Largest expected number of birthday's in a single day I have 1200 friends in Facebook. What is the largest number of them I can (probabilistically) expect to have their birthday on the same day? In general, what would it be if I had N friends?
 A: Here is a partial (asymptotic) answer.
Let's make some simplifying assumptions:

*

*There are 365 days in every year

*Every birthday is independent and uniformly distributed throughout the year

Let $X_1, \dots, X_N$ be the birthdays of your $N$ friends, so according to the above assumptions the $X_i$ are iid distributed uniformly on the set $\{1,2,\dots, 365\}$.  Define $D_j := \frac{1}{N} \#\{ n : X_n = j \}$ to be the proportion of birthdays that fall on day $j$.  The goal is to calculate $\mathbb{E}(\max_{1 \leq j \leq 365} D_j)$.
Proposition: $\mathbb{E}(\max_j D_j) \to \frac{1}{365}$ as $N \to \infty$.
Proof: First note that $\max_j D_j \geq \frac{1}{365}$ always holds, essentially by something like the pigeonhole principle.  Next apply a union bound to conclude
$$
\mathbb{P}\left(\max_j D_j > \frac{1}{365} + \epsilon\right) \leq \mathbb{P}\left(D_1 > \frac{1}{365} + \epsilon\right) + \dots + \mathbb{P}\left(D_{365} > \frac{1}{365} + \epsilon\right).
$$
Notice now that we can also express $D_j$ as the average $\frac{1}{N} \left( 1_{X_1 =j} + \dots 1_{X_N = j} \right)$.  By assumption 2, these indicators are iid with mean $\frac{1}{365}$.  Therefore by the weak LLN, $\mathbb{P}\left(D_{j} > \frac{1}{365} + \epsilon\right) \to 0$ as $N \to \infty$ for any $\epsilon > 0$ and for each $j$.
Thus we have shown that $\mathbb{P}\left(\max_j D_{j} > \frac{1}{365} + \epsilon\right) \to 0$ as $N \to \infty$ for any fixed $\epsilon > 0$.  We already know that $1 \geq\max_j D_j \geq \frac{1}{365}$ deterministically, so this implies that $\mathbb{E}(\max_j D_j) \to \frac{1}{365}$ for the following reason: we know that $\max_j D_j$ is a random variable which takes values in $[\frac{1}{365}, \frac{1}{365} + \epsilon]$ with probability $1-o(1)$ and values in $(\frac{1}{365} + \epsilon, 1]$ with probability $o(1)$.  Therefore
$$
\frac{1}{365} \leq \mathbb{E} (\max_j D_j) \leq (\frac{1}{365} + \epsilon)(1+o(1)) + 1 \cdot o(1)
$$
Taking $N \to \infty$ and then $\epsilon \to 0$ gives the result. $\square$
Note: It's possible to get a more quantitative estimate for finite $N$ by applying a large deviation inequality like Hoeffding's inequality instead of the weak LLN.
