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Comment by oneeyedpigeon | Hacker News Reader

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oneeyedpigeon·

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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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.

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?

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.

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.

If the host picks randomly, the chances must sum to 2/3, not 1.

Right.

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