If we keep drawing iid samples from a discrete uniform distribution, $\operatorname{unif}\{a,b\}$, eventually we are guaranteed to get a value that we already got. What is the expected value for the number of samples that must be drawn from the distribution for us to get a repeated value? Is there a closed form solution for this?
Progress I've made
For a distribution $\operatorname{unif}\{a,b\}$, let $N = b - a + 1$.
Obviously the expectation should be somewhere between $2$ and $N+1$. For a given $n$ such that $1 \leq n \leq N+1$, lets try to find the probability that the first repetition happens on the $nth$ sample. In the other words, we want the probability that the first $n-1$ samples DID NOT have a repetition AND the $nth$ sample DID have a repetition.
Probability that the first $n$ samples did not have a repetition is $$\frac{N}{N} \frac{N-1}{N}\frac{N-2}{N} ... \frac{N-(n-1)}{N} = \frac{\frac{N!}{(N-n)!}}{N^n} = \frac{N!}{N^n(N-n)!}$$
So, probability that the first $n - 1$ samples did not have a repetition is $$\frac{N!}{N^{n-1}(N-(n-1))!} = \frac{N!}{N^{n-1}(N-n+1)!}$$
Probability that the $nth$ sample did have a repetition, is simply $\frac{n-1}{N}$, since there are $n-1$ values already drawn.
This gives us a probability mass function for the first repetition $$\operatorname{pmf}(n) = \frac{N!}{N^n(N-n+1)!} \cdot (n-1)$$
Finally, the expected value is $$ \sum_{n=1}^{N+1} \frac{N!}{N^n(N-n+1)!} \cdot (n-1) \cdot n$$
That's pretty much where I'm stuck. Does this have a closed form solution? Or is there a way to reformulate the approach to get a better answer.
