Let there be $23$ or more people in a room.

The probability that at least $2$ of them have the same birthday is greater than $50 \%$.

## Proof

Let there be $n$ people in the room.

Let $\map p n$ be the probability that no two people in the room have the same birthday.

For simplicity, let us ignore leap years and assume there are $365$ days in the year.

Let the birthday of person $1$ be established.

The probability that person $2$ shares person $1$'s birthday is $\dfrac 1 {365}$.

Thus, the probability that person $2$ does not share person $1$'s birthday is $\dfrac {364} {365}$.

Similarly, the probability that person $3$ does not share the birthday of either person $1$ or person $2$ is $\dfrac {363} {365}$.

And further, the probability that person $n$ does not share the birthday of any of the people indexed $1$ to $n - 1$ is $\dfrac {365 - \paren {n - 1} } {365}$.

Hence the total probability that none of the $n$ people share a birthday is given by:

$\map p n = \dfrac {364} {365} \dfrac {363} {365} \dfrac {362} {365} \cdots \dfrac {365 - n + 1} {365}$
 $\ds \map p n$ $=$ $\ds \dfrac {364} {365} \dfrac {363} {365} \dfrac {362} {365} \cdots \dfrac {365 - n + 1} {365}$ $\ds$ $=$ $\ds \dfrac {365!} {365^n} \binom {365} n$

Setting $n = 23$ and evaluating the above gives:

$\map p {23} \approx 0.493$

Hence the probability that at least $2$ people share a birthday is $1 = 0.492 = 0.507 = 50.7 \%$

$\blacksquare$

## Conclusion

Counter-intuitively, the probability of a shared birthday amongst such a small group of people is surprisingly high.

### $3$ People Sharing a Birthday
Let $n$ be a set of people.
Let the probability that at least $3$ of them have the same birthday be greater than $50 \%$.
Then $n \ge 88$.