Suppose $x$ is randomly sampled uniformly from $[0,N)$. Suppose we take $M$ samples. In my testing, I have been using $N=50000, M = 1000000$. The expected number of times each $i \in[0,N)$ was sampled is $M/N$. The shape of this distrubition is uniform.
However, suppose we took this random uniform distribution and counted, not the number of times $x$ was $0$, but the number of values which were sampled $0$ times. Let's call that the $\mathbf{freq}(0)$. We can get the $\mathbf{freq}(j)$ values for all $[0,M]$. This distrubution is not uniform. It looks kind of like a Poisson distrubution centered at $M/N$, but I'm not sure.
In case I wasn't totally clear, here is an example. Suppose we randomly sample uniformly from $[0,10)$ 54 times. The number of times each was sampled is a random uniform distribution looking something like $\{5, 4, 5, 5, 6, 4, 3, 5, 6, 7, 4\}$, meaning that among the 54 samples, 0 was sampled 5 times, 1 was sampled 4 times, 2 was sampled 5 times, etc. But the distribution of these frequencies is $\{0, 0, 0, 1, 3, 4, 2, 1,0, 0\dots\}$ since there is one value which was sampled three times, 3 values which were sampled four times, 4 values which were sampled five times, etc. What is the shape of this distribution?
The probability that any fixed number in $[0, N)$ gets sampled once is $\frac{1}{N}$, so the distribution of the number of times it gets sampled in total is a binomial distribution $\text{Bin} \left( M, \frac{1}{N} \right)$. If $M, N$ are large and about the same size this approaches a Poisson distribution $\text{Pois} \left( \frac{M}{N} \right)$ by the Poisson limit theorem. These Poisson distributions for different numbers in $[0, N)$ are, it turns out, also approximately independent, so we approximately get $N$ independent samples from this Poisson distribution.