Suppose that $n$ balls are randomly thrown into $N$ bins. We can compute the expected number of bins that contain at least one ball as $E(X) = N(1 - (1 - 1/N)^n)$.
Now, suppose that instead we are asked for the expected number $E(Y)$ of bins that contain more than one ball. Can we recover this from $n$ and $E(X)$? My intuition is suggesting that this ought to be $E(Y) = n - E(X)$, but I'm not sure one way or the other.
My intuition is coming from the pigeon hole principle, I think. If $E(X)$ bins are occupied, and there are $n$ balls, then $n - E(X)$ balls are in a bin with at least one more ball in it. Granted, that doesn't tell me how the $n - E(X)$ balls are distributed in the bins.
I am trying to compute the expected number of bins with exactly one ball directly. Let $N_i$ be the number of balls in the $i$th bin, and let $Z_i = 1$ if $N_i = 1$, and $Z_i = 0$ otherwise.
Each $N_i$ is binomial with parameters $n$ and $p = 1/N$. Then $E(Z_i) = P(Z_i = 1) = P(N_i = 1) = \frac{n}{N} \left(\frac{N-1}{N}\right)^{n-1}$. So $E(Z) = n\left(\frac{N-1}{N}\right)^{n-1}$.
The expected number of bins with at least one ball is
$$ E(\geq 1) = N(1-(1-1/N)^n) $$
and the expected number with precisely one ball is (from the binomial distribution)
$$ E(1)=N(1-1/N)^{n-1} $$
so
$$ E(>1)=E(\geq 1)-E(1) = N(1-(1-1/N)^{n-1}(2-1/N)) $$
which doesn't seem to equal $n-E(\geq 1)$.