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The problem as stated, in the article you pulled your quote from, puts no limits on how the host decides whether or not to offer a switch: > Imagine that you’r

The problem as stated, in the article you pulled your quote from, puts no limits on how the host decides whether or not to offer a switch: > Imagine that you’re on a television game show and the host presents you with three closed doors. Behind one of them, sits a sparkling, brand-new Lincoln Continental; behind the other two, are smelly old goats. The host implores you to pick a door, and you select door #1. Then, the host, who is well-aware of what’s going on behind the scenes, opens door #3, revealing one of the goats. > “Now,” he says, turning toward you, “do you want to keep door #1, or do you want to switch to door #2?” 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. The problem as stated allows the host to offer switches only when the contestant picked the door with the prize, or only when the moon is gibbous, or only when the tide is going out. Diaconis and Gardner are completely correct to point out that the problem as stated is under specified and that the intent of the host matters.

ryao 2y ago
The problem as stated has the host open an incorrect door and offer the player a chance to change his choice. Inferring that another possible variation might exist does not change the fact that we are discussing the variation that was presented. Both you and Diaconis are wrong.
- CrazyStat 2y ago
  > The problem as stated has the host open an incorrect door and offer the player a chance to change his choice. Correct, in this one particular instance. You cannot conclude from this particular instance that the host always opens the door and offers a change. > Inferring that another possible variation might exist is totally reasonable, while denying the possibility that the host might be able to choose his actions specifically to benefit or screw you over is an unwarranted leap. The problem statement does not put constraints on the host. You cannot solve the problem by assuming that those constraints exist and then attack those like Diaconis who point out that those constraints don’t exist and that the thing that is unconstrained matters.
  - ryao 2y ago
    The problem statement does put constraints on the host, by specifying that the host opened an unselected door with a goat behind it, only to ask if the player wants to change his choice. The answer to the question of whether the player should change the choice is well defined. Other variations are irrelevant since they are different problems. Your argument is equivalent to denying that 2 + 2 = 4 is correct because the author had the option to write something other than a 2 as an operand.
    - alexey-salmin 2y ago
      Nope. The probability theory doesn't work like that. When you argue that 2+2=4 you assume 2 and 2 are known and they are not. A=you picked the car at first B=the host opened the door P(A|B) can be anywhere between 0 and 1. In your calculations you assume that P(A|B)=P(A) which is correct ONLY if A and B are independent. Independence of A and B is not in the problem statement, you invented this clause yourself.
      - ryao 2y ago
        This is an excellent example of what I am saying. 2 + 2 = 4 was already written and you are insisting that it was not. That said, the source material is this: https://web.archive.org/web/20130121183432/http://marilynvossavant.com/game-show-problem/ https://web.archive.org/web/20130121183432/http://marilynvos... The problem is well defined in the source material and what others are interjecting here is another problem.
        alexey-salmin 2y ago
        Where exactly does it state the independence of these two variables in the problem definition in the source material?
  - Nevermark 2y ago
    There is a genuine human language problem here, NOT A MATH PROBLEM, which accounts for two differing but self-consistent views. It is a legitimate difference in views because human language is genuinely ambiguous. "You pick a door, the host opens another door and reveals a goat. Should you switch?" Does this mean you are in one particular situation where the host opened a 2nd door, with a goat? Or does it mean the host always opens a 2nd door with a goat? If the host always opens a 2nd door, showing a goat, you should switch to the third unopened door. If all you know, is this time you picked a door, then the host revealed a goat, you don't know what to do. Maybe this host only opens goat doors after you pick the right door, in order to trick you into switching? In that case switching would be the worst thing you could do. A host with that strategy is a special case, but special cases where a potential general solution (always switch) doesn't work, are all you need to disprove the general solution. It cannot be a general solution if their is even one special case it doesn't work. Most people interpret the problem to mean the host always reveals goats. But if the language isn't clear on that, then you do have a different problem, whose solution is really impossible to optimize for without some more information on general host behavior or strategies. Without that information, all you can do is flip a coin. Or always stay, or always switch. You have no means to improve your odds whatever you do.
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.
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ryao 2y ago

The problem as stated has the host open an incorrect door and offer the player a chance to change his choice. Inferring that another possible variation might exist does not change the fact that we are discussing the variation that was presented. Both you and Diaconis are wrong.

CrazyStat 2y ago
> The problem as stated has the host open an incorrect door and offer the player a chance to change his choice. Correct, in this one particular instance. You cannot conclude from this particular instance that the host always opens the door and offers a change. > Inferring that another possible variation might exist is totally reasonable, while denying the possibility that the host might be able to choose his actions specifically to benefit or screw you over is an unwarranted leap. The problem statement does not put constraints on the host. You cannot solve the problem by assuming that those constraints exist and then attack those like Diaconis who point out that those constraints don’t exist and that the thing that is unconstrained matters.
- ryao 2y ago
  The problem statement does put constraints on the host, by specifying that the host opened an unselected door with a goat behind it, only to ask if the player wants to change his choice. The answer to the question of whether the player should change the choice is well defined. Other variations are irrelevant since they are different problems. Your argument is equivalent to denying that 2 + 2 = 4 is correct because the author had the option to write something other than a 2 as an operand.
  - alexey-salmin 2y ago
    Nope. The probability theory doesn't work like that. When you argue that 2+2=4 you assume 2 and 2 are known and they are not. A=you picked the car at first B=the host opened the door P(A|B) can be anywhere between 0 and 1. In your calculations you assume that P(A|B)=P(A) which is correct ONLY if A and B are independent. Independence of A and B is not in the problem statement, you invented this clause yourself.
    - ryao 2y ago
      This is an excellent example of what I am saying. 2 + 2 = 4 was already written and you are insisting that it was not. That said, the source material is this: https://web.archive.org/web/20130121183432/http://marilynvossavant.com/game-show-problem/ https://web.archive.org/web/20130121183432/http://marilynvos... The problem is well defined in the source material and what others are interjecting here is another problem.
      - alexey-salmin 2y ago
        Where exactly does it state the independence of these two variables in the problem definition in the source material?
- Nevermark 2y ago
  There is a genuine human language problem here, NOT A MATH PROBLEM, which accounts for two differing but self-consistent views. It is a legitimate difference in views because human language is genuinely ambiguous. "You pick a door, the host opens another door and reveals a goat. Should you switch?" Does this mean you are in one particular situation where the host opened a 2nd door, with a goat? Or does it mean the host always opens a 2nd door with a goat? If the host always opens a 2nd door, showing a goat, you should switch to the third unopened door. If all you know, is this time you picked a door, then the host revealed a goat, you don't know what to do. Maybe this host only opens goat doors after you pick the right door, in order to trick you into switching? In that case switching would be the worst thing you could do. A host with that strategy is a special case, but special cases where a potential general solution (always switch) doesn't work, are all you need to disprove the general solution. It cannot be a general solution if their is even one special case it doesn't work. Most people interpret the problem to mean the host always reveals goats. But if the language isn't clear on that, then you do have a different problem, whose solution is really impossible to optimize for without some more information on general host behavior or strategies. Without that information, all you can do is flip a coin. Or always stay, or always switch. You have no means to improve your odds whatever you do.

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.

deleted 2y ago

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