(The Monty Hall problem)
Suppose there are three doors, labeled A, B, and C. A new car is behind one of the three doors, but you don't know which. You select one of the doors, say, door A. The host then opens one of doors B or C, as follows: If the car is behind B, then they open C; if the car is behind C, then they open B; if the car is behind A, then they open either B or C with probability 1/2 each. (In any case, the door opened by the host will not have the car behind it.) The host then gives you the option of either sticking with your original door choice (i.e., A), or switching to the remaining unopened door (i.e., whichever of B or C the host did not open). You then win (i.e., get to keep the car) if and only if the car is behind your final door selection. (Source: Parade Magazine, "Ask Marilyn" column, September 9, 1990.) Suppose for definiteness that the host opens door B.
(a) If you stick with your original choice (i.e., door A), conditional on the host having opened door B, then what is your probability of winning? (Hint: First condition on the true location of the car. Then use Theorem 1.5.2.)
(b) If you switch to the remaining door (i.e., door C), conditional on the host having opened door B, then what is your probability of winning?
(c) Do you find the result of parts (a) and (b) surprising? How could you design a physical experiment to verify the result?
(d) Suppose we change the rules so that, if you originally chose A and the car was indeed behind A, then the host always opens door B. How would the answers to parts (a) and (b) change in this case?
(e) Suppose we change the rules so that, if you originally chose A, then the host always opens door B no matter where the car is. We then condition on the fact that door B happened not to have a car behind it. How would the answers to parts (a) and (b) change in this case?