A woman has two children, one of which is a girl. What is the probability that the other child is also a girl?

This is a little puzzle that I’ve been stewing over for two days now. I encountered it in the autobiography that I’m reading, Born on a Blue Day, by Daniel Tammet. In his book, Tammet, a mathematical savant, recounts how he’s always had a talent for understanding probability, and that most people find the subject unintuitive. I agreed with his sentiment when I read it. I didn’t count myself among the confused, however, until after I read his example problem, and subsequently answered it wrong.

I thought, as probably many do, that the answer was 1-in-2. That was incorrect. The actual answer is 1-in-3, which I still find rather confounding.

I reasoned that there would be a 50-50 chance. All things being equal, any given child has a 1/2 probability of being either a boy or a girl. Previous events have no effect on the probability of future ones, so I thought the odds would be 1/2 regardless. A sizable segment of the population believes otherwise, though. They think that if your first child is male then your next child is more likely to be female. This is false, of course, so I assumed Tammet’s puzzle would only trick people who made this sort of mistake.

Not so! The answer really is 1-in-3, and here’s why: Imagine that you’ve had two children. There are four possible arrangements of their genders, and each of those four possibilities is equally likely:

1. Boy, Boy
2. Boy, Girl
3. Girl, Boy
4. Girl, Girl

Of the four possibilities, three contain at least one girl. And of those three, only one combination contains two girls. So, to answer the original question: of all the women who have two children, and who have at least one girl, one in three will have two girls.

Q.E.D., right?

That explanation took me a full day to come up with, actually, although I have no doubt that most people would have worked it out much more quickly. In fact, Tammet provides an explanation in his text, although I didn’t want to spoil the puzzle for myself. And it still confused me for another day, because in spite of the demonstrable logic, the answer still didn’t make any sense to me. If I have one daughter, the odds of my next child being a girl are not 1-in-3. They are definitely 1-in-2. As I mentioned above, past events have no effect on future probabilities. It doesn’t matter if I have one daughter, or one son. My next child will always have a 50% chance of being a girl. So how could both of these statements be true?

As it turns out, they’re slightly different statements, although they appear very much alike. The first, the answer to Tammet’s question, gives the probability of there being two daughters, given the knowledge of at least one. The second, my interpretation, gives the probability of having a second daughter when you already have a first. Do you see the difference? My interpretation made the mistake of thinking that this was a question about specific daughters. But it’s not.

It doesn’t ask for the probability of a second child’s gender being female, given the gender of a first. It asks for the probability of a given family’s makeup, given the presence of at least one daughter. The odds of either specific daughter being a girl are still, of course, 1-in-2. Accordingly, the odds of there being a family with two children who are girls are 1-in-4. But given that we can eliminate all families which have all boys (i.e. we know there is at least one girl), those odds rise to 1-in-3. So we never really ask about specific children, and the world of probability makes sense to me once again. Mostly.