# Birthday Problem: Finding a Probability Function of an Event

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.

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

$$\begin{split}P(\text{someone shares your birthday} &= 1-P(\text{no one shares your birthday})\\ &=1-\left(\frac{364}{365}\right)^n\end{split}$$
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 $$1-{n\choose 0}(1-p)^n$$.