Door1) = P(car Door2) = P(car Door3) = 1/3.
Monty opens Door3 so P(car Door3) = 0. so the car is either behind Door1 or Door2 (with equal prior probabilities)
A) What is the probabilitiy the Car is behind Door1 given Monty opens Door3? B) What is the probabilitiy the Car is behind Door2 given Monty opens Door3?
The reason the answer to A) is 1/3 and the answer to B) is 2/3 is because if the car is behind Door2 the probability Monty opens Door3 = 1, whereas if the Car is behind Door1 the probability Monty opens Door3 = 1/2 (since he is just as likely to open Door2). So it is twice as likely he opened Door3 because he had to then because he chose chose to, therefore it's twice as likely the car is behind Door2 than Door1. However, if he opens Door3 with a probability of 1 when the car is behind Door1then the answer to A) and B) are the same and = 1/2.
Play the game 300 times (always picking Door1, and Monty always opens Door3 if possible).
In 100 trials the car is behind Door1 and Monty opens Door3.
In 100 trials the car is behind Door2 and Monty opens Door3.
In 100 trials the car is behind Door3 and Monty opens Door2,
In the 200 trials where Monty opens Door3 you win 100 by switching and 100 by staying (there is no benefit in switching).
Now change the scenario to where Monty opens a goat door at random
In 100 trials the car is behind Door1, Monty opens Door3 50 times and Door2 50 times
In 100 trials the car is behind Door2 and Monty opens Door3.
In 100 trials the car is behind Door3 and Monty opens Door2,
In the 150 trials where Monty opens Door3 you win 100 by switching and 50 by staying
I'm sorry, what? Let's try this with a table.
You Car Verdict 1 1 Stay 1 2 Switch 1 3 Switch 2 1 Switch 2 2 Stay 2 3 Switch 3 1 Switch 3 2 Switch 3 3 Stay
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@PalmerEldritch said:
The prior probabilities are P(Car Door1) = P(car Door2) = P(car Door3) = 1/3.
Note that having reached this point the prior probabilities are no longer the only information available, and note in particular that the fact of the prior probabilities P(car Door1) and P(car Door2) being equal does not magically upgrade them from 1/3 to 1/2. The conditional probabilties CP(car Door1) and CP(car Door2) do sum to 1 but we have no justification yet for assuming those are equal (and in fact they're not).@PalmerEldritch said:Monty opens Door3 so P(car Door3) = 0. so the car is either behind Door1 or Door2 (with equal prior probabilities)
A) What is the probabilitiy the Car is behind Door1 given Monty opens Door3? B) What is the probabilitiy the Car is behind Door2 given Monty opens Door3?
He can't open Door2 because in the case you're analyzing he's already opened Door3, and he doesn't get to open both.@PalmerEldritch said:The reason the answer to A) is 1/3 and the answer to B) is 2/3 is because if the car is behind Door2 the probability Monty opens Door3 = 1 whereas if the Car is behind Door1 the probability Monty opens Door3 = 1/2 (since he is just as likely to open Door2).
So it is twice as likely he opened Door3 because he had to then because he chose chose to, therefore it's twice as likely the car is behind Door2 than Door1.
Correct apart from the "therefore". Both of those things are true, but the second doesn't follow from the first; both follow from the prior probabilities associated with the contestant's initial choice of door.@PalmerEldritch said:However, if he opens Door3 with a probability of 1 when the car is behind Door1 then the answer to A) and B) are the same and = 1/2.
No, because when the car is behind Door1 the answer to A) is 1 and the answer to B) is 0 regardless of Monty's choice.
I think you're confusing yourself by conflating a mental model where all the doors are labelled by function (Door1 is the one the contestant picks, Door3 is whichever one Monty opens, Door2 is the remaining one) with one where the doors available to Monty have fixed labels (which puts Door2 into play but might also take Door3 out).
