I am stumped by the below confusion:
Question: How many people do we need in a class to make the probability that (at least) two people have the same birthday more than 1/2? (For simplicity, assume 365 days a year.)
I know that the answer to this question is $n=23$ people, but I was wondering why the following method does not work using the counting rule:
Out of $n$ people, choose any 2 people to share the same birthday. For that same birthday, there are 365 choices. For the other $n-2$ people, each one has 365 choices of birthdays. Therefore, the probability that at least two people have the same birthday is $$ \frac{\binom{n}{2} 365 \times 365^{n-2}}{365^n} = \binom{n}{2} \times \frac{1}{365} = 0.5 $$ which solving for gives us roughly $n=20$.
Where did I go wrong here? Thank you very much!

  • $\begingroup$ You're overcounting the number of ways two people out of $n$ can share the same birthday. $\endgroup$ Commented Jul 4, 2023 at 23:09
  • 2
    $\begingroup$ You are double counting cases where there are two pairs of people sharing a birthday and triple counting cases where three people share a birthday, once for each pair of the three. $\endgroup$ Commented Jul 4, 2023 at 23:11
  • $\begingroup$ Now a question: Does it make a difference that this is a (school) class, which could mean a non-linear distribution of dates of birth? $\endgroup$
    – gnasher729
    Commented Jul 4, 2023 at 23:16
  • $\begingroup$ That will depend on how the school class is determined. If, for instance, first-graders are determined by all students born in one specific calendar year, and if we pretend that all birthdates are equally likely (in real life, they are not), then the standard solution to the birthday problem applies. But most school classes are not determined this way, at least where I live. $\endgroup$
    – Dan Asimov
    Commented Jul 4, 2023 at 23:42

1 Answer 1


To elaborate on Joshua Wang's comment, your approach considers each possible pair of people, and correctly calculates the probability that they share a birthday to be $\frac{1}{365}$. However, when you sum over all pairs, you may be overcounting the number of possible situations where this occurs.

In particular, if three people $A,B,C$ all share the same birthday, then this situation is triple-counted: once for the pair $A,B$, once for the pair $B,C$ and once for the pair $A,C$.


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