Problem: Ignoring leap days, the days of the year can be numbered $1$ to $365$. Assume that birthdays are equally likely to fall on any day of the year. Consider a group of $n$ people, of which you are not a member. An element of the sample space $Ω$ will be a sequence of n birthdays (one for each person).

Event $A$: “someone in the group shares your birthday”

Find an exact formula for $P(A)$.

Solution provided:

It’s easier to calculate $P(A^c)$. There are $364^n$ outcomes in $A^c$ since there are $364$ choices for each birthday. So $ P(A)=1 − P(A^c)=1 − \frac{364^n}{365^n}$

My Solution:

Say $n=5$. First we can assume the first person has the same birthday as me. This leaves the other four people to each have a choice of $364$ days for their birthdays. So there are $364^4$ sequences per person. $5(364)^4$ can more generally be written as $n(364)^{n-1}$. My formula ends up as $\frac{n(364)^{n-1}}{365^n}$ which ends up being wrong. I cannot figure out why.

  • 2
    $\begingroup$ "Someone in the group shares your birthday" isn't the same as "there's only one person in the group sharing your birthday". $\endgroup$
    – Brian
    Jun 20, 2022 at 3:50

1 Answer 1


It is definitely easier to use the complement rule here.

$$\begin{split}P(\text{someone shares your birthday} &= 1-P(\text{no one shares your birthday})\\ &=1-\left(\frac{364}{365}\right)^n\end{split}$$

If you do it your way, you will have to consider all the cases. Let $p=1/365$ be the probability of sharing a birthday.

$$\begin{split}P(\text{someone shares your birthday}) &= P(one shares)+P(2shares)+...P(n shares)\\ &={n\choose1}p(1-p)^{n-1}+{n\choose 2}p^2(1-p)^{n-2}+...+{n\choose n}p^n\end{split}$$

Since we know that these are binomial probabilities, from the distribution Binomial(n, p), the remaining probability must be complemented from 0. That is, it equals $1-{n\choose 0}(1-p)^n$.


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