Distribution of the variance of simple random samples Suppose $X_1, \cdots, X_n$ are i.d.d. samples from population $X \sim B(1,p)$, i.e., 
$$ P(X_i = 1) = p, P(X_i = 0) = 1-p$$
And the sample variance is denoted by $S_n$, where 
$$ S_n = \sum_{i = 1}^n \frac{(X_i - \overline{X})^2}{n}$$
where $\overline{X}$ is the sample mean:
$$\overline{X} = \frac{1}{n}\sum_{i=1}^n X_i$$
Then what is the distribution of $S_n$?
My solution is as follows: 
Clearly $X_i = X_i^2$, for that the only two possible values of $X_i$ are 1 and 0. Now, by expanding $S_n$, we get
$$S_n = \frac{1}{n}\sum_{i=1}^n X_i^2 - {\overline{X}}\ ^2 = \frac{1}{n}\sum_{i=1}^n X_i - {\overline{X}}\ ^2 = \overline{X}(1-\overline{X})$$
By the additivity of $X_i$'s, $n\overline{X}\sim B(n,p)$, and $n(1-\overline{X}) \sim B(n,1-p)$
How do I proceed from here to get the distribution of $S_n$, or is there any mistake?
 A: When $\overline X=0$ or $\overline X=1$, $\overline X(1-\overline X)=0$. 
Then $$\mathbb P(S_n=0)=\mathbb P(\overline X=0)+\mathbb P(\overline X=1)=(1-p)^n+p^n.
$$
When $\overline X=\frac1n$ or $\overline X=\frac{n-1}n$, $\overline X(1-\overline X)=\frac{n-1}{n^2}$. 
Then $$\mathbb P\left(S_n=\frac{n-1}{n^2}\right)=\mathbb P\left(\overline X=\frac1n\right)+\mathbb P\left(\overline X=\frac{n-1}n\right)=np(1-p)^{n-1}+n(1-p)p^{n-1}.
$$
When $\overline X=\frac2n$ or $\overline X=\frac{n-2}n$, $\overline X(1-\overline X)=\frac{2(n-2)}{n^2}$. 
Then $$\mathbb P\left(S_n=\frac{2(n-2)}{n^2}\right)=\mathbb P\left(\overline X=\frac2n\right)+\mathbb P\left(\overline X=\frac{n-2}n\right)=\binom{n}{2}p^2(1-p)^{n-2}+\binom{n}{2}(1-p)^2p^{n-2}.
$$
And so on. If $n$ is odd then for all $k=0,1,\ldots,\frac{n-1}{2}$
$$\mathbb P\left(S_n=\frac{k(n-k)}{n^2}\right)=\binom{n}{k}p^k(1-p)^{n-k}+\binom{n}{k}(1-p)^kp^{n-k}.$$
If $n$ is odd then for all $k=0,1,\ldots,\frac{n}{2}-1$ probabilities are the same
$$\mathbb P\left(S_n=\frac{k(n-k)}{n^2}\right)=\binom{n}{k}p^k(1-p)^{n-k}+\binom{n}{k}(1-p)^kp^{n-k}$$
and for $k=\frac{n}2$ it is only one possibility to get value $S_n=\frac14$
$$\mathbb P\left(S_n=\frac{k(n-k)}{n^2}=\frac14\right)=\binom{n}{n/2}p^{n/2}(1-p)^{n/2}.$$
