Let’s Make a Deal

Posted by jns on 4 August 2008

This is David Flannery, a lecturer in mathematics at the Cork Institute of Technology, Cork, Ireland. He is shown with his daughter Sarah Flannery, author of the book In Code : A Mathematical Journey (New York : Workman Publishers, 2001); David Flannery is listed as her coauthor.

The book is a fascinating, entertaining, and instructive one, blurbed on the jacket as “a memoir with mathematics”. It’s Sarah’s autobiographical account of her route to winning some prestigious young scientist awards in 1998 and 1999 with a good dose of mathematical fun and really good writing about the mathematical ideas at the core of public-key cryptography, if you can imagine such a thing. I enjoyed reading it very much and I think a lot of other people might enjoy it too. Here’s my book note.

Okay, so her coauthor and father happens to have a very lovely beard, but it’s all just a pretext for passing along an interesting problem and its lovely solution from an appendix of the book. I’m sure this surprises no one here.

In America this statistical brain-teaser tends to be known as the “Let’s Make a Deal” problem. Imagine you’re playing that game. Host Monty presents you with three doors: behind one is a wonderful and expensive new car, behind the other two nothing of much value. You chose one door at random. Host Monty then opens one of the two remaining doors, revealing an item of little value.

Do you change your initial choice?

Most of us who’ve done some statistics are suckered into saying “no” at first because we think that the probability of our having made the correct choice has not changed. However, it has indeed changed and a more careful analysis says we should always choose the other door after Monty opens the door of his choice. One can perhaps go along with that by realizing that Monty did not choose his door at random–he knows where the car is and he chose not to reveal the car, so the information we have at our disposal to make our “random” choice has changed.

However, seeing this clearly and calculating the attendant probabilities is neither easy nor obvious–usually. I’ve read several solutions that typically say something like “look, it’s simple, all you have to do….” and they go on to make inscrutable statements that don’t clarify anything.

The answer in Flannery’s book is the clearest thing I’ve ever read on the subject; the reasoning and calculation itself by one Erich Neuwirth is breathtaking in its transparency. From “Appendix B: Answers to Miscellaneous Questions”, this is her complete response to the question “Should you switch?”

Yes. In order to think clearly about the problem get someone to help you simulate the game show with three mugs (to act as the doors) and a matchstick (to act as the car). Close your eyes and get your helper to hide the matchstick under one of the mugs at random. Then open your eyes, choose a mug and let your helper reveal a mug with nothing under it. Play this game many times, and count how often you win by not switching and how often by switching.

The following explanation of the answer “Yes, you should switch” is from Erich Neuwirth of Bad Voeslau, Austria, and appears on page 369 of the Mathematical Association of America’s The College Mathematics Journal, vol. 30, no. 5, November 1999:

“Imagine two players, the first one always staying with the selected door and the second one always switching. Then, in each game, exactly one of them wins. Since the winning probability for the strategy “Don’t switch” is 1/3, the winning probability for the second one is 2/3, and therefore switching is the way to go.” [pp. 302--303]

Do you see the reason? Because one player always wins, the individual probabilities of each player’s winning must add up to 1. The probability of a win for the “Don’t switch” player is obviously 1/3 (1 of 3 doors chosen at random), so the probability of a win for the “Always switch” player must be 1 – 1/3 = 2/3. Brilliant!