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@Ben L. said:
I'm sorry, what? Let's try this with a table.
You Car Verdict 1 1 Stay 1 2 Switch 1 3 Switch 2 1 Switch 2 2 Stay 2 3 Switch 3 1 Switch 3 2 Switch 3 3 Stay I'm not sure what your point is. In the problem as stated, you pick Door1 and Monty opens Door3, and the question is "What is the probsbility the car is behind Door2?" .
That is is not the same question as "What is the probability of winning the car if you always switch?"
The answer to the 1st question depends on the method Monty uses to open Door3 if the car is behind Door1. The answer to the 2nd question doesn't, on average you'll win 2 out of 3 times.
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@PalmerEldritch said:
Play the game 300 times (always picking Door1, and Monty always opens Door3 if possible).
In 100 trials the car is behind Door1 and Monty opens Door3.
In 100 trials the car is behind Door2 and Monty opens Door3.
In 100 trials the car is behind Door3 and Monty opens Door2,
Play the game 300 times (always picking Door1, and Monty always opens Door3 if possible)
In 100 trials the car is behind Door1 and Monty opens Door3. If I switch 50 times and stay 50 times then I win the 50 times I stay
In 100 trials the car is behind Door2 and Monty opens Door3. If I switch 50 times and stay 50 times then I win the 50 times I switch
In 100 trials the car is behind Door3 and Monty opens Door2. If I switch 50 times and stay 50 times then I win the 50 times I switch@PalmerEldritch said:
Now change the scenario to where Monty opens a goat door at random.
In 100 trials the car is behind Door1 and Monty opens Door3.
In 100 trials the car is behind Door2 and Monty opens Door3.
In 100 trials the car is behind Door3 and Monty opens Door2,
Now change the scenario to where Monty opens a goat door at random
In 100 trials the car is behind Door1, Monty opens Door3 50 times and Door2 50 times. If I switch 50 times and stay 50 times then I win the 50 times I stay
In 100 trials the car is behind Door2 and Monty opens Door3. If I switch 50 times and stay 50 times then I win the 50 times I switch
In 100 trials the car is behind Door3 and Monty opens Door2. If I switch 50 times and stay 50 times then I win the 50 times I switchAgain, this example shows that if I switch I win 100 times and if I stay I win 50 times.
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@RTapeLoadingError said:
@PalmerEldritch said:
Play the game 300 times (always picking Door1, and Monty always opens Door3 if possible).
In 100 trials the car is behind Door1 and Monty opens Door3.
In 100 trials the car is behind Door2 and Monty opens Door3.
In 100 trials the car is behind Door3 and Monty opens Door2,
Play the game 300 times (always picking Door1, and Monty always opens Door3 if possible)
In 100 trials the car is behind Door1 and Monty opens Door3. If I switch 50 times and stay 50 times then I win the 50 times I stay
In 100 trials the car is behind Door2 and Monty opens Door3. If I switch 50 times and stay 50 times then I win the 50 times I switch
In 100 trials the car is behind Door3 and Monty opens Door2. If I switch 50 times and stay 50 times then I win the 50 times I switch@PalmerEldritch said:
Now change the scenario to where Monty opens a goat door at random.
In 100 trials the car is behind Door1 and Monty opens Door3.
In 100 trials the car is behind Door2 and Monty opens Door3.
In 100 trials the car is behind Door3 and Monty opens Door2,
Now change the scenario to where Monty opens a goat door at random
In 100 trials the car is behind Door1, Monty opens Door3 50 times and Door2 50 times. If I switch 50 times and stay 50 times then I win the 50 times I stay
In 100 trials the car is behind Door2 and Monty opens Door3. If I switch 50 times and stay 50 times then I win the 50 times I switch
In 100 trials the car is behind Door3 and Monty opens Door2. If I switch 50 times and stay 50 times then I win the 50 times I switchAgain, this example shows that if I switch I win 100 times and if I stay I win 50 times.
