@talah: But the probabilities are at the heart of this problem. That part is inescapable.
Your initial choice does affect the outcome. If you pick the wrong door (2/3 options, so good chance of that), you are basically forcing the presenter to tell you which door is the good one by not opening it. The other 1/3 of the time, you are letting him do his WIFOM.
Another way to look at it is as follows: Imagine that right as you are about to pick a door magical lightning strikes you, and there are now three near-identical universes. The only difference in each of them is the arrangement of your "select door" neurones, so you pick a different door in each universe. Now suppose that you actually are aware of this arrangement, but nothing else. Only one of those yous picks the right door. The other two don't. But you don't know which one got the right door. Two of the yous need to switch, the other one has to not do so, but they don't know who they are. But since there's three of you, and only one benefits from not switching, it's more reasonable for all 3 to assume they should switch. Then 2 of them win, and only 1 loses.
Monty Hall Bsuinss
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Your initial decision forces Monty to not open the door you picked. This may seem obvious, but it's crucial.
If you pick first, then there is a 2/3 chance that you pick a goat door. All of a sudden, he is forced to open the other goat door; he has no choice in the matter. By picking a goat door, you are basically forcing him to tell you where the car is.
1/3 of the time, you pick the car and he chooses which door to open. Similarly, if Monty opens a door first, he always has a choice. In these cases you technically gain no information about the remaining doors.
Incidentally, imagine if when you pick a door Monty was allowed to open your door, to show you it's a goat. You would know for a fact that you want to 100% switch doors, but not which door to switch to.You don't have ambiguity; you haveoptions. - Mitillos
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