Originally Posted by

**Moo** So I'd like to know how to compute the probability that out of n people, 3 (or more) have the same birthday ?

Hi,

If you want an exact formula, this comes from simple combinatorics, but there's no "nice-looking" answer...: I get, for at least 3 simultaneous birthdays (let $\displaystyle N=365$)

$\displaystyle p=1-\frac{1}{N^n}\sum_{0\leq 2k \leq n}\frac{n!N!}{2^k k!(n-2k)!(N-(n-k))!}$.

I did not check the number 88 with Maple, it would be safer to try...

How I got the formula: we have to compute the number of $\displaystyle n$-uplets from $\displaystyle \{1,\ldots,N\}$ where the same value is not taken three times. The index $\displaystyle k$ in the formula stands for the number of pairs of matching birthdays (k=0 gives the usual case). For a given number $\displaystyle k$, and given the positions of the pairs, we have $\displaystyle N(N-1)\cdots(N-(n-k)+1)$ possibilities (we choose $\displaystyle k$ values for the pairs and $\displaystyle n-(2k)$ values for the others, all different). And for a given k, the number of positions of these k pairs is: $\displaystyle {n\choose 2k}(2k-1)(2k-3)\cdots 3\cdot 1$ (choice of the subset made by the pairs, and choice of the matchings). This gives a total number:

$\displaystyle \sum_{0\leq 2k\leq n}{n\choose 2k}(2k-1)(2k-3)\cdots5\cdot 3\cdot N(N-1)\cdots(N-(n-k)+1)$.

The previous formula follows after a mild rewriting.

I let you guess how ugly a general formula might look like...

PS. I forgot to mention: "Happy Birthday, Moo!"