Consider a group of $n$ people. Assume that each person's birthday is drawn uniformly at random from the $365$ possibilities. What is the smallest value
of such that the expected number of pairs of distinct people with the same birthday is at least one?.
My approach using indicator random variable:
$X_{\bf i, j, i \neq j} = \begin{cases} &1 \qquad \text{ if ith and jth person having same birthday} \\ &0 \qquad \text{ if ith and jth person NOT having same birthday } \ \end{cases} \\\\$
$ \begin{align*} & E(X_{i,j}) = 1*P(X_{i,j}) + 0*\left ( 1 - P(X_{i,j}) \right ) \\ & \Rightarrow E(X_{i,j}) = P(X_{i,j}) \\ & \Rightarrow \text{Overall E} = \sum E_{i,j} \\ \\ \hline \\ &\text{ We need } E \geq 1 \\ &\Rightarrow \left [ \sum_{i=1}^{k}\sum_{j=i+1}^{k}E_{i,j} \right ] \geq 1 \\ &\Rightarrow \left [ \sum_{i=1}^{k}\sum_{j=i+1}^{k}P_{i,j} \right ] \geq 1 \\ &\Rightarrow \left [ \sum_{i=1}^{k}\sum_{j=i+1}^{k}\left ( \frac{1}{365} \right ) \right ] \geq 1 \\ &\Rightarrow \left ( \frac{1}{365} \right ) \left [ \sum_{i=1}^{k}\sum_{j=i+1}^{k}1 \right ] \geq 1 \\ & \Rightarrow \left ( \frac{1}{365} \right )\left [ \sum_{i=1}^{k-1}1 \right ] \geq 1\\ & \Rightarrow \left ( \frac{1}{365} \right )\left [ \frac{k.(k-1)}{2} \right ] \geq 1\\ & \Rightarrow k^2 - k - 730 \geq 0\\ & \Rightarrow k = \left \lceil \frac{1+ \sqrt{1+4*730}}{2} \right \rceil = \left \lceil 27.52 \right \rceil = 28 \\ \\ \end{align*}$
I hope there may be other ways to do this problem and there are other variations to this problem as well (a few are in Kenneth Rosen Book). Please help if there exist other methods.
Thanks!