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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 r

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.

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.