Now change the scenario to where Monty opens a goat door at random
In 100 trials the car is behind Door1, Monty opens Door1 33.(3) times, Door2 33.(3) times, and Door3 33.(3) times. WHAT THE FUCK MONTY YOU JUST SHOWED ME THE CAR NOW I CAN'T WIN IT
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@Ben L. said:
Now change the scenario to where Monty opens a goat door at random
In 100 trials the car is behind Door1, Monty opens Door1 33.(3) times, Door2 33.(3) times, and Door3 33.(3) times. WHAT THE FUCK MONTY YOU JUST SHOWED ME THE CAR NOW I CAN'T WIN ITThen there's the scenario where the goat and car are both behind door 2. You win the car 50% of the time but there's a 1 in 3 chance the goat shit in it.
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In reply to 'RTapeLoadingError'
What's your point? In the problem as stated Monty opens Door3, he doesn't open Door2.
In those caes where he opens Door3 you win 50% of the time by switching and 50% by staying (unless he picks a goat door at random when he has a choice, whcih is what I've been saying all along is a necessary condition in order to get an answer of 2/3 for switching to the problem as stated)
In the ignorant Monty version where Monty opens a door at random, in those cases where he opens a goat door you win 50% of the time by switching and 50% of the time by staying.
There's no difference in the analysis really, if Monty opens Door2 in the 1st example you win 100% of the time by switching (and overall switching wins 2 times out of 3), in the ignorant Monty version if Monty opens the car door you win 0% of the by switching.(and overall switching wins 1 time out of 3).
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@PalmerEldritch said:
There's no difference in the analysis really because I got them both wrong as seen above
FTFY
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@Ben L. said:
@PalmerEldritch said:
There's no difference in the analysis really because I got them both wrong as seen above
FTFYI got nothing wrong. Do you dispute the fact that in the ignorant Monty version if Monty opens a goat door it's 50/50?
Go and learn probability theory, have a look at Bayes Theory and see how it applies to the MHP , it's not that complicated.
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@PalmerEldritch said:
@Ben L. said:
@PalmerEldritch said:
There's no difference in the analysis really because I got them both wrong as seen above
FTFYI got nothing wrong. Do you dispute the fact that in the ignorant Monty version if Monty opens a goat door it's 50/50?
Go and learn probability theory, have a look at Bayes Theory and see how it applies to the MHP , it's not that complicated.
@TFA said:Initially, the car is equally likely behind any of the three doors: the odds on door 1, door 2, and door 3 are 1:1:1. This remains the case after the player has chosen door 1, by independence. According to Bayes' rule, the posterior odds on the location of the car, given the host opens door 3, are equal to the prior odds multiplied by the Bayes factor or likelihood, which is by definition the probability of the new piece of information (host opens door 3) under each of the hypotheses considered (location of the car). Now, since the player initially chose door 1, the chance the host opens door 3 is 50% if the car is behind door 1, 100% if the car is behind door 2, 0% if the car is behind door 3. Thus the Bayes factor consists of the ratios 1/2 : 1 : 0 or equivalently 1 : 2 : 0, while the prior odds were 1 : 1 : 1. Thus the posterior odds become equal to the Bayes factor 1 : 2 : 0. Given the host opened door 3, the probability the car is behind door 3 is zero, and it is twice more likely to be behind door 2 than door 1.
