If I put $M$ balls into $N$ boxes at random, what is the average number of balls in the boxes that are not empty?

Question

I have a very brief question: if I put $M$ balls into $N$ boxes at random, what is the average number of balls in the boxes that are not empty?

Answer 1

Let $X$ denotes the number of non-empty boxes.

Then $P(X=r)={N\choose r}\left(\frac{1}{2}\right)^r\left(\frac{1}{2}\right)^{N-r}={N\choose r}\left(\frac{1}{2}\right)^N$ (assuming binomial distribution)

Let $E(Y)$ denotes the average number of balls in non-empty boxes,

then , $E(Y)|(X=r)=\frac{M}{r}$ (assuming uniform distribution of balls in non-empty boxes)

Then $E(Y)=\sum_{r=1}^NP(X=r)E(Y)|(X=r)=\frac{M}{2^N}\sum_{r=1}^N\frac{{N\choose r}}{r}$

Answer 2

Let $A$ be the number of non-empty boxes. Then the average number of balls in each box=$\displaystyle{\frac{M}{A}}$.

In random distribution, the value of $A$ may vary.

Probability of $A$ boxes being selected= $\displaystyle{\frac{\binom{N}{A}}{\binom{N}{1}+\binom{N}{2}+\dots \binom{N}{M}}}$

Hence, expected value of average=$\displaystyle{\sum_{A=1}^{M} \frac{M}{A}.{\frac{\binom{N}{A}}{\binom{N}{1}+\binom{N}{2}+\dots \binom{N}{M}}}}$