I have the following problem: n balls are dropped, independently, in n different urns and we want to calculate $\mathbb{E}V_{n}$, where $V_{n}$ is the number of urns with zero balls.
I know that the zero count problem is generally difficult, but I think that this case can be simple because the distribution of the number of balls in the urns is multinomial with the same $p$ for each urn, which is $1/n$. So the probability that a given urn is empty, after dropping the balls, is $ q=(n-1)^{n} / n^{n} $.
My first thought was that the distribution of $V_n$ is binomial$(n-1,q)$, and thus $\mathbb{E}V_n = (n-1)q$. I did a lot of computational simulations of this problem to check my hunch and got a numerical value very very close to $np$ as if its distribution were binomial$(n,p)$.
This has me very confused becasuse $V_n$ can not be $n$. So I want to know if the distribution of $V_n$ is in fact some of the familiar distributions like the binomial or on the other hand, if there is some way to calculate $ \mathbb {E} V_n $ without worrying about its distribution.
Thank you very much.
For each of the urns $i = 1,2,...,n,$ let $X_i$ be the random variable which takes the value 1 if urn $i$ is empty, and 0 otherwise. Then $$\sum_{i=1}^n X_i = V_n$$ By linearity of expectation, $$\mathbb{E}[V_n] = \sum_{i=1}^n \mathbb{E}[X_i].$$ As you point out in your problem description, the probability that a particular urn is empty is $\frac{(n-1)^n}{n^n}.$ Therefore, for all $i \in \{1,2,...,n\},$ $\mathbb{E}[X_i] = \frac{(n-1)^n}{n^n}.$ It follows that $$\mathbb{E}[V_n] = n \frac{(n-1)^n}{n^n} = \frac{(n-1)^n}{n^{n-1}}.$$