Throwing stones into $N$ pots with uniform probability over the pots If I have $N$ pots, and for $k$ iterations I randomly select a pot and throw a stone in it, what distribution do I have for the number of stones in a particular pot?  We can assume $k \geq 100$ or so, and that the number of pots $N$ is $\leq \frac{k}{10}$ or so.  I'd also be open to tightening these assumptions a bit in a direction that lends itself to a nice distribution.
 A: The random number of stones in each pot $X_1, X_2, \ldots, X_N$ follows a multinomial distribution with parameters $p_1 = p_2 = \ldots = p_N = 1/N$ and $k$ trials:  $$\Pr[\boldsymbol X = \boldsymbol x] = \binom{k}{x_1, x_2, \ldots, x_N} p_1^{x_1} p_2^{x_2} \ldots p_N^{x_N} = \frac{k!}{x_1! x_2! \ldots x_N!} \frac{1}{N^k}, $$ where $0 \le x_1, x_2, \ldots x_N \le k$, $\sum x_i = k$, $p_1 = p_2 = \ldots = p_N = \frac{1}{N}$.
So for example, if $N = 5$ and $k = 100$, then the probability that there are $10$, $15$, $20$, $25$, and $30$ stones in the first, second, third, fourth, and fifth pots respectively is $$\Pr[\boldsymbol X = (10, 15, 20, 25, 30)] = \frac{100!}{10!15!20!25!30!} \cdot \frac{1}{5^{100}} \approx 2.49064 \times 10^{-7}.$$
If, however, you want the distribution of the number of stones for a single pot, disregarding the outcome of all the other pots, then that follows a binomial distribution, with parameter $p = 1/N$:  $$\Pr[X = x] = \binom{k}{x} (1/N)^x (1-1/N)^{k-x},$$ where $X$ is the number of stones in the pot of interest.  For instance, in the above example, if we were interested in the probability of seeing $10$ stones in the first pot, then this is $$\Pr[X_1 = 10] = \binom{100}{10} (1/5)^{10} (4/5)^{100-10} \approx 0.00336282.$$
In the case for $N$ and $k$ large, such that $k/N$ is assumed to be some constant rate $\lambda$, then the probability distribution of the number of stones in a single distinguished pot (e.g., the first pot) is Poisson with parameter $\lambda = k/N$:  $$\Pr[X = x] = e^{-\lambda} \frac{\lambda^x}{x!}.$$  Thus with $k = 100$ and $N = 5$ as above, $\lambda = 20$ and $$\Pr[X = 10] = e^{-20} \frac{20^{10}}{10!} \approx 0.00581631.$$  This approximation is better when $\lambda$ is small and $k$ is very large, so comparing $k = 1000$, $N = 250$, $\lambda = 4$, we get $\Pr[X = 10] \approx 0.00529248$ under Poisson, and $\Pr[X = 10] \approx 0.00522368$ under the Binomial distribution.
When $k$ is large but $N$ is small, a Normal approximation is more appropriate, with mean $\mu = k/N$ and variance $\sigma^2 = k/N(1-1/N)$.
