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
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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:
- $\map p {53} \approx \dfrac 1 {53.01697}$
$\blacksquare$
Sources
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $53$