# Facebook Birthday Problem: Part 2

Recently, I devised the following problem:

Suppose that you have n friends, and you always say “Happy Birthday” to each friend on his/her birthday. On how many days of the year will you not say “Happy Birthday” to one of your friends?

Until somebody tells me otherwise, I’m calling this the Facebook birthday problem in honor of Facebook’s daily alerts to say “Happy Birthday” to friends.

Here’s how I solved this problem. Let $I_k$ be an indicator random variable for “no friend has a birthday on day $k$, where $k = 366$ stands for February 29 and $k = 1, \dots, 365$ stand for the “usual” 365 days of the year. Therefore, the quantity $N$, representing the number of days of the year on which no friend has a birthday, can be written as

$N = I_1 + \dots + I_{365} + I_{366}$

Let’s start with any of the “usual” days. In any four-year span, there are $4 \times 365 + 1 = 1461$ days, of which only one is February 29. Assuming the birthday’s are evenly distributed (which actually doesn’t happen in real life), the chance that someone’s birthday is not on day $k$ is

$\displaystyle 1 - \frac{4}{1461} = \displaystyle \frac{1457}{1461}$.

Therefore, the chance that all $n$ friends don’t have a birthday on day $k$ is

$\displaystyle \left( \frac{1457}{1461} \right)^n$.

Since the expected value of an indicator random variable is the probability of the event, we see that

$E(I_k) = \displaystyle \left( \frac{1457}{1461} \right)^n$

for $k = 1, \dots, 365$. Similarly, the expected value for the indicator for February 29 is

$E(I_{366}) = \displaystyle \left( \frac{1460}{1461} \right)^n$.

Since $E(X+Y) = E(X) + E(Y)$ even if $X$ and $Y$ are dependent, we therefore conclude that

$E(N) = E(I_1) + \dots + E(I_{365}) + E(I_{366}) = 365 \displaystyle \left( \frac{1457}{1461} \right)^n + \left( \frac{1460}{1461} \right)^n$.

This function is represented by the red dots on the graph below.

In tomorrow’s post, I’ll calculate of the standard deviation of $N$.