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The vital part about this problem is that the presenter knows which box is the right one. By opening an empty box, he's giving you extra information. And that m

by mcv 4y ago

The vital part about this problem is that the presenter knows which box is the right one. By opening an empty box, he's giving you extra information. And that means there's 2/3 chance of the remaining box having the prize. If, on the other hand, the presenter didn't know which box is the right one, and just opens one of the other two boxes at random (with a 1/3 chance of opening the box with the prize), then, if the opened box turns out to be empty, the chance of the remaining box being the right one drops to 50%. The difference between these two scenarios becomes obvious if we expand it to 100 boxes: You pick a box, 1% chance of being right. Of the other 99, at least 98 are empty. The presenter knows which, opens the 98 empty boxes, and now there's a 99% chance of the remaining box being the right one. Other scenario: The presenter opens 98 boxes, not knowing which are empty, so he has a 98% chance of opening the box with the prize. On the unlikely chance that all are empty, there prize must have been in either the box you picked, or the remaining box, but we still have no information about which it is, so there's a 50% that you're holding the right box. Of course if you don't know whether the presenter knows, and you don't know if he was lucky that the 98 boxes he opened were all empty, or that he knew, then the situation becomes quite a bit more complicated, but the chance of switching being the best option is going to be larger than 50%. How much? Let's say there's an a priori 50% chance that he knows or doesn't know. If he doesn't know, then opening 98 empty boxes is pretty unlikely, so it's pretty likely that he knows. It's probably possible to calculate those odds, but I'm not going to try that now. And if the presenter's behaviour changes depending on whether you picked the right box or not, for example he uses his knowledge to actively tempt you away from the right box, then all bets are off. Or maybe him opening another box is proof that you've got the right one. Or maybe that's what he wants you to think...

firstlink 4y ago
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.

by mcv 4y ago

firstlink 4y ago

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.