Probability that no two people are born on the same day: $n$ people and $k$ days

birthdayprobabilitystatistics

Say we have $n$ people and $k$ days. What is the probability that NO two of the $n$ people are born on the same day?

This is of course assuming that $n \leq k$ (otherwise the answer is $0$ by the pigeonhole principle!).
I said the answer was: $k \choose n$$\cdot n! \cdot \frac{1}{k^n}$ because in order for this to happen, n distinct birthdays must be chosen, they can be arranged in any way amongst the $n$ people, and there are $k^n$ total arrangements. Is this correct?

Best Answer

Your ${k \choose n} \cdot n! \cdot \frac{1}{k^n}$ is correct assuming equal independent probabilities for each day and person

To illustrate it with $k=365$, you get

  • with $n=22$, the probability the birthdays are all different of about $0.5243047$
  • with $n=23$, the probability the birthdays are all different of about $0.4927028$

and so the solution to the birthday problem that the median number of people to have at least one matching birthday is $23$

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