I've pondered the Monty Hall problem before, and have now rethought it in what I think is a more concise and hopefully understandable way.
For simplicity, label the door that leads to the prize P, and the doors that lead to empty rooms E1 and E2. If the original guess is P, then switching answers after either E1 or E2 are opened will lose; i.e., this will mean switching from P to E2 or E1. On the other hand, if the original choice is E1, Monty must then open E2, and switching from E1 to P wins. Similarly, if the original choice is E2, then Monty must open E1, and switching from E2 to P wins. Because it is twice as likely for the contestant's original guess to be E1 or E2 than the lone P, the probability of winning by switching doors is 2/3.
posted by Harvey Leff
August 25, 2008
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