4 ms·

That actually doesn't matter. What does matter is if he knows he is going to reveal a door that is empty. If he does know you should switch. If he reveals a doo

That actually doesn't matter. What does matter is if he knows he is going to reveal a door that is empty. If he does know you should switch. If he reveals a door arbitrarily, and it just so happens to be empty, it doesn't matter if you switch. Edit: Since I have been downvoted, here is a citation: http://probability.ca/jeff/writing/montyfall.pdf http://probability.ca/jeff/writing/montyfall.pdf (see Monty Fall problem)

maxerickson 11y ago
I agree that my phrasing is not superiorally precise, but the meaning is clear enough, the choice of the door isn't arbitrary.
kgwgk 11y ago
To be fully pedantic, what matters is not "if he knows he is going to reveal a door that is empty". Maybe he has no idea and he just reads his lines off a teleprompter. What matters is if you know that he is going to reveal a door that is empty, because then you can reason according to that information.
baddox 11y ago
The notion that he always randomly opens a door without knowing what's behind it but the door is always a losing door is fairly bizarre and doesn't make much sense when you try to analyze probability in the context of repeated experiments. Something must make him always choose a losing door. Either he knows what's behind the doors, or there's some external force ("fate," "God," or whatever) that causes the door to always be a losing door. Regardless, the problem as stated has Monty Hall open a losing door. Thus it's clear that you should switch after Monty Hall does so, and a very simple computer program can show that you tend to win by switching. Another way to phrase the problem is that you choose one door, then Monty Hall (without revealing anything) gives you the option of taking the door you chose, or taking both the other doors. That's an equivalent problem (assuming the goat has zero utility), and it makes it very obvious that you should switch.
- dllthomas 11y ago
  "Something must make him always choose a losing door." The problem is not always presented in a way that makes it clear that the game always progresses this way, and it's not just a description of the particular circumstance you find yourself in one particular play. Consider: "You have $X, your opponents have $Y and $Z. You select Potent Potables for $400 and it's a daily double. How much should you wager?" I think you'll agree that is going to be read near-universally as a statement of a particular situation that could arise in Jeopardy, not a statement of how Jeopardy games always go.
  - baddox 11y ago
    > The problem is not always presented in a way that makes it clear that the game always progresses this way, and it's not just a description of the particular circumstance you find yourself in one particular play. I'm not sure how that matters. The question is whether, in this scenario, you choose to switch after Monty Hall reveals the door. The terms of the thought experiment dictate that Monty Hall will reveal a losing door. Regardless of what force is actually causing the revealed door to be a losing door, you should switch, because you're essentially being given the option to take your original one door, or to take both of the other doors (at least one of which is a losing door).
    - dllthomas 11y ago
      The terms of the thought experiment are intended to dictate that Monty will always reveal a losing door. However, they are often worded such that a reading that Monty merely has this time revealed a losing door is a valid interpretation. In that case, the problem is under-specified, as we don't know how Monty picks (randomly or always a losing door) and that changes the answer.
      - baddox 11y ago
        That does sound underspecified, although I would think that it still makes sense to switch. If Monty reveals a winning door, it obviously doesn't matter whether you switch (both other doors are losers), but if he reveals a losing door you're better off switching because you're still essentially getting 2 doors instead of the 1 door you originally selected.

maxerickson 11y ago

I agree that my phrasing is not superiorally precise, but the meaning is clear enough, the choice of the door isn't arbitrary.

kgwgk 11y ago

To be fully pedantic, what matters is not "if he knows he is going to reveal a door that is empty". Maybe he has no idea and he just reads his lines off a teleprompter. What matters is if you know that he is going to reveal a door that is empty, because then you can reason according to that information.

baddox 11y ago

The notion that he always randomly opens a door without knowing what's behind it but the door is always a losing door is fairly bizarre and doesn't make much sense when you try to analyze probability in the context of repeated experiments. Something must make him always choose a losing door. Either he knows what's behind the doors, or there's some external force ("fate," "God," or whatever) that causes the door to always be a losing door. Regardless, the problem as stated has Monty Hall open a losing door. Thus it's clear that you should switch after Monty Hall does so, and a very simple computer program can show that you tend to win by switching. Another way to phrase the problem is that you choose one door, then Monty Hall (without revealing anything) gives you the option of taking the door you chose, or taking both the other doors. That's an equivalent problem (assuming the goat has zero utility), and it makes it very obvious that you should switch.

dllthomas 11y ago
"Something must make him always choose a losing door." The problem is not always presented in a way that makes it clear that the game always progresses this way, and it's not just a description of the particular circumstance you find yourself in one particular play. Consider: "You have $X, your opponents have $Y and $Z. You select Potent Potables for $400 and it's a daily double. How much should you wager?" I think you'll agree that is going to be read near-universally as a statement of a particular situation that could arise in Jeopardy, not a statement of how Jeopardy games always go.
- baddox 11y ago
  > The problem is not always presented in a way that makes it clear that the game always progresses this way, and it's not just a description of the particular circumstance you find yourself in one particular play. I'm not sure how that matters. The question is whether, in this scenario, you choose to switch after Monty Hall reveals the door. The terms of the thought experiment dictate that Monty Hall will reveal a losing door. Regardless of what force is actually causing the revealed door to be a losing door, you should switch, because you're essentially being given the option to take your original one door, or to take both of the other doors (at least one of which is a losing door).
  - dllthomas 11y ago
    The terms of the thought experiment are intended to dictate that Monty will always reveal a losing door. However, they are often worded such that a reading that Monty merely has this time revealed a losing door is a valid interpretation. In that case, the problem is under-specified, as we don't know how Monty picks (randomly or always a losing door) and that changes the answer.
    - baddox 11y ago
      That does sound underspecified, although I would think that it still makes sense to switch. If Monty reveals a winning door, it obviously doesn't matter whether you switch (both other doors are losers), but if he reveals a losing door you're better off switching because you're still essentially getting 2 doors instead of the 1 door you originally selected.