So maybe i missed something simple, but I could not find material related to the following problem:
Assume we have an urn with $N$ balls of which $k$ are colored black. Assume that $k$ is some fixed number $k \ll N$. I am interested in the number of balls required to draw (without replacement) to have some (fixed) probability $p$ (e.g. $p=\frac{1}{2}$) of drawing at least one black ball. If we let $f_k$ be a function describing the number of samples needed to achieve probability $p$, it seems that roughly $$f_k(N) \approx g(k)\cdot N.$$ For example, for $k=5$, calculating the probability yields the following for $N=100,1000,10000,100000:$ $$f_5(100) = 13, f_5(1000) = 130, f_5(10000) = 1295, f_5(100000) = 12945.$$ So in this case it seems that $g(5) \approx 0.12945$, i.e. we need to draw approximately $13\%$ of the balls to achieve a probability of at least $\frac{1}{2}$.
I am now mainly interested whether, for fixed $p$, there is a closed form (or strong approximation) for this function $g$, i.e. given fixed $k$ and $p$, i want to know what fraction of balls i have to draw to achieve probability at least $p$ of drawing a black ball.
Now from the probabilistic viewpoint, it seems that this problem is somewhat related to hypergeometric distributions. For fixed number of samples $n$, the probability of drawing at least one black ball is given by $$\mathbb{P}["\text{Atleast one black ball in $n$ draws}"] = 1 - \prod_{j=0}^{n-1} \frac{N-k-j}{N-j} = 1 - \frac{{N-k \choose n}}{{N \choose n}},$$ but it is unclear to me how to go from here to finding the number of draws required to achieve some set probability.
Thanks for your help!
If we apply heavy approximation we might get
$1-p=\frac{{N-k \choose n}}{{N \choose n}} = \frac{(N-k)!(N-n)!}{(N-k-n)!N!} \approx {\frac{(N-k)^{N-k+1/2}(N-n)^{N-n+1/2}}{(N-k-n)^{N-k-n+1/2}N^{N+1/2}}} \approx {\frac{(N-n)^{k}}{N^{k}}} =\left(1-\frac nN\right)^k$
with the first approximation using Stirling's approximation and the second assuming that $k \ll N-n$ and cancelling similarly sized terms
That might suggest that $$n \approx N\left(1-(1-p)^{1/k}\right)$$
If $p=\frac12$ and $k=5$, as in your examples, then this suggests $n \approx 0.1294494\, N$ which is essentially what you found with rounding up. In other cases the approximation may not be quite as good.