Math Guy: The Birthday Problem

toggle caption Bruce Cook

The Weekend Edition Saturday Math Guy, Stanford professor Keith Devlin, has a problem. In fact, he has more than one... which he's happy to share with Scott Simon.

Up for a Challenge? If you think the two examples at right are tricky, take a look at two of the most notorious probability puzzles of all: The Two Envelopes Paradox The Monty Hall Problem

What is the probability that in a room filled with 23 people at least two of them have the same birthday? (It's more than half!)

Devlin explains:

The birthday problem asks how many people you need to have at a party so that there is a better-than-even chance that two of them will share the same birthday. Most people think the answer is 183, the smallest whole number larger than 365/2. In fact, you need just 23. The answer 183 is the correct answer to a very different question: How many people do you need to have at a party so that there is a better-than-even chance that one of them will share YOUR birthday? If there is no restriction on which two people will share a birthday, it makes an enormous difference. With 23 people in a room, there are 253 different ways of pairing two people together, and that gives a lot of possibilities of finding a pair with the same birthday.

Here is the precise calculation. To figure out the exact probability of finding two people with the same birthday in a given group, it turns out to be easier to ask the opposite question: what is the probability that NO two will share a birthday, i.e., that they will all have different birthdays? With just two people, the probability that they have different birthdays is 364/365, or about .997. If a third person joins them, the probability that this new person has a different birthday from those two (i.e., the probability that all three will have different birthdays) is (364/365) x (363/365), about .992. With a fourth person, the probability that all four have different birthdays is (364/365) x (363/365) x (362/365), which comes out at around .983. And so on. The answers to these multiplications get steadily smaller. When a twenty-third person enters the room, the final fraction that you multiply by is 343/365, and the answer you get drops below .5 for the first time, being approximately .493. This is the probability that all 23 people have a different birthday. So, the probability that at least two people share a birthday is 1 - .493 = .507, just greater than 1/2.

The Children Puzzle

I tell you that a couple has two children and that (at least) one of them is a boy. I ask you what is the probability that their other child is a boy. Most people think the answer is 1/2, arguing that it is equally likely that the other child is a boy or a girl. But that's not the right answer for the question I have asked you. Here's why. In terms of order of birth, there are four possibilities for the couple's children: BB, BG, GB, GG. When I tell you that at least one child is a boy, I rule out the possibility GG. That leaves three possibilities: BB, BG, GB. With two of these, the other child is a girl; so the probability of the other child being a girl is 2/3. Leaving the probability of the other child being a boy at 1/3.