Numb3rs: CBS brings Math to Prime Time

The setup for the series is quite simple: two brothers, one FBI agent, other math professor at Caltech. Professor brother sometimes helps agent brother solve crimes. For example, in last night’s episode a fugitive was on the lose. They put this on the news and people can call in saying they’ve seen him, but there are a lot of false sightings. The professor brother finds out about the false positive problem, makes a comment about “Markov Chains”, and sets to work. Eventually, he’s able to filter out the good reports from the bad ones (based of the timing and assuming that the fugitive’s movements satisfy the Markov property). Know the fugitive’s movements helps the agent brother track him down.

Later in the episode, the professor goes over the Monty Hall problem with his class. Looking for resources on that problem lead me to another which I found even more counterintuitive.

Suppose there’s family with two children. If I tell you the older child is a girl, what are the chances that the younger child is a girl? Presuming that gender is independent of birth order and the gender of one sibling doesn’t affect the gender of the other, then the answer is 50%. Likewise, if I tell you that the younger child is a girl, the chances that the older child is a girl are 50%. Now what if I tell you that one of children is a girl (could be older or younger), what is the probability that the other child is a girl? One third. Here’s why.

Suppose that there are 4000 families with two children. Since each gender is equally likely for any birth, about 2000 families will have a boy first, and 2000 will have a girl first. Of the 2000 that had a boy first, about 1000 will have a boy second and 1000 will have a girl second. So we get the following:

• About 1000 have two boys.
• About 1000 have two girls.
• About 1000 have a girl and then a boy.
• About 1000 have a boy and then a girl.

To find the probability that the other child is a girl given one is, we divide the number of families where both children are girls by the number in which at least one is. The quotient is 1000 / 3000, so the probability is one third.

Now, for the unintuition. Suppose that I tell you one of the children is a girl and that her name is Mary. Does the probability that the other child is a girl change? If so, how? In general, it does change, and if we make “reasonable” assumptions, the answer is one half. Here’s why.

Suppose that we 4000 families as before. Furthermore, let’s suppose that four out of every thousand girls is named Mary. (Estimate comes from the social security department.) Then of the 2000 families with one girl, the girl is named Mary in about 8. Of the 1000 families with two girls, about 4 will name the older Mary. Of those four, we can assume that none will name their younger child Mary. As for the 996 that didn’t name their oldest Mary, about 4 will name their younger child Mary. So in total, of the 4000 families, about 16 will have a girl named Mary. And of those, half will be families with two girls. So the probability that one child is a girl given the the other is a girl named Mary is one half.