A car is equally likely to be behind one of three doors. You select one of the three doors (say, Door #1). The host, who knows where the car is, then reveals one non-selected door (say, Door #3) which does not contain the car. You can then 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?The surprising answer is that you have probability just 1/3 of winning the car if you stick, and 2/3 if you switch.
I have written a book which discusses this problem, and written an article about this problem and its variants, and even appeared in a television skit about it. There are also numerous other discussions of this problem available on the web. However, it seems that some people are not convinced despite all this evidence. (Indeed, I have received numerous enquiries about this problem, including several aggressive e-mails insisting that the answer must be 1/2 and I am a fool to say otherwise. I now know how vos Savant must have felt back in 1990.)
So, what follows is the absolute simplest, most direct explanation of the Monty Hall problem that I know.
If the car is behind Door #1, then after you pick Door #1, the host will open another door (either #2 or #3), and you will then switch to the remaining door (either #3 or #2), thus LOSING.
If the car is behind Door #2, then after you pick Door #1, the host will be forced to open Door #3, and you will then switch to Door #2, thus WINNING.
If the car is behind Door #3, then after you pick Door #1, the host will be forced to open Door #2, and you will then switch to Door #3, thus WINNING.
Hence, in 2 of the 3 (equally-likely) possibilities, you will win. Ergo, the probability of winning by switching is 2/3.
Still not convinced? Well, there are many other explanations in my book Struck By Lightning, in my article about this problem, and elsewhere on the web. Hopefully one of them will finally convince you!