Three doors hide one car. You pick door 1. The host knows the location, always opens a goat door other than yours, and when he has a free choice he opens door 3 with probability 3/4. He opens door 3. What is your probability of winning if you switch to door 2?

Three doors hide one car. You pick door 1. The host knows the location, always opens a goat door other than yours, and when he has a free choice he opens door 3 with probability 3/4. He opens door 3. What is your probability of winning if you switch to door 2?

Approach: Write the likelihood of the host opening door 3 under each car location, then apply Bayes. The only branch where the host has a free choice is the one where your first pick was already right.

4/7. Each door holds the car with prior 1/3. If the car is behind door 1 the host has a free choice, so the likelihood of him opening door 3 is 3/4. If the car is behind door 2 he is forced, likelihood 1. If the car is behind door 3 the likelihood is 0. Bayes rule gives the posterior probability for door 2 as (1/3)/((1/3)(3/4) + 1/3) = 1/(3/4 + 1) = 4/7. Had the host bias sent him to door 2 instead, switching would win 1/(1/4 + 1) = 4/5. The general rule is 1/(1 + q) with q the probability he opens the door he actually opened when free, so switching is never worse than 1/2.

Follow-up: What value of q makes the two possible host actions equally informative about your original pick?

Key concepts: bayes rule, likelihood, posterior probability, host bias.