What’s the probability that in a room of n people, at least two have the same birthday? It’s higher than you think.
Let $$p(n)$$ be the probability of at least two people having the same birthday. We’ll find $$\overline{p}(n)$$, the probability that no two people share the same birthday. $$p(n) = 1 - \overline{p}(n)$$.
$$ \begin{aligned} \overline{p}(n) &= 1 \cdot (1 - \frac{1}{365}) \cdot (1 - \frac{2}{365}) \cdot \cdots \cdot (1 - \frac{n - 1}{365}) \ &= \frac{365 \cdot 364 \cdot \cdots \cdot (365 - n + 1)}{365n}\ &= \frac{365!}{365n(365 - n)!}\ &= \frac{n! \cdot {365 \choose n}}{365n} \end{aligned} $$
So $$p(n) = 1 - \frac{n! \cdot {365 \choose n}}{365n}$$. The table below shows that we have more than a 50% chance when there are just 23 people. When n = 100, it’s almost certain to have two people sharing the same birthday.
| n | p(n) |
|---|---|
| 10 | 11.7% |
| 20 | 41.1% |
| 23 | 50.7% |
| 30 | 70.6% |
| 50 | 97.0% |
| 57 | 99.0% |
| 100 | 99.99997% |
| 200 | 99.9999999999999999999999999998% |
| 300 | (100 − (6×10−80))% |
| 350 | (100 − (3×10−129))% |
| 365 | (100 − (1.45×10−155))% |
| 366 | 100% |