Focus on one particular box

Firstly, we focus on one box only:

For each thrown stone, the probability that a stones falls in this particular box is $\frac{1}{M}$. The probability that it falls on another box is $1-\frac{1}{M}=\frac{M-1}{M}$.

Figure 4: Probability tree leads to a binomial distribution.

If we throw $N$ stones, we can build a binary tree of height $N$. In each tree's node, we choose the left path if the stone falls into the particular box we focus. The right path is chosen if it falls in another one. The probability that no stone falls in our box after $N$ throws is $p_{none} = \frac{M-1}{M}^N$ and follows the rightmost arm of the tree with height $N$.

In general, the probability to have $k$ stones into one box follows the binomial distribution. With $p=\frac{1}{M}$, we state

$$p(k) = \binom{N}{k} p^{k} (1-p)^{N-k}.$$

In N binomial k, we select all paths in the tree with $k$ stones inside the box and we assign to each sub path the correct probability depending if it was a success or a failure. In short, we have a $B(n,p)$ probability distribution for each box of our problem where $n=N$ and $p=\frac{1}{M}$. $E[B(n,p)]$ is $\frac{N}{M}$.

Figure 5: Dependence versus independence

To compute the standard deviation, one can choose a Poisson distribution as approximation for $B(n,p)$. This is because $n$ is big ($N$) and $p$ is small ( $\frac{1}{M}$). For the Poisson distribution, $\lambda$ is then $\lambda=n P=\frac{N}{M}$. The variance of the Poisson distribution is $\lambda=\frac{N}{M}$, The standard deviation for Poisson is therefore $\sqrt(\frac{N}{M})$. Alternatively, one can compute $\sigma$ for $B(n,p)$ using $Var(k)=E(k^2)-E(k)^2$ and a differentiation trick. For large $M$s, standard deviation and variance are the same for both distributions, $N (1-\frac{1}{M}) \frac{1}{M}$ for Binomial and $\frac{N}{M}$ for Poisson.

Here, it is important to notice the following: while $p_{none}=(\frac{M-1}{M})^{N}$ is the probability that one particular box is empty, it is not possible to generalize with $(1-p_{none})^{M}$ to the problem where all boxes have at least one stone; the boxes are not independent from each other: a stone that does not land into a box does not fall apart but lands in another box, as in Figure 5.