Monty Hall with a host who has a favorite door

Question

Standard three-door setup: you pick Door 1, and the host always opens a goat door. But this host has a quirk: whenever he has a choice of which goat to reveal (which happens only when the car is behind your Door 1), he opens Door 3 with probability $p$ and Door 2 with probability $1 - p$ . You see him open Door 3.

What is your probability of winning if you switch to Door 2, in terms of $p$ ? Evaluate it at $p = 1$ .

Accepted Answer

Condition on the observed event "host opens Door 3." The car is equally likely behind each door. - Car behind Door 1 (prob 	frac13): the host has a free choice and opens Door 3 with probability p. Contribution 	frac13 p. - Car behind Door 2 (prob 	frac13): the host cannot open Door 2 (the car) or Door 1 (your pick), so he must open Door 3. Contribution 	frac13\cdot 1. - Car behind Door 3 (prob 	frac13): he would have to open Door 2, so he never opens Door 3 here. Contribution 0. Total probability of opening Door 3: 	frac13 p + 	frac13 = 	frac{p+1}{3}. Switching wins exactly when the c…

Monty Hall with a host who has a favorite door

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