# Probability of no 2 People out of 53 Sharing the Same Birthday

## Theorem

Let there be $53$ people in a room.

The probability that no $2$ of them have the same birthday is approximately $\dfrac 1 {53}$.

## 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 $n = 53$ and evaluating the above gives:

$\map p {53} \approx 0.01887$

or:

$\map p {53} \approx \dfrac 1 {53.01697}$

$\blacksquare$