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Comment by HALtheWise | Hacker News Reader

Fair warning: I haven't rigorously verified the math on this, and tfa is all about the importance of vigorously verifying the math.

In the case where the host revealing the car means you restart, there are three equally likely situations after the host opens a door: - You picked the car, and the host revealed a goat - You picked a goat, and the host revealed the other goat - You picked a goat, and the host reveals the car (restart)

Of the two terminal cases, one gets you the car if you switch, and one gets you the car if you don't switch. In the non-terminal case, it doesn't matter what policy you have because you don't even get a chance to apply it.

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One way to think about it is to imagine Monty picking randomly in either case, but then (internally and invisibly) correcting his pick.

Assuming, WLOG, that we choose door 1, that leaves us with 6 equally likely cases just before that final correction (or lack thereof):

    A) Car 1, Monty 2
    B) Car 1, Monty 3
    C) Car 2, Monty 2
    D) Car 2, Monty 3
    E) Car 3, Monty 2
    F) Car 3, Monty 3

If Monty doesn't correct in cases C and F, then when he shows us a goat behind (say) 2 then we learn we are in either A or E - it's 50/50. If Monty does correct himself, then we might have been in A or E or F.

Ah, yep, that’s right. Another way to see it is that we’re interested in the probability that your door has a goat behind it, given that you didn’t need to start over:

P(you chose goat | host didn’t choose car) = P(you chose goat, host didn’t choose car) / P(host didn’t choose car).

The numerator is 2/3 * 1/2, and the denominator is 2/3, so the ratio is indeed 1/2.

(A rejection sampling loop, where you repeatedly simulate a process until a condition holds, has the same distribution over final outcomes as the conditional distribution—so repeatedly restarting the game if the host chooses the car induces the same distribution on final results as simply conditioning on the host not choosing the car.)

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