## Thursday, October 13, 2005

### Monty Hall was a Friend of Mine

Michael Shackleford, who also has the marvelous Wizard of Odds page, has an old set of math problems on a web site he no longer updates. Problem 16 on that website has been the source of more mail for him than any other problem. Here's how he states it:

A box contains two coins. One coin is heads on both sides and the other is heads on one side and tails on the other. One coin is selected from the box at random and the face of one side is observed. If the face is heads what is the probability that the other side is heads?

Now Mr. Shackleford has way more math expertise than I'll ever have, and his websites are usually excellent. However, in this case he does not seem to realize that his otherwise correct solution of 2/3 contains an unstated assumption. If he made just a minor change to the problem statement, his answer would be completely correct. As it stands, his answer assumes too much. The correct answer, given the problem as stated, is "possibly 1/2, possibly 2/3, depending..." The part that it depends upon is whether a random face of the selected coin is observed. If, once a coin is randomly selected, then a random face of the coin is observed, then beyond a doubt, his answer of 2/3 is correct. You can see his solution to work it out yourself.

However, when I read the problem I was reminded of the classic Monty Hall problem, problem 186 on this page: http://mathproblems.info/group10.html. In the Monty Hall problem, the fact that someone else has more information than you do makes a critical difference in the outcome. In the case of this problem, if someone else pulled the coin out of the box, and then decided which face to show you, then the solution of 1/2 is correct.

Considering Mr. Shackleford's own statement of Bayes' theorm of conditional probability as applied to this problem:

1/2 * 1 / [ (1/2 * 1) + (1/2 * 1/2) ] = 1/2 / (1/2 + 1/4) = (1/2)/(3/4) = (1/2)*(4/3) = 2/3.

The difference is that if someone else chooses the face on the coin to show me, then Pr(choosing heads given that one headed coin was chosen) = 1, not 1/2. The resulting math is:

1/2 * 1 / [ (1/2 * 1) + (1/2 * 1)] = 1/2 / (1/2 + 1/2) = (1/2)/(1) = (1/2)*(1) = 1/2

The correct wording of the problem is then:

A box contains two coins. One coin is heads on both sides and the other is heads on one side and tails on the other. One coin is selected from the box at random and a random face of the coin is observed. If the face is heads what is the probability that the other side is heads?

Now, Mr. Shackleford tries to justify his wording of the problem with the following explanation:
Some other people have questioned my exact wording of this problem, mainly objecting over the tense of the verb I use for when the coin is chosen. What appears now was taken almost exactly from a similar problem in Probability and Statistics (second edition) by Morris H. Degroot on page 63, problem number 5. I changed cards to coins and eliminated what would be the coin with two tails. The book is a commonly used college text on the subject and should be above reproach.

It really seems to me that Mr. Degroot has made the same mistake, however, and I don't think Mr. Shackleford is well served by perpetuating this mistake. When I encountered problem 16 on Mr. Shackleford's page initially, I immediately realized the ambiguity and came up with the two solutions as the possible results, depending on the randomness of the presented face.

Unfortunately, Mike Shackleford is no longer willing to read any emails relating to problem 16. I can understand why this might be, due to the large number of people who have evidently written in arguing about his math. His math is correct as far as it goes, but he really ought to make the problem statement clearer. I have emailed him about this post on my blog, but I don't expect he'll read the email. If he responds, I'll let you know.

I hope you found this interesting.