4 ms·

> All you know is that in this particular instance the host has opened a door and offered a switch. You cannot conclude that the host always opens a door and of

by the_af 2y ago

> All you know is that in this particular instance the host has opened a door and offered a switch. You cannot conclude that the host always opens a door and offers a switch. And in this particular instance, it makes sense to switch. I'm sorry, but the problem is well-formed and well-specified.

penteract 2y ago
> And in this particular instance, it makes sense to switch. Are you are accepting that the host might be someone who only opens a door with a goat when your first choice was the door with a car behind it, and still arguing that you should switch?
- ryao 2y ago
  The question asks what the best choice in this situation is, not a different situation. The answer does not depend on whether any other situations exist.
  - toast0 2y ago
    We don't really know the situation if all we were told is that we picked a door, and the host showed us a goat behind another door. If we know that the host will always show us a goat behind another door, then yes, we should clearly switch. If the host typically lets us just open the door, but will show us a goat before we open the door if the show is running too fast and they need to kill time, then we should switch if offered. If the host typically lets us open the door, but will show us a goat if the show is running too fast, or if the prize budget is running low and we picked the car, then we should switch if we think the previous games went quickly, but not if there were some slow games already. If the host only shows a goat when the contestent picks the car, then we should never switch. Many problem statements include that the host always shows a goat; and if it doesn't you can kind of assume it, because it's a well-known problem, but if it's a novel problem and unsaid, then how are you supposed to know? I haven't watched enough Let's Make a Deal to know if they always give a second choice. Reading the NYT article linked elsewhere in the thread, I am reminded that Monty Hall could offer cash to not open doors too, so with the problem as stated and Let's Make a Deal being referenced, I have to assume an antagonistic host, unless provided with more information on their behavior. As stated, assuming unknown behavior of the host, we can't put a number on the probability of switching. Also, to address another point you made elsewhere in the thread. In addition to specifying the host behavior, it should also be specified in the problem statement that the car and goat positions were determined randomly, or at least the car was random, and the two goats are considered equal and assigned as convenient.
    - ryao 2y ago
      Here is what Marilyn vos Savant had to say: > So let’s look at it again, remembering that the original answer defines certain conditions, the most significant of which is that the host always opens a losing door on purpose. (There’s no way he can always open a losing door by chance!) Anything else is a different question. https://web.archive.org/web/20130121183432/http://marilynvossavant.com/game-show-problem/ https://web.archive.org/web/20130121183432/http://marilynvos... What you are discussing is a different problem.
      - toast0 2y ago
        Sure, the answer to the question states that the host always opens a losing door. The question does not state that. It's an assumption that was made in the answering. If that's a premise, then yes, always switch. If you only go by the question, there's not enough information.
        ryao 2y ago
        That was how everyone interpreted the original column (according to thousands of letters sent to Marilyn vos Savant), and it was made explicit in a clarification made in a follow up to the column. What you are discussing is a different problem.
alexey-salmin 2y ago
No, not in this version of it. A=you picked the car at first B=the host opened the door P(A|B) can be anywhere between 0 and 1. When you say "it makes sense to switch" you assume that P(A|B)=P(A) which is correct only if A and B are independent. Their independence is not given in the problem statement.
- the_af 2y ago
  That they are independent is the only reasonable assumption, everything else is going out of your way to complicate the problem. If they are NOT independent because Monty only shows you a goat behind the door if you've picked the wrong (or right) door, this is giving the game away. You don't need to guess, you always know what to pick with 100% certainty based on Monty's algorithm. (Also, the game show doesn't work like this. And the text doesn't mention Monty's motivations, which in standard logic puzzle formulation must mean they are irrelevant, just as the phase of the Moon is also irrelevant and you must not take it into consideration) If Monty picks randomly instead of always a goat, and he shows a car, the game has ended and no probabilities are involved, because you don't get to switch anymore; you've lost. If Monty opens a door and there's a goat, we're within the parameters of the problem as stated (and you should switch!).
  - alexey-salmin 2y ago
    > That they are independent is the only reasonable assumption No, this is not true. From the mathematical viewpoint Monty can have any strategy as long as it satisfies the problem statement. Which is, he DID open the door for whatever reason, the rest is uncertain. This literally what Diaconis means when he says "the strict argument would be that the question cannot be answered without knowing the motivation of the host" -- yes, in the strict sense he is indeed correct. This thread started because ryao stated that Diaconis is wrong [1]. Now even if you try to play the card of "reasonable assumptions" and rule out "boring" strategies because they are "giving the game away" this still won't eliminate all "non-independent" cases. The space of possible probability distributions here is way bigger than your list above. I can come up with an infinite number of "reasonable non-independent" strategies for Monty. For example: 1) He rolls a dice before the game in his dressing room, secretly from the audience. 2) If he gets 6: he will open a door if you guessed incorrectly. If you guessed correctly he won't open the door. 3) If he gets 1-5: he will open a door if you guessed correctly. If you guessed incorrectly he won't open the door. The situation is still the same: you've made your guess, then Monty opened the door with a goat and now you need to figure out whether to switch or not. It matches the problem definition stated above: the door was opened but we don't know why. Let's see your chances if we assume Monty follow the dice approach: event A: you've guessed correctly from the first try event B: Monty opened the door P(A|B): probability that you've guessed correctly given that Monty opened the door -- if it's less than 50% you should switch P(A) = 1/3 P(B) = (1/6)x(2/3) + (5/6)x(1/3) = 7/18 P(AB) = (5/6)x(1/3) = 5/18 P(A|B) = P(AB)/P(B) = 5/7 So, in this case Monty doesn't "give up the game" -- there's still a significant random aspect to it. However in this setup for you it's better for you to stay (5/7 of winning) rather than switch (2/7). [1] https://news.ycombinator.com/item?id=43005371 https://news.ycombinator.com/item?id=43005371 UPD fixed a typo, it's 5/7 not 5/8

by the_af 2y ago

penteract 2y ago

> And in this particular instance, it makes sense to switch. Are you are accepting that the host might be someone who only opens a door with a goat when your first choice was the door with a car behind it, and still arguing that you should switch?

