Sharing a Birthday (56/365)

I think most have heard something like you only need suprisingly few people in a room before two people in the room end up sharing a birthday. But I never bothered to work it out. Let me do that.

First, forget leap years. Let a year have $365$ days. Note that if you have 366 people in the room you are guaranteed that someone will share a birthday. On the other end of the spectrum, if there are only two people in the room then the probability that the two of them share a birthday is given by one minus the number of ways they cannot share a birthday divided by the number of ways we can assign them a birthday: $\frac{(365)(364)}{(365)(365)}$.

In general, let’s say there are $k$ people in the room. The following gives the number of ways to assign different birthdays to each of the $k$ people.

$\displaystyle D(k) = (365)(365-1)(365-2)\dots(365-k+1) = (365)_k \\ \text{Note that when } k > 365 \text{ we get } D_k = 0$

The number of ways to assign any birthday to each of the $k$ people.

$\displaystyle N(k) = 365^k$

So, the probability that at least one pair out of $k$ will share a birthday is given by

$\displaystyle \frac{D(k)}{N(k)} = 1 - \frac{(365)_k}{365^k}$

Graphing it below. By $k=41$ the probability is already at $0.9$.