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Exactly. It made perfect sense to me once I realized that the host knows where the prize is and thus is giving you information by opening a door.

kangnkodos 7y ago
Yes. If the host's algorithm is to pick a door randomly, then your odds change. In this other scenario, your odds do not improve by switching. If the problem does not state what the host's algorithm is, and you have no knowledge of the host's algorithm (example: you have never seen the TV show before), then the logical action is to assume the host chooses a door to reveal randomly.
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- Kiro 7y ago
  > If the host's algorithm is to pick a door randomly, then your odds change. In this other scenario, your odds do not improve by switching. Isn't it the other way round?
twiceaday 7y ago
I don't think you are correct. The host does not give you any useful information by opening a goat door. Opening a goat door only serves to confuse you into not realizing that the switch-choice is actually between one of the doors and a grouping of the remaining two doors. Opening a goat door in the grouping of the other two only changes the groups appearance, making it look like the single door that you picked. It does not change the functional state of that group in terms of its probability of containing a car.
- 0x5345414e 7y ago
  It doesn't change the state of the group of the two remaining doors, correct (ie. the probability remains at 2/3). But you can't choose two doors, you can only choose one. The host is effectively telling you which door to switch to--the one he didn't pick. The only scenario in which the host doesn't give you any useful information by opening a door is when the host chooses the door at random. Assuming the host didn't choose the door with the prize, your odds are 50:50. Of course all of this hinges on you being aware of the host's decision making paradigm, and that you want a car more than you want a goat.
  - twiceaday 7y ago
    I'm not sure what your point is. Assuming the host always opens a goat door, you can simply restate the problem as being allowed to pick two doors at the same time. You are literally choosing two doors, because if the car was behind the first he would have opened the second, and if the car was behind the second he would have opened the first. It would have looked like you are switching to one other door but you can simply look at it as switching to the other two doors, where the elimination of the dud door can be done before or after.
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elihu 7y ago
You also have to trust the host to behave fairly. For instance, a malicious host might decide to open a door with a goat only when your first choice was correct. If the rules of the game are such that the host must always open a door with a goat regardless of whether your first guess was right or wrong, then it makes sense to switch doors. If not, it might be a trap; probably the best strategy is just to flip a coin.
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kangnkodos 7y ago

Yes. If the host's algorithm is to pick a door randomly, then your odds change. In this other scenario, your odds do not improve by switching. If the problem does not state what the host's algorithm is, and you have no knowledge of the host's algorithm (example: you have never seen the TV show before), then the logical action is to assume the host chooses a door to reveal randomly.

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Kiro 7y ago
> If the host's algorithm is to pick a door randomly, then your odds change. In this other scenario, your odds do not improve by switching. Isn't it the other way round?

twiceaday 7y ago

I don't think you are correct. The host does not give you any useful information by opening a goat door. Opening a goat door only serves to confuse you into not realizing that the switch-choice is actually between one of the doors and a grouping of the remaining two doors. Opening a goat door in the grouping of the other two only changes the groups appearance, making it look like the single door that you picked. It does not change the functional state of that group in terms of its probability of containing a car.

0x5345414e 7y ago
It doesn't change the state of the group of the two remaining doors, correct (ie. the probability remains at 2/3). But you can't choose two doors, you can only choose one. The host is effectively telling you which door to switch to--the one he didn't pick. The only scenario in which the host doesn't give you any useful information by opening a door is when the host chooses the door at random. Assuming the host didn't choose the door with the prize, your odds are 50:50. Of course all of this hinges on you being aware of the host's decision making paradigm, and that you want a car more than you want a goat.
- twiceaday 7y ago
  I'm not sure what your point is. Assuming the host always opens a goat door, you can simply restate the problem as being allowed to pick two doors at the same time. You are literally choosing two doors, because if the car was behind the first he would have opened the second, and if the car was behind the second he would have opened the first. It would have looked like you are switching to one other door but you can simply look at it as switching to the other two doors, where the elimination of the dud door can be done before or after.
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elihu 7y ago

You also have to trust the host to behave fairly. For instance, a malicious host might decide to open a door with a goat only when your first choice was correct. If the rules of the game are such that the host must always open a door with a goat regardless of whether your first guess was right or wrong, then it makes sense to switch doors. If not, it might be a trap; probably the best strategy is just to flip a coin.

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