Solution: 2016 Fall Final - 20

Author: Michiel Smid

Question

Let $n \geq 2$ be an integer and consider a group $P_1,P_2,\dots,P_n$ of $n$ people. Each of these people has a uniformly random birthday, which is independent of the birthdays of the other people. We ignore leap years; thus, the year has 365 days.
Define the random variable $X$ to be the number of unordered pairs $\{P_i,P_j\}$ of people that have the same birthday.
What is the expected value $\mathbb{E}(X)$ of $X$?
Hint: Use indicator random variables.

(a)

$\frac{1}{365} \cdot n^2$

(b)

$\frac{1}{365} \cdot {n \choose 2}$

(c)

$\left( \frac{1}{365} \right)^2 \cdot {n \choose 2}$

(d)

$\left( \frac{1}{365} \right)^2 \cdot n^2$

Solution

We take all possible unordered pairs by choosing 2 out of the n people: $ \binom{n}{2} $

$P_i$‘s birthday can be any day: $ \frac{365}{365} $

$P_j$‘s birthday must be the same day: $ \frac{1}{365} $

$ \mathbb{E}(X) = \binom{n}{2} \cdot \frac{1}{365} $