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@Ben L. said:
@PalmerEldritch said:
[/quote]
[quote user="TFA"]Initially, the car is equally likely behind any of the three doors: the odds on door 1, door 2, and door 3 are 1:1:1. This remains the case after the player has chosen door 1, by independence. According to Bayes' rule, the posterior odds on the location of the car, given the host opens door 3, are equal to the prior odds multiplied by the Bayes factor or likelihood, which is by definition the probability of the new piece of information (host opens door 3) under each of the hypotheses considered (location of the car). Now, since the player initially chose door 1, the chance the host opens door 3 is 50% if the car is behind door 1, 100% if the car is behind door 2, 0% if the car is behind door 3. Thus the Bayes factor consists of the ratios 1/2 : 1 : 0 or equivalently 1 : 2 : 0, while the prior odds were 1 : 1 : 1. Thus the posterior odds become equal to the Bayes factor 1 : 2 : 0. Given the host opened door 3, the probability the car is behind door 3 is zero, and it is twice more likely to be behind door 2 than door 1.@Ben L. said:
@PalmerEldritch said:
There's no difference in the analysis really because I got them both wrong as seen above
FTFYI got nothing wrong. Do you dispute the fact that in the ignorant Monty version if Monty opens a goat door it's 50/50?
Go and learn probability theory, have a look at Bayes Theory and see how it applies to the MHP , it's not that complicated.
And the bit highlighted and underlined is what I've beein saying all along is a necessary condition to arrive at a solution of 2/3, and is what 'flabdablet' has been saying all along is irrelevant.
Thank you for proving my point.
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@PalmerEldritch said:
This thread gives me a headache; I can't believe I'm actually joining the argument.In those caes where he opens Door3 you win 50% of the time by switching and 50% by staying (unless he picks a goat door at random when he has a choice, whcih is what I've been saying all along is a necessary condition in order to get an answer of 2/3 for switching to the problem as stated)
I wrote a simple script that implements the problem. The script randomly picks a winning door, and the "contestant" picks a door. Monty then opens a goat door, and the contestant then switches. I tried all of what I think PalmerEldritch's cases are.
- Contestant chooses a random door, and Monty opens a random goat door. Result: in 1E6 iterations, the contestant won 66.6571% of the time.
- Contestant always chooses door 1, and Monty always opens door 3, unless door 3 has the car, in which case he opens door 2. Result: In 1E6 iterations, the contestant won 66.6473% of the time
- Contestant always chooses door 1, and Monty chooses randomly between door 2 and door 3 unless one of them has the car, in which case he opens the only goat door available. Result: In 1E6 iterations, the contestant won 66.7309% of the time.
Clearly, there is no statistically significant difference in the percentages between the cases. Whether Monty prefers to open a specific door when he has a choice, or flips a coin, makes no difference in the outcome.
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@HardwareGeek said:
@PalmerEldritch said:
This thread gives me a headache; I can't believe I'm actually joining the argument.In those caes where he opens Door3 you win 50% of the time by switching and 50% by staying (unless he picks a goat door at random when he has a choice, whcih is what I've been saying all along is a necessary condition in order to get an answer of 2/3 for switching to the problem as stated)
I wrote a simple script that implements the problem. The script randomly picks a winning door, and the "contestant" picks a door. Monty then opens a goat door, and the contestant then switches. I tried all of what I think PalmerEldritch's cases are.
- Contestant chooses a random door, and Monty opens a random goat door. Result: in 1E6 iterations, the contestant won 66.6571% of the time.
- Contestant always chooses door 1, and Monty always opens door 3, unless door 3 has the car, in which case he opens door 2. Result: In 1E6 iterations, the contestant won 66.6473% of the time
- Contestant always chooses door 1, and Monty chooses randomly between door 2 and door 3 unless one of them has the car, in which case he opens the only goat door available. Result: In 1E6 iterations, the contestant won 66.7309% of the time.
Clearly, there is no statistically significant difference in the percentages between the cases. Whether Monty prefers to open a specific door when he has a choice, or flips a coin, makes no difference in the outcome.
In situation 2, only count the cases where Monty opens Door3 (since that is the door opened in the problem posted). You'll find switching is 50%.
I'm not denying that on average you win 66.67% by switching. I'm arguing that in order to arrive at answer of 2/3 to the example in the problem as stated, Monty needs to pick a goat door at random when he has the choice (your situation 1 )
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Okay, you've clearly lost your mind, since those sentences share only the number 0.5 and nothing else.
