Solution: 2017 Winter Final - 19

Author: Michiel Smid

Question

Let $n \geq 3$ 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 triples $\{P_i,P_j,P_k\}$ of people (i.e., subsets consisting of three people) that have the same birthday.
What is the expected value $\mathbb{E}(X)$ of $X$?
Hint: Use indicator random variables.

(a)

${\frac{1}{365^2}} \cdot n^3$

(b)

${\frac{1}{365^3}} \cdot {n \choose 3}$

(c)

${\frac{1}{365^3}} \cdot n^3$

(d)

${\frac{1}{365^2}} \cdot {n \choose 3}$

Solution

Let’s check for the probability that 3 people have the same birthday for every subset of 3 people

We choose 3 people out of the n people: $ \binom{n}{3} $

The first dude in the subset can have birthday on any day of the year: $ \frac{365}{365} $

The second bro in the subset must have the same birthday as the first dude: $ \frac{1}{365} $

The thid mate in the subset must have the same birthday as the first dude: $ \frac{1}{365} $

The probability that 3 people have the same birthday is $ \frac{1}{365^2} $

$ \mathbb{E}(X) = \binom{n}{3} \cdot \frac{1}{365^2} $