I’ve been reading Leonard Mlodinow’s The Drunkard’s Walk: How Randomness Rules our Lives. The book covers the Monty Hall problem, Bayes’s Theorem, availability bias, the illusion of control and so forth. If these are unfamiliar, look no further for an entertaining account.

On the other hand, I can’t say that I learned much I didn’t already know. Nevertheless, I still enjoyed reading the book – it’s well written and filled with interesting nuggets (Did you know that the great mathematician Paul Erdos refused to believe that you should switch doors?). If you teach probability theory or intro stats you will find lots of good examples to brighten up your lectures.

One problem did intrigue me. Suppose that a family has two children. What is the probability that both are girls? Ok, easy. Probability of a girl is one half, probabilities are independent thus probability of two girls is 1/2*1/2=1/4.

Now what is the probability of having two girls if at least one of the children is a girl? A little bit harder. Temptation is to say that if one is a girl the probability of the other being a girl is 1/2 so the answer is 1/2. That’s wrong because you are not told which of the two children is a girl and that makes a difference. Better approach is to note that without any additional information there are four possibilities of equal likelihood for the sex of two children (B,B), (G,B), (B,G), (G,G). If we know that at least one is a girl we can remove (B,B) so three equally likely possibilities, (G,B), (B,G), (G,G), remain and of these 1 has two girls so the answer is 1/3.

Ok, now here is the stumper. What is the probability of a family having two girls if one of the children is a girl named Florida?

At first it seems impossible that knowing the name should make a difference. Surely, the answer is 1/3 just as before? After all, every child has a name. But knowing the name does make a difference. Here’s a hint, Florida is a rare name.