Figuring out the variance of the birthday paradox Update to reflect what I think is the calculation for $E[X_1\cdot X_2]$
Given n people, if I want to estimate how many of them are likely to have an overlapping birthday with any other person, how do I calculate the variance?
So far I have $E[X]=n\cdot p$ with $n$ as the number of people and $p$ the probability of any single person having an overlapping birthday
I think $E[X_1\cdot X_2]$ is as follows: 
Using total probability+the fact that $X_i$ is an indicator variable:
$E[X_1\cdot X_2]=P_{X_1,X_2}(1)=P(X_1\cdot X_2 |X_1=1)+P(X_1\cdot X_2 |X_1=0)$
Now again as $X_i$ is an indicator variable:
$P(X_1\cdot X_2 |X_1=0)=0$
Now from Bayes conditioning:
$P(X_1\cdot X_2 |X_1=1)=P(X_1 $and$ X_2)\cdot P(X_2)$
This is: $P(X_1\cdot X_2 |X_1=1)\cdot (1-(\frac{364}{365})^n)$
What I'm not sure about is whether: $P(X_1 $and$ X_2)=(1-(\frac{363}{364})^{n-3}+ \frac{1}{365})$
If I'm right:$P(X_1\cdot X_2 |X_1=1)\cdot (1-(\frac{364}{365})^n)=(1-(\frac{363}{364})^{n-3}+ \frac{1}{365})\cdot (1-(\frac{364}{365})^n)=E[X_1\cdot X_2]$
 A: For $i=1,\dots,n$ let $X_i$ take value $1$ if person $i$ shares his birthday with some of the other persons and let $X_i$ take value $0$ otherwise.
Then $X=X_1+\cdots+X_n$ is the number of persons that share their birthday with someone else of the persons.
For finding variance you can make use of the bilinearity of the covariance and of symmetry:$$\mathsf{Var}(X)=\mathsf{Cov}(X,X)=\sum_{i=1}^n\sum_{j=1}^n\mathsf{Cov}(X_i,X_j)=n\mathsf{Var}(X_1)+n(n-1)\mathsf{Cov}(X_1,X_2)$$
If $p$ is the probability of any singular person to share his birthday with someone else then $$\mathbb EX_1=P(X_1=1)=p=1-\left(\frac{364}{365}\right)^{n-1}$$Further we have:
$$\mathsf{Var}(X_1)=p(1-p)$$and: $$\mathsf{Cov}(X_1,X_2)=\mathbb EX_1X_2-\mathbb EX_1\mathbb EX_2=\mathbb EX_1X_2-p^2$$
So there is one thing left to find now which is: $$\mathbb EX_1X_2=P(X_1=1=X_2)$$
Can you find that yourself?

addendum:
$\begin{aligned}P\left(X_{1}=1=X_{2}\right) & =P\left(X_{1}=1\right)+P\left(X_{2}=1\right)+P\left(X_{1}=0=X_{2}\right)-1\\
 & =2p-1+P\left(X_{1}=0=X_{2}\right)\\
 & =2\left(1-\left(\frac{364}{365}\right)^{n}\right)-1+\frac{364}{365}\left(\frac{363}{365}\right)^{n-2}\\
 & =1-2\left(\frac{364}{365}\right)^{n}+\frac{364}{365}\left(\frac{363}{365}\right)^{n-2}
\end{aligned}
$
