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This alternate problem (commonly known as "Monty Fall") has always infuriated me more than the original. I do not believe it is possible that it matters whether

This alternate problem (commonly known as "Monty Fall") has always infuriated me more than the original. I do not believe it is possible that it matters whether the host "knows" or doesn't. The problem asks for the probability of switching resulting in a win, conditioned on the host doesn't reveal the prize. It does not matter mathematically why the host doesn't reveal the prize (i.e., whether P(host doesn't reveal the prize) < 1 or = 1), the conditional probability and thus the answer will be the same. Arguing Monty Fall is 50/50 is just the fallback position of unrepentant Monty Hall halfers, CMV. ETA: I also believe the 50/50 result hinges upon an implausible interpretation of the problem. It usually goes like this: if the host accidentally opens the prize, then the game is void. Under this interpretation, 50/50 is correct, as can be easily verified through simulation. However this isn't the Monty Fall problem as stated! The problem states that the host does not reveal the prize, not that the game is void if he does. The difference is, in the voided game version, the games where you originally pick the prize (and thus are destined to lose by switching) are never voided, whereas the games where you stand a chance of winning (because you didn't originally pick the prize) are sometimes voided. This unequal voiding probability skews the game against you, reducing your 2/3 win rate to 1/2. But again, this is the wrong game, it is not the one which was actually described. If the host "does not" reveal the prize, then it really doesn't matter why.

mcv 4y ago
> I do not believe it is possible that it matters whether the host "knows" or doesn't. I think it matters a lot, and I suspect this is the issue that Monty Hall nonbelievers (who think the chance is 50% even if the host knows) are struggling with, but from the opposite direction of you. Let's try to work out the probabilities: You randomly pick box (let's call it A), 1/3 chance of being correct. The host randomly randomly opens one of the other two (let's call it B). 1/3 chance he opens the box with the prize, invalidating the game. 2/3 chance he opens an empty box. If he picks an empty box (2/3 chance), each of the other boxes (A and the remaining box, let's call it C) still has 1/2 chance of holding the prize, so 1/2 * 2/3 = 1/3 chance each a priori, but 1/2 after eliminating the 1/3 chance of invalidating the game. So now we've got 1/3 chance of A having the prize, 1/3 of the game being invalid, 1/3 of the third box holding the prize. After eliminating the invalid game, there's still an equal chance of A and C holding the prize. When the host knows which box holds the prize and uses that information to always pick an empty box (which he always can, because there will always be at least one unpicked empty box), there's no chance of invalidating the game, so the remaining box has a 2/3 chance of holding the prize. > The problem states that the host does not reveal the prize, not that the game is void if he does. Fair point, but the host can only reliable pick an empty box if he knows which boxes are empty. If he doesn't, there's always a chance of him accidentally opening the box with the prize. If he opens an empty box by luck, then you dodged that 1/3 chance of invalidating the game. You're now in the changed probability space given that the host didn't accidentally invalidate the game. Each box had an a 1/3 priori chance of being right, but given the 2/3 chance that the host didn't invalidate the game, they now each have a (1/3)/(2/3)=1/2 chance of being the correct box. It is the knowledge of the host that eliminates the chance of an invalid game and puts the 2/3 chance on the remaining box. > If the host "does not" reveal the prize, then it really doesn't matter why. It does. You dodged a bullet, and that has an influence on the remaining probabilities. In the original Monty Hall problem, the host provides extra information, in Monty Fall, he doesn't, but you dodge a bullet. That difference matters.
- firstlink 4y ago
  > Fair point, but the host can only reliable pick an empty box if he knows which boxes are empty You're ignoring the difference by posing a non mathematical objection to the statement of the problem. This is like having a physics problem about frictionless spherical cows and objecting that cows aren't really frictionless and spherical, and giving an answer for a different problem. Sure you might come up with a valid answer for your different problem, but it isn't a valid answer for your original problem. Likewise 1/2 is a valid answer to the invalidated trials version of the problem here, but not a valid answer to the problem as stated, which remains 2/3.
  - EGPRC 4y ago
    The Monty Fall problem asks you about the case when the host has managed to reveal a goat. That only includes a subset of the total attempts that you would start selecting a door, not all. Once a goat is revealed, you could only be in the 1/3 case of when the player's choice is correct, or in the 1/3 case of when the other door randomly left closed is correct, so each represents 1/2 of this subset, not one 1/3 and the other 2/3.
    - firstlink 4y ago
      > That only includes a subset of the total attempts that you would start selecting a door, not all. In the trial invalidation version of the problem, but not in the problem as stated. The problem as stated provides, the host does not reveal a car (or your original door). Let me put it another way. If the question had asked, "Conditional on the host not revealing a car (or your original door), what is the probability of winning if you switch?", then that too would be 1/2. My problem is that this is not what is asked. We don't get to answer an easier or different version of the problem. And the reason it matters is because people start spouting woo about how the host's knowledge or intentions are what mattered, when that isn't true at all. What matters, as you have demonstrated and I have tried to clarify (we aren't really disagreeing), is whether we are exploring the conditional probability in the uniform distribution over a sample space where the host might open another door (1/2), or the probability in the uniform distribution over a sample space where he cannot (2/3). If one denies the difference between these versions of the problem, one ends up in woo-space where the host's knowledge or intentions matter. They don't.
EGPRC 4y ago
You can get that the answer must be 50% once the host just reveals a goat by chance using reductio ad absurdum. Notice that since the host is doing it randomly, it would be the same if you (the contestant) were who made the revelation instead of the host. By the end both are doing it without knowledge so the results shouldn't tend to be different. For example, you could pick door 1 and then decide to reveal door 2. But in this way what you are doing is basically selecting which two doors will remain closed for the second part. I mean, selecting door 1 and then revealing door 2 is the same as picking both door 1 and door 3 at once, and then discarding the other. So, if door 2 results to have a goat, which one do you think is which should have 2/3 probabilities of having the car, door 1 or door 3? Consider that the doors don't "know" if you picked them at once or one after the other. The result would have been the same if you had first declared door 1 as your staying option and number 3 as your switching one, or viceversa. So, neither of them can be more likely than the other.
EGPRC 4y ago
Moreover, to understand why with a random revelation of the goat the chances of each option are 50%, you first need to understand the real reason why they are not 50% in standard Monty Hall problem. It is because when the player has picked a goat door, the host is restricted to reveal specifically which has the only other goat, but when the player has picked the car door, the host is free to reveal any of the other two, we don't know which in advance, they are equally likely for us, because both would have goats in that case. For example, if you select door 1 and he reveals door 2, it was 100% sure that he would have taken #2 if the correct were #3, as he wouldn't have had another option. Instead, we couldn't ensure that he would have opened door 2 in case the correct were yours, as he could have preferred to reveal door 3 in that case (each of them would have had 50% chance of being removed). So, from the times that you start selecting door 1, it tends to happen with twice the frequency that he opens door 2 once the correct is #3 than once the correct is #1. Instead, if the host does not know the locations and his revelation is random, he cannot make that distinction of revealing one door more than the other depending on the prize location, precisely because he does not know where it is. If you pick door 1, you cannot say that he will open specifically door 2 with more frequency when door 3 is the correct than when door 1 is the correct. If he decides to open door 2, that choice is independent of where the car is.

