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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

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).
    - jjnoakes 3y ago
      > The problem is not about the game overall > The Monty Hall problem is not about "the whole game". The Monty Hall problem is about the whole game. That's the difference, and that's why the 50/50 claim doesn't make much sense here. > The Monty Hall problem is quite explicitely about what would you do in the following situation: [...] After all those things have happened you're being offered to switch from your initial pick to the remaining door. Steps (3) and (4) in the Monty Hall problem are a bit more nuanced - the host opens a door guaranteed to be a goat, because the host knows where the goat is. At this point, when you choose to switch or stay with your door, you have a 1/3 chance of winning if you stay, and a 2/3 chance of winning if you switch. But the key point that you are missing is that this is the same for every single time you play the entire game. It doesn't matter where the car is in step (1), or which door you choose in step (2), or which door the host opens is step (3) - if you always switch doors, no matter what, every single game you double your odds. And in the 50/50 scenario posed, that just doesn't hold. It isn't analogous.

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).
  - jjnoakes 3y ago
    > The problem is not about the game overall > The Monty Hall problem is not about "the whole game". The Monty Hall problem is about the whole game. That's the difference, and that's why the 50/50 claim doesn't make much sense here. > The Monty Hall problem is quite explicitely about what would you do in the following situation: [...] After all those things have happened you're being offered to switch from your initial pick to the remaining door. Steps (3) and (4) in the Monty Hall problem are a bit more nuanced - the host opens a door guaranteed to be a goat, because the host knows where the goat is. At this point, when you choose to switch or stay with your door, you have a 1/3 chance of winning if you stay, and a 2/3 chance of winning if you switch. But the key point that you are missing is that this is the same for every single time you play the entire game. It doesn't matter where the car is in step (1), or which door you choose in step (2), or which door the host opens is step (3) - if you always switch doors, no matter what, every single game you double your odds. And in the 50/50 scenario posed, that just doesn't hold. It isn't analogous.