Suppose there are $N$ marbles in an urn. $K \ll N$ of those marbles are red, the rest are black. How can I determine the expected number of draws without replacement before I draw a red marble?
I think this is related to a hyper-geometric distribution, but I'm not sure how to rearrange the formula to get what I need.
I've also identified that the result I want can be given by the recurrence $$ g(n) = \begin{cases} 1 & n \leq K \\ \frac{K}{n}\cdot 1 + (1 - \frac{K}{n}) \cdot g(n - 1) & \text{otherwise} \end{cases} $$ evaluated at $N$. Unfortunately, I need a closed form, if one exists.
Thanks!
The probability that you get $k$ black marbles followed by a red marble is:
$$\dfrac{(N-K)(N-K-1)\cdots (N-K-k+1)}{N(N-1)\cdots (N-k+1)}\cdot \dfrac{K}{N-k}$$
Rewriting in terms of factorials, we have:
$$\dfrac{\tfrac{(N-K)!}{(N-K-k)!}}{\tfrac{N!}{(N-k)!}}\cdot \dfrac{K}{N-k} = \dfrac{(N-K)!(N-k-1)!}{N!(N-K-k)!}\cdot K$$
Since this occurs after $k+1$ draws of marbles, and each outcome is mutually exclusive (you cannot get the first red marble on draws $k_1$ and $k_2$ unless $k_1=k_2$), you can write:
$$g(N) = \sum_{k=0}^{N-K}(k+1)\cdot K\cdot \dfrac{(N-K)!(N-k-1)!}{N!(N-K-k)!}$$
According to Wolframalpha, a closed form appears to be:
$$g(N) = \dfrac{N+1}{K+1}$$