mcv 4y ago

> I do not believe it is possible that it matters whether the host "knows" or doesn't. I think it matters a lot, and I suspect this is the issue that Monty Hall nonbelievers (who think the chance is 50% even if the host knows) are struggling with, but from the opposite direction of you. Let's try to work out the probabilities: You randomly pick box (let's call it A), 1/3 chance of being correct. The host randomly randomly opens one of the other two (let's call it B). 1/3 chance he opens the box with the prize, invalidating the game. 2/3 chance he opens an empty box. If he picks an empty box (2/3 chance), each of the other boxes (A and the remaining box, let's call it C) still has 1/2 chance of holding the prize, so 1/2 * 2/3 = 1/3 chance each a priori, but 1/2 after eliminating the 1/3 chance of invalidating the game. So now we've got 1/3 chance of A having the prize, 1/3 of the game being invalid, 1/3 of the third box holding the prize. After eliminating the invalid game, there's still an equal chance of A and C holding the prize. When the host knows which box holds the prize and uses that information to always pick an empty box (which he always can, because there will always be at least one unpicked empty box), there's no chance of invalidating the game, so the remaining box has a 2/3 chance of holding the prize. > The problem states that the host does not reveal the prize, not that the game is void if he does. Fair point, but the host can only reliable pick an empty box if he knows which boxes are empty. If he doesn't, there's always a chance of him accidentally opening the box with the prize. If he opens an empty box by luck, then you dodged that 1/3 chance of invalidating the game. You're now in the changed probability space given that the host didn't accidentally invalidate the game. Each box had an a 1/3 priori chance of being right, but given the 2/3 chance that the host didn't invalidate the game, they now each have a (1/3)/(2/3)=1/2 chance of being the correct box. It is the knowledge of the host that eliminates the chance of an invalid game and puts the 2/3 chance on the remaining box. > If the host "does not" reveal the prize, then it really doesn't matter why. It does. You dodged a bullet, and that has an influence on the remaining probabilities. In the original Monty Hall problem, the host provides extra information, in Monty Fall, he doesn't, but you dodge a bullet. That difference matters.

