How big does a group of people need to be before there is at least a 50% chance that two people share a birthday.
23 people.
Proof
For simplicity, ignore leap years. This won’t significantly affect the result.
Assume that all birthdays are equally likely. If birthdays are unequally distributed, then it only makes it more likely that people share a birthday.
We will determine the probability that no one shares a birthday with anyone. Adding people one at a time:
- The first person won’t share a birthday (as they are the only person). i.e. they have a \(1 = \frac{365}{365}\) chance of being unique.
- The second person only has to avoid the first person’s birthday for a \(\frac{364}{365}\) chance of being unique.
- The \(n\)th person has to avoid \(365-(n-1)\) days for a \(\frac{366-n}{365}\) chance of being unique.
Then the probability \(P'(n)\) that no one shares a birthday is the product that each individual has a unique birthday:
\[P'(n) = \frac{365 \times 364 \times ... \times (366-n)}{365^n} = \frac{365!}{(365-n)! 365^n}\]The probability that there are two people who share a birthday \(P(n)\) is the complement:
\[P(n) = 1 - \frac{365!}{(365-n)! 365^n}\]23 is the smallest number for which \(P(n) > 0.5\).
Other values
By the pigeonhole principle, you need 366 people to be guaranteed to be two people who share a birthday (367 to include leap years).
However chance climbs very quickly to be close to 1 with much smaller groups:
| n | P(n) |
|---|---|
| 1 | 0 |
| 10 | .117 |
| 23 | .507 |
| 50 | .970 |
| 70 | .999 |
Birthday line
At a movie theater, the manager announces that they will give a free ticket to the first person in line whose birthday is the same as someone who has already bought a ticket. You have the option of getting in line at any time. What position in line gives you the greatest chance of being the first duplicate birthday?
Assume that you don’t know anyone else’s birthday, and that birthdays are distributed randomly throughout the year.
Position 20.
Proof
Your probability of getting a free ticket when you are the nth person is line is (probability that none of the first n-1 people share a birthday) \(\times\) (probability that you share a birthday with one of the first n-1 people):
\[P(n) = \frac{365 \times 364 \times ... \times (365-(n-2))}{365^{n-1}} \frac{n-1}{365}\]We want the first \(n\) such that \(P(n) > P(n+1)\):
\[\begin{align} \frac{365 \times 364 \times ... \times (365-(n-2))}{365^{n-1}} \frac{n-1}{365} & > \frac{365 \times 364 \times ... \times (365-(n-2)) \times (365-(n-1))}{365^n} \frac{n}{365} \\ 365 (n-1) & > (365 - (n-1)) n \\ n^2 - n - 365 & > 0 \end{align}\]Solving for \(n\) we find roots at \(n \approx -18.6\) and \(n \approx 19.6\). The negative root is invalid for this problem.
Thus we need \(n > 19.6\) giving a minimum of \(n = 20\).