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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 prop

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
    What about the envelopes problem? Do you think that claiming that you should switch the envelopes “but only in the situation where you happened to” find $1 in your first pick is not relevant? The question asked is precisely what to do in that particular situation! Knowing that if you switch you get more that if you don’t seems useful… I’m also curious about your answer to the following question, really. “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 do you think that the probabilities would be if you you were to find yourself in that situation? A hint: the probability that the car is behind the open door is zero. What about the others? Don’t you have any answer at all?
    - jjnoakes 3y ago
      I'm not sure why you keep asking the same question. The answer to the second one is here: https://news.ycombinator.com/item?id=36752870 https://news.ycombinator.com/item?id=36752870 and the answer to the first one is basically alluded to here: https://news.ycombinator.com/item?id=36758795 https://news.ycombinator.com/item?id=36758795 In your hypothetical situation, the rules are something like: "Pick a door, then the host will randomly select one of the other doors. If the host doesn't select your door (1/3 odds) AND the host doesn't open a door with a car behind it (2/3 odds) then you can choose to switch or keep your door." The odds of winning _after_ the host already beats 1/3 and 2/3 odds is indeed 50/50, but the odds of playing the game _overall_ is 33/66. The only real way I can see of making this situation analogous to the Monty Hall problem is to consider the odds of the whole game (like Monty Hall does, which is always 66% chance of winning by switching, every time, no matter the path the game takes from start to end). If you want to cherry-pick just the situation where the host ALREADY beat some odds, then you do get 50/50, but it really is a non-sequitur in this thread (or in any thread about Monty Hall).
      - kgwgk 3y ago
        > The odds of winning _after_ the host already beats 1/3 and 2/3 odds is indeed 50/50, but the odds of playing the game _overall_ is 33/66. The problem is not about the game overall anymore than the question about whether you'd like to switch the envelope after you got $1 is about the game overall. The question is whether you want to switch conditional on the situation where you got $1!. That's not cherry-picking - it's considering the exact situation described in the problem. > The only real way I can see of making this situation analogous to the Monty Hall problem is to consider the odds of the whole game The Monty Hall problem is not about "the whole game". The Monty Hall problem is quite explicitely about what would you do in the following situation: 1) you were given the choice of three doors: one car and two goats 2) you picked a door 3) the host opened another door 4) there was a goat behind that door After all those things have happened you're being offered to switch from your initial pick to the remaining door. What is the probability - after all those things have happened - that the car is behind each of the doors? If you don't agree that this describes the problem there is no point in reading further. We can just agree that we have very different understandings of what the problem is about. ----- If you agree with the description above the answer to the question depends your assumptions about how the host's choice in step 3) was done. There are many different assumptions that could be made. If he was avoiding the door you picked and the door with the car - as the original formulation of the problem implies - the probababilities are 1/3 and 2/3 (for the door you picked and the remaining one respectively). If he was completely ignoring the location of the car when he made the choice the probabilities are 1/2 and 1/2 (it doesn't matter if he was avoiding your door or not).

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
  What about the envelopes problem? Do you think that claiming that you should switch the envelopes “but only in the situation where you happened to” find $1 in your first pick is not relevant? The question asked is precisely what to do in that particular situation! Knowing that if you switch you get more that if you don’t seems useful… I’m also curious about your answer to the following question, really. “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 do you think that the probabilities would be if you you were to find yourself in that situation? A hint: the probability that the car is behind the open door is zero. What about the others? Don’t you have any answer at all?
  - jjnoakes 3y ago
    I'm not sure why you keep asking the same question. The answer to the second one is here: https://news.ycombinator.com/item?id=36752870 https://news.ycombinator.com/item?id=36752870 and the answer to the first one is basically alluded to here: https://news.ycombinator.com/item?id=36758795 https://news.ycombinator.com/item?id=36758795 In your hypothetical situation, the rules are something like: "Pick a door, then the host will randomly select one of the other doors. If the host doesn't select your door (1/3 odds) AND the host doesn't open a door with a car behind it (2/3 odds) then you can choose to switch or keep your door." The odds of winning _after_ the host already beats 1/3 and 2/3 odds is indeed 50/50, but the odds of playing the game _overall_ is 33/66. The only real way I can see of making this situation analogous to the Monty Hall problem is to consider the odds of the whole game (like Monty Hall does, which is always 66% chance of winning by switching, every time, no matter the path the game takes from start to end). If you want to cherry-pick just the situation where the host ALREADY beat some odds, then you do get 50/50, but it really is a non-sequitur in this thread (or in any thread about Monty Hall).
    - kgwgk 3y ago
      > The odds of winning _after_ the host already beats 1/3 and 2/3 odds is indeed 50/50, but the odds of playing the game _overall_ is 33/66. The problem is not about the game overall anymore than the question about whether you'd like to switch the envelope after you got $1 is about the game overall. The question is whether you want to switch conditional on the situation where you got $1!. That's not cherry-picking - it's considering the exact situation described in the problem. > The only real way I can see of making this situation analogous to the Monty Hall problem is to consider the odds of the whole game The Monty Hall problem is not about "the whole game". The Monty Hall problem is quite explicitely about what would you do in the following situation: 1) you were given the choice of three doors: one car and two goats 2) you picked a door 3) the host opened another door 4) there was a goat behind that door After all those things have happened you're being offered to switch from your initial pick to the remaining door. What is the probability - after all those things have happened - that the car is behind each of the doors? If you don't agree that this describes the problem there is no point in reading further. We can just agree that we have very different understandings of what the problem is about. ----- If you agree with the description above the answer to the question depends your assumptions about how the host's choice in step 3) was done. There are many different assumptions that could be made. If he was avoiding the door you picked and the door with the car - as the original formulation of the problem implies - the probababilities are 1/3 and 2/3 (for the door you picked and the remaining one respectively). If he was completely ignoring the location of the car when he made the choice the probabilities are 1/2 and 1/2 (it doesn't matter if he was avoiding your door or not).