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@PalmerEldritch said:
In reply to 'RTapeLoadingError'
What's your point
I guess my point is that I don't understand your point. You've described a very specific scenario (i.e. I choose door 1, Monty chooses door 3 if possible, Monty doesn't open door 2) and then noted that you see different probabilities when you perform 50 more door 3 opens in a "stay wins" scenario than if you don't.
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@PalmerEldritch said:
In situation 2, only count the cases where Monty opens Door3 (since that is the door opened in the problem posted). You'll find switching is 50%.
That's not the MHP. One of the reasons this thread is so hard to follow is that people keep arguing about different problems as if they were the MHP. Different problem, different answer. Of course the winning percentage is only 50%, since you've arbitrarily chosen to ignore 1/3 of the potentially winning cases.
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This is what was posted:
"Suppose you're on a game show, and you're given the choice of 3 doors. Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?"
Quite clear: you pick Door1. Monty opens Door3. Is to your advantage to switch to Door2 ? And the answer as I pointed out in my very 1st post is that the question is ambiguous and the solution depends on:
1. Monty has to offer you chance to switch whether you've picked a goat or the car.
2. Monty must reveal a goat door
3. Given a choice of goat doors to open, Monty picks a door at random.
If those 3 conditions are met then the answer is yes, it is to your advantage to switch.
'flabdablet' repeatedly said condition 3 was superfluous. I pointed out that if condition 3 is NOT met then the probability of winning the car by switching to Door2 is NOT = 2/3, and that if in fact Monty has a preference for opening Door3 whenever possible then there is no advantage in switching to Door2. That is correct and mathemativcally proven by Bayes Theory.
I don't know what you're all arguing about. Everything I've posted has been in response the problem as stated above, and everything I''ve posted has been accurate.
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@PalmerEldritch said:
'flabdablet' repeatedly said condition 3 was superfluous.
Not only did I repeatedly say that, I repeatedly explained my reasoning for saying that. You have yet to address any of that reasoning.@PalmerEldritch said:I pointed out that if condition 3 is NOT met then the probability of winning the car by switching to Door2 is NOT = 2/3, and that if in fact Monty has a preference for opening Door3 whenever possible then there is no advantage in switching to Door2.
You have repeatedly claimed that condition 3 is necessary, but you have yet to address any of the critiques of that claim.@PalmerEldritch said:That is correct and mathemativcally proven by Bayes Theory.
Bayes's Theorem gives you a formula for deriving the probability of event A given that event B has occurred, when what you have is an independent probability for event A, an independent probability for event B, and the probability of event B given that event A has occurred. It's not the same thing as the formula for conditional probability. And since positing conditions 1 and 2 means that the probability of Monty revealing a goat is 1, there's not even any need to invoke the conditional probability formula: switching doors becomes logically equivalent to picking the best prize available from two doors you didn't pick originally, doubling your chance of winning. Probabilities associated with Monty's choice of goat door simply don't enter the calculation.
The only time you need to apply conditional probability to work this problem is if you don't posit condition 2. In that case, for a definite solution to exist you do need to specify the probability that the door Monty opens will reveal a car. But even that is not a choice of goat doors; condition 3 is still redundant.
If your dogged insistence that condition 3 is necessary is trolling rather than simple dogmatic unwillingness to admit error, then well done I suppose.