ryao 2y ago
The question asks what the best choice in this situation is, not a different situation. The answer does not depend on whether any other situations exist.
- toast0 2y ago
  We don't really know the situation if all we were told is that we picked a door, and the host showed us a goat behind another door. If we know that the host will always show us a goat behind another door, then yes, we should clearly switch. If the host typically lets us just open the door, but will show us a goat before we open the door if the show is running too fast and they need to kill time, then we should switch if offered. If the host typically lets us open the door, but will show us a goat if the show is running too fast, or if the prize budget is running low and we picked the car, then we should switch if we think the previous games went quickly, but not if there were some slow games already. If the host only shows a goat when the contestent picks the car, then we should never switch. Many problem statements include that the host always shows a goat; and if it doesn't you can kind of assume it, because it's a well-known problem, but if it's a novel problem and unsaid, then how are you supposed to know? I haven't watched enough Let's Make a Deal to know if they always give a second choice. Reading the NYT article linked elsewhere in the thread, I am reminded that Monty Hall could offer cash to not open doors too, so with the problem as stated and Let's Make a Deal being referenced, I have to assume an antagonistic host, unless provided with more information on their behavior. As stated, assuming unknown behavior of the host, we can't put a number on the probability of switching. Also, to address another point you made elsewhere in the thread. In addition to specifying the host behavior, it should also be specified in the problem statement that the car and goat positions were determined randomly, or at least the car was random, and the two goats are considered equal and assigned as convenient.
  - ryao 2y ago
    Here is what Marilyn vos Savant had to say: > So let’s look at it again, remembering that the original answer defines certain conditions, the most significant of which is that the host always opens a losing door on purpose. (There’s no way he can always open a losing door by chance!) Anything else is a different question. https://web.archive.org/web/20130121183432/http://marilynvossavant.com/game-show-problem/ https://web.archive.org/web/20130121183432/http://marilynvos... What you are discussing is a different problem.
    - toast0 2y ago
      Sure, the answer to the question states that the host always opens a losing door. The question does not state that. It's an assumption that was made in the answering. If that's a premise, then yes, always switch. If you only go by the question, there's not enough information.
      - ryao 2y ago
        That was how everyone interpreted the original column (according to thousands of letters sent to Marilyn vos Savant), and it was made explicit in a clarification made in a follow up to the column. What you are discussing is a different problem.

alexey-salmin 2y ago

No, not in this version of it. A=you picked the car at first B=the host opened the door P(A|B) can be anywhere between 0 and 1. When you say "it makes sense to switch" you assume that P(A|B)=P(A) which is correct only if A and B are independent. Their independence is not given in the problem statement.

the_af 2y ago
That they are independent is the only reasonable assumption, everything else is going out of your way to complicate the problem. If they are NOT independent because Monty only shows you a goat behind the door if you've picked the wrong (or right) door, this is giving the game away. You don't need to guess, you always know what to pick with 100% certainty based on Monty's algorithm. (Also, the game show doesn't work like this. And the text doesn't mention Monty's motivations, which in standard logic puzzle formulation must mean they are irrelevant, just as the phase of the Moon is also irrelevant and you must not take it into consideration) If Monty picks randomly instead of always a goat, and he shows a car, the game has ended and no probabilities are involved, because you don't get to switch anymore; you've lost. If Monty opens a door and there's a goat, we're within the parameters of the problem as stated (and you should switch!).
- alexey-salmin 2y ago
  > That they are independent is the only reasonable assumption No, this is not true. From the mathematical viewpoint Monty can have any strategy as long as it satisfies the problem statement. Which is, he DID open the door for whatever reason, the rest is uncertain. This literally what Diaconis means when he says "the strict argument would be that the question cannot be answered without knowing the motivation of the host" -- yes, in the strict sense he is indeed correct. This thread started because ryao stated that Diaconis is wrong [1]. Now even if you try to play the card of "reasonable assumptions" and rule out "boring" strategies because they are "giving the game away" this still won't eliminate all "non-independent" cases. The space of possible probability distributions here is way bigger than your list above. I can come up with an infinite number of "reasonable non-independent" strategies for Monty. For example: 1) He rolls a dice before the game in his dressing room, secretly from the audience. 2) If he gets 6: he will open a door if you guessed incorrectly. If you guessed correctly he won't open the door. 3) If he gets 1-5: he will open a door if you guessed correctly. If you guessed incorrectly he won't open the door. The situation is still the same: you've made your guess, then Monty opened the door with a goat and now you need to figure out whether to switch or not. It matches the problem definition stated above: the door was opened but we don't know why. Let's see your chances if we assume Monty follow the dice approach: event A: you've guessed correctly from the first try event B: Monty opened the door P(A|B): probability that you've guessed correctly given that Monty opened the door -- if it's less than 50% you should switch P(A) = 1/3 P(B) = (1/6)x(2/3) + (5/6)x(1/3) = 7/18 P(AB) = (5/6)x(1/3) = 5/18 P(A|B) = P(AB)/P(B) = 5/7 So, in this case Monty doesn't "give up the game" -- there's still a significant random aspect to it. However in this setup for you it's better for you to stay (5/7 of winning) rather than switch (2/7). [1] https://news.ycombinator.com/item?id=43005371 https://news.ycombinator.com/item?id=43005371 UPD fixed a typo, it's 5/7 not 5/8