Speaking of the Monty Hall problem, I recently came across this terrific essay (PDF format), by Jeffrey Rosenthal, a professor in the Department of Statistics at the University of Toronto. Rosenthal discusses several variants of the Monty Hall problem, and shows how a clever application of Bayes’ Theorem helps to distinguish between them. Here are the variants he considers:
- Monty Hall Problem: A car is equally likely to be behind any one of three doors. You select one of the three doors (say, Door #1). The host then reveals one nonselected door (say, Door #3) which does not contain the car. At this point, you choose whether to stick with your original choice (i.e. Door #1), or switch to the remaining
door (i.e. Door #2). What are the probabilities that you will win the car if you stick, versus if you switch?
- Monty Fall Problem: In this variant, once you have selected one of the three doors, the host slips on a banana peel and accidentally pushes open another door, which just happens not to contain the car. Now what are the probabilities that you will win the car if you stick with your original selection, versus if you switch to the remaining door?
- Monty Crawl Problem: As in the original problem, once you have selected one of the three doors, the host then reveals one non-selected door which does not contain the car. However, the host is very tired, and crawls from his position (near Door #1) to the door he is to open. In particular, if he has a choice of doors to open (i.e., if your original selection happened to be correct), then he opens the smallest number available door. (For example, if you selected Door #1 and the car was indeed behind Door #1, then the host would always open Door #2, never Door #3.) What are the probabilities that you will win the car if you stick versus if you switch?
- Monty Small Problem: In this variant, the host is only somewhat tired. If he has a choice of doors to open, then he has a small probability p of opening the largest number available door, otherwise (with probability 1-p) he opens the smallest number available door. What is the probability that you will win the car if you then switch to the third door? (The case p = 1/2 is the original problem, while p = 0 is Monty Crawl.)
Incidentally, variants on the Monty Hall problem are something of a hobby of mine. Feel free to suggest other variants in the comments.