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A letter regarding Brian Hayes's review of Digital Dice

I read with a great deal of interest Brian Hayes's review of the book Digital Dice in the July-August 2008 issue (vol. 96, no. 4, pp. 334-336). In the review, titled "Programs and Probabilities," Hayes discusses the Monty Hall problem. I had read about the problem in an article by John Tierney in the New York Times on April 8, 2008, and wondered at the time why such a simple problem would generate so much controversy.

Hayes’s Monte Carlo results suggest that switching doors doubles your chances of winning, but he does not really explain why that is happening. The probabilistic explanation offered by "the other side" in the review is correct but does not explain in simple terms why that result occurs.

What if one wanted to explain what is happening to a bright high-school sophomore who has never studied probability? There is a direct approach that is similar to what happens in our thinking in a chess game: If I do this, then this happens, but if I do this other thing, then that could happen. And the problem here is much simpler than those contained in any real chess game.

Here's a direct approach:

Let's assume without any loss of generality that my initial choice is always door 1. There are three different cases to consider: The prize itself is behind door 1; it is behind door 2; or it is behind door 3.

Let's now assume the prize is behind door 1. Monty Hall opens either door 2 or door 3. In either case I have two choices: I can stay with my original choice, door 1, or I can switch to the door that is still closed (to door 2 if Monty opens door 3 or to door 3 if Monty opens door 2). If I don't switch I win the prize, and if I switch I lose.

Now assume the prize is behind door 2. Monty opens door 3 and shows me that it is empty. If I don't switch (stay with door 1) I lose, but if I switch to door 2, I win.

Finally, assume the prize is behind door 3. Monty opens door 2 and shows me that it is empty. If I don't switch (stay with door 1), I lose; but if I switch to door 3, I win.

Thus if I don't switch I win only when the prize is behind the door of my original choice (door 1), but if I switch, I win in two cases, when the prize is behind door 2 or is behind door 3.

The switching strategy does twice as well as the "don't switch" strategy.

Of course this is closely related to the probabilistic approach, but if the problem had been approached in this direct way, I don't think there would have been any controversy.

One more thought: Sticking with the original choice neglects the additional information given to us by Monty when he opens a door. The additional information given to us by Monty when he opens door 3 is that he didn't open door 2. The additional information given to us by Monty when he opens door 2 is that he didn't open door 3. The switching strategy takes advantage of this additional information and may help the reader understand that there is no paradox involved here.

Ethan Aronoff
Los Angeles, California

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