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Wow - I think that's the most straightforward explanation I've ever heard. I think, to rephrase it, you're saying: a) because a goat is revealed, swapping will

Wow - I think that's the most straightforward explanation I've ever heard. I think, to rephrase it, you're saying: a) because a goat is revealed, swapping will always give you the opposite of what you had b) you're more likely to pick a goat in the first place, so you're more likely to switch to a car

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jonsen 12y ago
The most straightforward explanation I've heard is this: The chance is 1/3 the car is behind the door chosen at first. The chance is 0 for the opened door. As the three chances must sum to 1, the last door has a chance of 2/3.
- dllthomas 12y ago
  That's great for defusing the "there are two options and it's random so it must be 50/50" initial of some, but it doesn't quite work for the "Monty chooses randomly, and happens to have revealed a goat" situation. I think the next point is that, when the host picks randomly, it is then correspondingly less likely that you've found yourself in that situation.
  - jonsen 12y ago
    If the host picks randomly, the chances must sum to 2/3, not 1.
    - dllthomas 12y ago
      Right.
hamburglar 12y ago
The notion of revealing the goat before you make your second choice is a bit of a distraction. It becomes a lot more clear if you realize that in your first choice, you're dividing the doors into two groups: the group you choose has a 1/N chance of having a car in it, and the group you didn't choose has a (N-1)/N chance of having a car in it. Now, imagine that without opening any doors, Monty said: if you're willing to switch to the N-1 group, you win if there's a car in any of them. Do you take it? Of course.
- lukasb 12y ago
  Yeah this is it I think. If I can restate more simply for the 3-door case: You pick. The doors are now in two groups: your door (1/3 chance) and the other two doors (2/3 chance). The host picks. The second group is now reduced to a single door, still with 2/3 chance. Now you decide - stick with the door with 1/3 chance, or switch to the door with 2/3 chance?

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jonsen 12y ago

The most straightforward explanation I've heard is this: The chance is 1/3 the car is behind the door chosen at first. The chance is 0 for the opened door. As the three chances must sum to 1, the last door has a chance of 2/3.

dllthomas 12y ago
That's great for defusing the "there are two options and it's random so it must be 50/50" initial of some, but it doesn't quite work for the "Monty chooses randomly, and happens to have revealed a goat" situation. I think the next point is that, when the host picks randomly, it is then correspondingly less likely that you've found yourself in that situation.
- jonsen 12y ago
  If the host picks randomly, the chances must sum to 2/3, not 1.
  - dllthomas 12y ago
    Right.

hamburglar 12y ago

The notion of revealing the goat before you make your second choice is a bit of a distraction. It becomes a lot more clear if you realize that in your first choice, you're dividing the doors into two groups: the group you choose has a 1/N chance of having a car in it, and the group you didn't choose has a (N-1)/N chance of having a car in it. Now, imagine that without opening any doors, Monty said: if you're willing to switch to the N-1 group, you win if there's a car in any of them. Do you take it? Of course.

lukasb 12y ago
Yeah this is it I think. If I can restate more simply for the 3-door case: You pick. The doors are now in two groups: your door (1/3 chance) and the other two doors (2/3 chance). The host picks. The second group is now reduced to a single door, still with 2/3 chance. Now you decide - stick with the door with 1/3 chance, or switch to the door with 2/3 chance?