Having just spent three hours explaining the value of trigonometric substitutions and partial fraction expansions to not very enthusiastic calculus students, I’m not really in the mood for a lengthy post today. So how about yet another variation on the Monty Hall problem.

In this version the contestant is shown 10 identical doors. One contains a valuable prize, the other nine contain goats. The contestant chooses one door. The host then opens a door he knows to be empty and gives the contestant the choice of switching to one of the remaining eight unopened doors. After the contestant makes her decision, the host opens another door he knows to be empty. Again the contestant is given the option of switching. This continues until only two doors remain. The contestant makes her final decision, and the door she has now chosen is opened. The question is: What strategy should the contestant follow to maximize her chances of winning?

Be sure to justify your answer. Enjoy!