firstlink 4y ago
> Fair point, but the host can only reliable pick an empty box if he knows which boxes are empty You're ignoring the difference by posing a non mathematical objection to the statement of the problem. This is like having a physics problem about frictionless spherical cows and objecting that cows aren't really frictionless and spherical, and giving an answer for a different problem. Sure you might come up with a valid answer for your different problem, but it isn't a valid answer for your original problem. Likewise 1/2 is a valid answer to the invalidated trials version of the problem here, but not a valid answer to the problem as stated, which remains 2/3.
- EGPRC 4y ago
  The Monty Fall problem asks you about the case when the host has managed to reveal a goat. That only includes a subset of the total attempts that you would start selecting a door, not all. Once a goat is revealed, you could only be in the 1/3 case of when the player's choice is correct, or in the 1/3 case of when the other door randomly left closed is correct, so each represents 1/2 of this subset, not one 1/3 and the other 2/3.
  - firstlink 4y ago
    > That only includes a subset of the total attempts that you would start selecting a door, not all. In the trial invalidation version of the problem, but not in the problem as stated. The problem as stated provides, the host does not reveal a car (or your original door). Let me put it another way. If the question had asked, "Conditional on the host not revealing a car (or your original door), what is the probability of winning if you switch?", then that too would be 1/2. My problem is that this is not what is asked. We don't get to answer an easier or different version of the problem. And the reason it matters is because people start spouting woo about how the host's knowledge or intentions are what mattered, when that isn't true at all. What matters, as you have demonstrated and I have tried to clarify (we aren't really disagreeing), is whether we are exploring the conditional probability in the uniform distribution over a sample space where the host might open another door (1/2), or the probability in the uniform distribution over a sample space where he cannot (2/3). If one denies the difference between these versions of the problem, one ends up in woo-space where the host's knowledge or intentions matter. They don't.

EGPRC 4y ago

You can get that the answer must be 50% once the host just reveals a goat by chance using reductio ad absurdum. Notice that since the host is doing it randomly, it would be the same if you (the contestant) were who made the revelation instead of the host. By the end both are doing it without knowledge so the results shouldn't tend to be different. For example, you could pick door 1 and then decide to reveal door 2. But in this way what you are doing is basically selecting which two doors will remain closed for the second part. I mean, selecting door 1 and then revealing door 2 is the same as picking both door 1 and door 3 at once, and then discarding the other. So, if door 2 results to have a goat, which one do you think is which should have 2/3 probabilities of having the car, door 1 or door 3? Consider that the doors don't "know" if you picked them at once or one after the other. The result would have been the same if you had first declared door 1 as your staying option and number 3 as your switching one, or viceversa. So, neither of them can be more likely than the other.

EGPRC 4y ago

Moreover, to understand why with a random revelation of the goat the chances of each option are 50%, you first need to understand the real reason why they are not 50% in standard Monty Hall problem. It is because when the player has picked a goat door, the host is restricted to reveal specifically which has the only other goat, but when the player has picked the car door, the host is free to reveal any of the other two, we don't know which in advance, they are equally likely for us, because both would have goats in that case. For example, if you select door 1 and he reveals door 2, it was 100% sure that he would have taken #2 if the correct were #3, as he wouldn't have had another option. Instead, we couldn't ensure that he would have opened door 2 in case the correct were yours, as he could have preferred to reveal door 3 in that case (each of them would have had 50% chance of being removed). So, from the times that you start selecting door 1, it tends to happen with twice the frequency that he opens door 2 once the correct is #3 than once the correct is #1. Instead, if the host does not know the locations and his revelation is random, he cannot make that distinction of revealing one door more than the other depending on the prize location, precisely because he does not know where it is. If you pick door 1, you cannot say that he will open specifically door 2 with more frequency when door 3 is the correct than when door 1 is the correct. If he decides to open door 2, that choice is independent of where the car is.