2 ms·

But that is the situation, there are exactly two doors and one price left.

by codr7 5y ago

ColinWright 5y ago
We are either talking past each other, or you are missing the point. In the Monty Hall problem it's not 50:50. Really. It's not. * There are three door, one conceals a prize; * You choose one door; * The host, knowing where the prize is, opens one of the other doors to reveal a non-prize; * You have the option to stick with your choice, or to switch; * If you stick, your probability of having the prize is 1/3; * If you switch, your probability of having the prize is 2/3. There's enough discussion elsewhere, or you can write your own simulation (but make sure you simulate the right thing, and not something different). As others have pointed out, you need to know in advance that (a) the host will always open another door, (b) the host knows where the prize is located, and (c) the host will never reveal the prize. But given that, the odds really are not 50:50.
- codr7 5y ago
  I am aware, and at one point I accepted the whole thing as is; but the older I get the more I feel like it doesn't make any sense to me, theories and proofs be damned. I just can't see the point in switching; opening the door provides new information, but that information applies equally to both remaining doors. Sorry if that makes people upset, but what am I supposed to do?
  - ColinWright 5y ago
    > opening the door provides new information, but that information applies equally to both remaining doors. This is intuitively obvious, but where it goes wrong. You've chosen door A. Doors B and C remain, at least one hides a goat, and the host knows where everything is. They can always open one of B and C, so the fact that they do so tells you nothing about A. But if they open B it tells you they didn't open C, and vice versa. So when they open a door they are telling you nothing about the door you chose, and something about the other unopened door. > ...what am I supposed to do? You can try writing a simulation and watch it work, but another way to think about it is this. Suppose you choose one door, put your hand on it, then close your eyes. Now you're given the choice of sticking, or switching to both the other doors. Clearly it's to your advantage to switch. Now with your eyes still closed the host opens one of the other doors. You can still stick, or switch. In effect, switching lets you choose both of the other doors, but you can ignore the one the host opened. Does that help?

But that is the situation, there are exactly two doors and one price left.

by codr7 5y ago

But that is the situation, there are exactly two doors and one price left.

We are either talking past each other, or you are missing the point. In the Monty Hall problem it's not 50:50. Really. It's not. * There are three door, one conceals a prize; * You choose one door; * The host, knowing where the prize is, opens one of the other doors to reveal a non-prize; * You have the option to stick with your choice, or to switch; * If you stick, your probability of having the prize is 1/3; * If you switch, your probability of having the prize is 2/3. There's enough discussion elsewhere, or you can write your own simulation (but make sure you simulate the right thing, and not something different). As others have pointed out, you need to know in advance that (a) the host will always open another door, (b) the host knows where the prize is located, and (c) the host will never reveal the prize. But given that, the odds really are not 50:50.

codr7 5y ago
I am aware, and at one point I accepted the whole thing as is; but the older I get the more I feel like it doesn't make any sense to me, theories and proofs be damned. I just can't see the point in switching; opening the door provides new information, but that information applies equally to both remaining doors. Sorry if that makes people upset, but what am I supposed to do?
- ColinWright 5y ago
  > opening the door provides new information, but that information applies equally to both remaining doors. This is intuitively obvious, but where it goes wrong. You've chosen door A. Doors B and C remain, at least one hides a goat, and the host knows where everything is. They can always open one of B and C, so the fact that they do so tells you nothing about A. But if they open B it tells you they didn't open C, and vice versa. So when they open a door they are telling you nothing about the door you chose, and something about the other unopened door. > ...what am I supposed to do? You can try writing a simulation and watch it work, but another way to think about it is this. Suppose you choose one door, put your hand on it, then close your eyes. Now you're given the choice of sticking, or switching to both the other doors. Clearly it's to your advantage to switch. Now with your eyes still closed the host opens one of the other doors. You can still stick, or switch. In effect, switching lets you choose both of the other doors, but you can ignore the one the host opened. Does that help?