Monty Fall, when the host opens a door by accident

Question

You pick Door 1 of three. This time the host does not know where the car is: he trips and knocks open one of the other two doors at random, and it happens to reveal a goat.

What is your probability of winning if you switch, and why does it differ from the classic answer?

Accepted Answer

The classic 	frac23 relies on the host always opening a goat on purpose. Here the reveal was accidental, so we must condition on the process that actually generated it. Let the car be equally likely behind each door, and the host opens Door 2 or Door 3 at random. Condition on the observed event "host opened Door 3, and it was a goat": - Car behind Door 1 (prob 	frac13): host opens Door 3 w.p. 	frac12, and it is a goat. Contribution 	frac13\cdot	frac12 = 	frac16. - Car behind Door 2 (prob 	frac13): host opens Door 3 w.p. 	frac12, a goat. Contribution 	frac16. - Car behind Door 3 (prob…

Monty Fall, when the host opens a door by accident

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