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That's the point though, no? The only way to justify the 50/50 claim made above that I can see is that the host must choose the door to open at random from amon

That's the point though, no? The only way to justify the 50/50 claim made above that I can see is that the host must choose the door to open at random from among all 3 doors. So scenarios do arise where the host opens your door and it is the car, or that the host opens some other unchosen door and it is the car. If you don't allow for those possibilities, then the host really isn't making their choice with equal probability. The 50/50 scenario doesn't make sense.

kgwgk 3y ago
> If you don't allow for those possibilities, then the host really isn't making their choice with equal probability. I don't understand your point. The host can make the choice with equal probability and all those possibilities were initially allowed. Once that a choice was made some possibilities are obviously no longer allowed. > The 50/50 scenario doesn't make sense. This is the original problem: https://web.archive.org/web/20130121183432/http://marilynvossavant.com/game-show-problem/ https://web.archive.org/web/20130121183432/http://marilynvos... Suppose you’re on a game show, and you’re given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say #1, and the host, who knows what’s behind the doors, opens another door, say #3, which has a goat. [It's implicit that his knowledge is used to pick a goat with certainty.] He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors? This is one possible 50/50 scenario: You pick a door, say #1, and the host picks a number between 1 and 3 from a hat, say #3, and opens that door, which has a goat. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors? The answer in the first variant is "yes, switching from #1 to #2 doubles my probability of winning". The answer in the second variant is "it doesn't matter, the probability of winning is the same". The scenarios that didn't happen are irrelevant. We are not been asked to commit to a strategy in advance. We are asked if we want to switch from door #1 to door #2 after we learn that door #3 was not the good one. That depends only the scenarios that still remain possible and their probabilities. A much simpler problem to stress that the question is whether we want to change conditional on what we know - not all the things that were possible a priory: You're given the choice between two evelopes. One contains $1 and the other $10. You open an envelope and find $1. Is it to your advantage to switch to the other envelope? The answer is obviously yes. If you found $10 the answer would obviously be no. But if you have to commit to switching or not before you see what was your first choice the answer would be "it doesn't change anything". Your answer to the question is conditional on what may have happened. You don't have to "allow for other possibilities" that could have happened but you know already that didn’t.
- jjnoakes 3y ago
  The original situation covers all scenarios. Any choice of door you make and all possible doors the host might open. The odds are the same every time. The proposed 50/50 scenario only makes sense for some possible situations - those where the host draws a door at random AND the door isn't the same one I chose AND the door opened isn't the car... once you put that situation on equal footing with the original one (all possible doors chosen and all possible doors opened) then it isn't 50/50 any more. Some of those situations are 100% wins (I chose door 3, host opens door 3, car is behind door 3) and some are 100% losses (I chose door 2, host opens door 1, car is behind door 1). Once you look at all outcomes, 50/50 doesn't apply. It becomes 33/66. If your point is that you can cherry-pick a subset of outcomes and get 50/50 odds then I can't really argue, but I'm also afraid I don't how that's useful.
  - kgwgk 3y ago
    > If your point is that you can cherry-pick a subset of outcomes and get 50/50 odds then I can't really argue, Why do you say “cherry-pick a subset of outcomes”? It’s the subset of outcomes relevant for the question. Consider the simple problem that I proposed: “You're given the choice between two envelopes. One contains $1 and the other $10. You open an envelope and find $1. Is it to your advantage to switch to the other envelope?” Do you agree that the answer is “yes” or would you argue that “once you look at all outcomes” the 100% probability of getting $10 by switching doesn’t apply? > but I'm also afraid I don't how that's useful. It’s useful to answer the following question: “You pick door #1. The host picks a number between 1 and 3 from a hat and opens the corresponding door #3. There is a goat behind the door. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors?” What’s your answer?
    - jjnoakes 3y ago
      That's about as relevant as this: "I claim the odds of flipping a fair coin 100 times in a row and getting heads every time is 50/50, but only in the situation where you happened to flip 99 heads in a row already". Of course it is true, but again, not useful in my opinion.

kgwgk 3y ago

> If you don't allow for those possibilities, then the host really isn't making their choice with equal probability. I don't understand your point. The host can make the choice with equal probability and all those possibilities were initially allowed. Once that a choice was made some possibilities are obviously no longer allowed. > The 50/50 scenario doesn't make sense. This is the original problem: https://web.archive.org/web/20130121183432/http://marilynvossavant.com/game-show-problem/ https://web.archive.org/web/20130121183432/http://marilynvos... Suppose you’re on a game show, and you’re given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say #1, and the host, who knows what’s behind the doors, opens another door, say #3, which has a goat. [It's implicit that his knowledge is used to pick a goat with certainty.] He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors? This is one possible 50/50 scenario: You pick a door, say #1, and the host picks a number between 1 and 3 from a hat, say #3, and opens that door, which has a goat. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors? The answer in the first variant is "yes, switching from #1 to #2 doubles my probability of winning". The answer in the second variant is "it doesn't matter, the probability of winning is the same". The scenarios that didn't happen are irrelevant. We are not been asked to commit to a strategy in advance. We are asked if we want to switch from door #1 to door #2 after we learn that door #3 was not the good one. That depends only the scenarios that still remain possible and their probabilities. A much simpler problem to stress that the question is whether we want to change conditional on what we know - not all the things that were possible a priory: You're given the choice between two evelopes. One contains $1 and the other $10. You open an envelope and find $1. Is it to your advantage to switch to the other envelope? The answer is obviously yes. If you found $10 the answer would obviously be no. But if you have to commit to switching or not before you see what was your first choice the answer would be "it doesn't change anything". Your answer to the question is conditional on what may have happened. You don't have to "allow for other possibilities" that could have happened but you know already that didn’t.

jjnoakes 3y ago
The original situation covers all scenarios. Any choice of door you make and all possible doors the host might open. The odds are the same every time. The proposed 50/50 scenario only makes sense for some possible situations - those where the host draws a door at random AND the door isn't the same one I chose AND the door opened isn't the car... once you put that situation on equal footing with the original one (all possible doors chosen and all possible doors opened) then it isn't 50/50 any more. Some of those situations are 100% wins (I chose door 3, host opens door 3, car is behind door 3) and some are 100% losses (I chose door 2, host opens door 1, car is behind door 1). Once you look at all outcomes, 50/50 doesn't apply. It becomes 33/66. If your point is that you can cherry-pick a subset of outcomes and get 50/50 odds then I can't really argue, but I'm also afraid I don't how that's useful.
- kgwgk 3y ago
  > If your point is that you can cherry-pick a subset of outcomes and get 50/50 odds then I can't really argue, Why do you say “cherry-pick a subset of outcomes”? It’s the subset of outcomes relevant for the question. Consider the simple problem that I proposed: “You're given the choice between two envelopes. One contains $1 and the other $10. You open an envelope and find $1. Is it to your advantage to switch to the other envelope?” Do you agree that the answer is “yes” or would you argue that “once you look at all outcomes” the 100% probability of getting $10 by switching doesn’t apply? > but I'm also afraid I don't how that's useful. It’s useful to answer the following question: “You pick door #1. The host picks a number between 1 and 3 from a hat and opens the corresponding door #3. There is a goat behind the door. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors?” What’s your answer?
  - jjnoakes 3y ago
    That's about as relevant as this: "I claim the odds of flipping a fair coin 100 times in a row and getting heads every time is 50/50, but only in the situation where you happened to flip 99 heads in a row already". Of course it is true, but again, not useful in my opinion.