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Condition 3 is necessary as shown by Bayes Theorem. Did you even read the earlier post by 'Ben L' where he quoted the Bayes entry from Wikipedia: (I've repeatred it below):
"Initially, the car is equally likely behind any of the three doors: the odds on door 1, door 2, and door 3 are 1:1:1. This remains the case after the player has chosen door 1, by independence. According to Bayes' rule, the posterior odds on the location of the car, given the host opens door 3, are equal to the prior odds multiplied by the Bayes factor or likelihood, which is by definition the probability of the new piece of information (host opens door 3) under each of the hypotheses considered (location of the car). Now, since the player initially chose door 1, the chance the host opens door 3 is 50% if the car is behind door 1, 100% if the car is behind door 2, 0% if the car is behind door 3. Thus the Bayes factor consists of the ratios 1/2 : 1 : 0 or equivalently 1 : 2 : 0, while the prior odds were 1 : 1 : 1. Thus the posterior odds become equal to the Bayes factor 1 : 2 : 0. Given the host opened door 3, the probability the car is behind door 3 is zero, and it is twice more likely to be behind door 2 than door 1".
The bit you need to concentrate on is the bolded part. If you change that 50% to something else you don't get an answer of 2/3 for switching, if you change it to "the chance the host opens door 3 is 100% if the car is behind door 1" (for example) you get an answer of 1/2 for switching.
The MHP is the classic conditional probability problem, it appears in hundreds of textbooks and papers on probability theory, and you use Bayes Theory to solve conditional probability problems mathematically.
If you Google "Monty Hall Problem Bayes Solution" you get 35,500 hits. How can you possibly say Bayes Theory doesn't apply? Give it up, you're wrong. Go and click on a few of those 35,500 web pages.
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Actually you're both wrong.
Condition 3 should be: "Given a choice of goat doors, it doesn't matter what method is used to choose between them, so long as the player does not know that method".
This is to ensure that if, for example,- Monty always chooses the lowest numbered goat
- Player selects door #3
- Monty opens door #2
If the player knows Monty's method he can deduce that the car MUST be behind door number 1.
If he doesn't know the method, he cannot make that deduction.
- Monty always chooses the lowest numbered goat
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@eViLegion said:
Actually you're both wrong.
Condition 3 should be: "Given a choice of goat doors, it doesn't matter what method is used to choose between them, so long as the player does not know that method".
This is to ensure that if, for example,
1) Monty always chooses the lowest numbered goat
2) Player selects door #3
3) Monty opens door #2
If the player knows Monty's method he can deduce that the car MUST be behind door number 1. If he doesn't know the method, he cannot make that deduction.It does matter what method is used if you want to calculate either:
a) the probability the car is behind Door2 (given you picked Door1 and Monty opened Door3) or
b) the probability the car is behind Door1 (given you picked Door1 and Monty opened Door3)
If the player doesn't know what method is used then he is not worse off by switching , but if he were to accept a betting proposition (say $100 bet wins $70) that the car is behind Door2, on the basis that he estimates he has a 66.67% of winning such a bet, then he'd be a fool if he didn't know what door selection method Monty used.
And the question asked in the original post was "Is it to your advantage to switch your choice?", and the answer is yes only if Monty doesn't have a preference for opening Door3 over Door2 when given the choice.
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But that's the point of this game:
We are NOT an external observer, who knows Monty's mind, trying to calculate any specific probabilities.
We ARE the player, trying to determine if we should stick or switch, [b]with only the information available to us.[/b]
Crucially, this means that we're actually only interested in that information. Therefore, only the appearance of randomness is required for
the player to conclude that switching gives him the 66.7% chance instead of the 33.3% chance.
Monty's "pick between two goat doors" decision process may affect the underlying probabilities, but without imparting any information
about that process, the player will not able to glean anything of use to affect his/her choice. From the players point of view it is apparently
random, so their choice should be made as though it is random.
Only if we play the game multiple times in succession, might the player start to suspect that Monty has not eliminated the doors randomly,
and then they might be able to deduce what the systematic choice being made is, and thus deduce a better tactic than simply sticking or
switching.
Arguably, though, Palmer is more correct than Flab... we DO need some kind of 3rd condition, and we DO need it to appear to the player as random.
![http://imgur.com/n4X6yS9.png[/img][/url]](http://imgur.com/n4X6yS9.png%5B/img%5D%5B/url%5D){kind=link}