People entering until two share a birthday

Question

People walk into a room one at a time, birthdays independent and uniform over $365$ days. You stop counting the instant someone shares a birthday with a person already in the room.

On average, how many people enter before that first shared birthday?

Accepted Answer

Let T be the number of people up to and including the one who first causes a match. As people enter with distinct birthdays, the chance the next person also avoids everyone shrinks, the birthday-problem product. The expected wait for a calendar of N days is E[T] \approx \sqrt{\frac{\pi N}{2}} = 1.253\sqrt{N}. With N = 365: E[T] \approx 1.253\sqrt{365} \approx 1.253 	imes 19.1 \approx 23.9, so about 24 people on average. This sits right next to the famous "23 for even odds" fact: the median wait is around 23, and the mean is around 24. Both come from the same square-root law. Takeaway. The av…

People entering until two share a birthday

Your answer

More Probability questions