4 ms·

I always thought of this as a bullshit sleight-of-language problem. Realistically, it doesn't make a difference if you switch, because the chance of the prize b

I always thought of this as a bullshit sleight-of-language problem. Realistically, it doesn't make a difference if you switch, because the chance of the prize being behind any door is 50%. The choice doesn't 'carry over' it's probability. You make a decision on one door out of 100. The odds are 1/100. Then all doors are eliminated besides one door and the one you already chose. the probability of the other door is 50%, BUT the probability of choosing the same door is 50% as well.

yayachiken 5y ago
But the probability is not 50%. It is 99% to win if you switch and 1% if you stick with the same door. Imagine you repeat the process multiple times, but the contestant chooses to pick door 1 every time. Would you still assume that staying with door 1 every time would give you better odds, that is you would win 50% of the time if you always pick door 1 out of 100?
- RubyRidgeRandy 5y ago
  Think about it this way: there are two doors as the starting condition, only one has the prize behind it. 50% odds. The host opens the wrong door, so only one remains. The prize MUST be behind the only remaining door. The host gives you the option to select that door or he can start a new game where there is only one door and one prize. Would anyone really argue that there is a difference in outcome here? But the logic I see is that people are saying you should switch because the probability in the first set is 50% even though its literally impossible in this scenario for it to be 50%. ----- I know mathematically it's wrong. But I hate this problem so much. I am convinced is a philosophical / semantical issue.
  - dkjaudyeqooe 5y ago
    Think of it this way: Each door has 1/3 chance of containing the prize, that never changes, so your choice has a 1/3 chance of being right and the unselected doors have a 2/3 chance. Again, these facts will never change, you'd surely agree. So once one of those unselected doors is opened, the remaining door has a 2/3 chance because you now know which of the unselected doors you must chose to capture the 2/3 chance (because the opened door has been revealed to be empty). Those pair of doors still contain the 2/3 chance, you just know which to choose now.
  - yayachiken 5y ago
    The problem formulation doesn't really work for n<3, as you cannot safely open n-2 doors anymore. Of course you get semantical problems then. But even then, I don't see your problem here. The invariant is that the chosen door has probability 1/n to contain the prize, and all other doors together have probability 1-1/n. This still holds for n=2: Both doors have a probability of 50% to contain the prize. Also if you do not believe or grok the mathematics, you can just whip up a program to simulate the problem, and verify the probabilities empirically.
8note 5y ago
The door that's left over went through many trials of not being selected by the host. The one you chose doesn't get those additional tests. If it's bad, that door had 98 chances where it could have been chosen to be opened by the host, and didn't get chosen. It's either a very lucky empty door, or the correct door. You can take this all the way to the limiting case where you pick one out of infinite doors, where the probability of you getting it right is ~0. Then, you're presented with a second door, knowing that there's definitely a prize between the two. You're prior knowledge that there's no chance that it's behind the door you picked doesn't go away because you've been given a second door. Hell, you can go even further, and say you pick between several doors known to be empty, then be presented with a second door and the knowledge that there's a winning door between the two. Your chances looking at the doors starting with two doors is still 50/50, unless you remember that your door is already empty

yayachiken 5y ago

But the probability is not 50%. It is 99% to win if you switch and 1% if you stick with the same door. Imagine you repeat the process multiple times, but the contestant chooses to pick door 1 every time. Would you still assume that staying with door 1 every time would give you better odds, that is you would win 50% of the time if you always pick door 1 out of 100?

RubyRidgeRandy 5y ago
Think about it this way: there are two doors as the starting condition, only one has the prize behind it. 50% odds. The host opens the wrong door, so only one remains. The prize MUST be behind the only remaining door. The host gives you the option to select that door or he can start a new game where there is only one door and one prize. Would anyone really argue that there is a difference in outcome here? But the logic I see is that people are saying you should switch because the probability in the first set is 50% even though its literally impossible in this scenario for it to be 50%. ----- I know mathematically it's wrong. But I hate this problem so much. I am convinced is a philosophical / semantical issue.
- dkjaudyeqooe 5y ago
  Think of it this way: Each door has 1/3 chance of containing the prize, that never changes, so your choice has a 1/3 chance of being right and the unselected doors have a 2/3 chance. Again, these facts will never change, you'd surely agree. So once one of those unselected doors is opened, the remaining door has a 2/3 chance because you now know which of the unselected doors you must chose to capture the 2/3 chance (because the opened door has been revealed to be empty). Those pair of doors still contain the 2/3 chance, you just know which to choose now.
- yayachiken 5y ago
  The problem formulation doesn't really work for n<3, as you cannot safely open n-2 doors anymore. Of course you get semantical problems then. But even then, I don't see your problem here. The invariant is that the chosen door has probability 1/n to contain the prize, and all other doors together have probability 1-1/n. This still holds for n=2: Both doors have a probability of 50% to contain the prize. Also if you do not believe or grok the mathematics, you can just whip up a program to simulate the problem, and verify the probabilities empirically.

8note 5y ago

The door that's left over went through many trials of not being selected by the host. The one you chose doesn't get those additional tests. If it's bad, that door had 98 chances where it could have been chosen to be opened by the host, and didn't get chosen. It's either a very lucky empty door, or the correct door. You can take this all the way to the limiting case where you pick one out of infinite doors, where the probability of you getting it right is ~0. Then, you're presented with a second door, knowing that there's definitely a prize between the two. You're prior knowledge that there's no chance that it's behind the door you picked doesn't go away because you've been given a second door. Hell, you can go even further, and say you pick between several doors known to be empty, then be presented with a second door and the knowledge that there's a winning door between the two. Your chances looking at the doors starting with two doors is still 50/50, unless you remember that your door is already empty