Expectation and Variance, number of balls Question:
n boxes are ordered in a row on a table , labeled 1 to n.
In every box there's a ball. In every step Nicole chooses a ball randomly (in uniform distribution) , puts it out of the box, then chooses a box and puts the ball in it (again, in uniform distribution, between the n boxes).
$X_1$ is the the number of balls in box number 1 on the 37th step.
What is $E[X_1], V[X_1]$
Thoughts:
I understood the solution is 1 for both, from "symmetry", But when i try to calculate them formally (using the definition of expectation), It doesn't work out ...
 A: Each ball starts in one of the boxes; think of the ball that starts in box $j$ as "ball $j$". Write
$$
X_1=\sum_{j=1}^{n}1_{\{\text{Ball $j$ is in box 1 after 37 steps}\}}.
$$
What is $P(\text{Ball $j$ is in box 1 after 37 steps})$? 
If $j\neq 1$, then 
$$
P(\text{Ball $j$ is in box 1 after 37 steps})=\left(1-\left(1-\frac{1}{n}\right)^{37}\right)\cdot\frac{1}{n}.
$$
Why?  For ball $j$ to end in box 1, it must have been chosen at least once; the probability that this happens is the first term.  Once we know ball $j$ has been chosen at least once, its position is uniformly random, hence the $\frac{1}{n}$.  This is true for all $j\neq 1$; for $j=1$, we have
$$
P(\text{Ball $1$ is in box 1 after 37 steps})=\left(1-\frac{1}{n}\right)^{37}+\left(1-\left(1-\frac{1}{n}\right)^{37}\right)\cdot\frac{1}{n};
$$
the second term is just like before, while the first term corresponds to the possibility that ball 1 has never been chosen - which leaves it in box 1.
Breaking up $\mathbb{E}[X_1]$ over summation, we have
$$
\mathbb{E}[X_1]=\left(1-\frac{1}{n}\right)^{37}+\left(1-\left(1-\frac{1}{n}\right)^{37}\right)=1,
$$
as you claimed.
Try using these same indicators to compute $\mathbb{E}[X_1^2]$ (or, even better, $\mathbb{E}[X_1(X_1-1)]$) to get at the variance.
