A random variable ${\displaystyle X}$ follows the hypergeometric distribution if its probability mass function (pmf) is given by
$p_{X}(k)=\operatorname{Pr}(X=k)=\frac{\left(\begin{array}{c} K \\ k \end{array}\right)\left(\begin{array}{l} N-K \\ n-k \end{array}\right)}{\left(\begin{array}{l} N \\ n \end{array}\right)}$
It is written ${\textstyle X\sim \operatorname {Hypergeometric} (N,K,n)}$.
I am trying to map the random variable $X$ to another binary random variable $Y$ as follows:
$Y=0 $ if $X=0 $ else $Y=1$.
Moreover, I want $Y$ to have a uniform distribution. I am trying to figure out what value of $K$ can lead to this.
In summary, my question can be stated as below:
What value of $K$ (exactly or approximately) in terms of $N$ and $n$ results in $p_{X}(0)=\operatorname{Pr}(X=0)=1/2?$
Using the binomial approximation to the hypergeometric distribution, $$ P(X=0)\approx P\big(\operatorname{Bin}(n,\tfrac{K}{N})=0\big)=(1-K/N)^n $$ For this to be equal to $1/2$, we would need $$ \boxed{K= N\cdot(1- 2^{-1/n}).} $$ More precisely, $K$ would need to be the closest integer to $N(1-2^{-1/n})$.
This only works if the binomial distribution is a good approximation to the hypergeometric distribution. According to [Holmes], the error in this approximation is at most $\frac{n-1}{N-1}$. Therefore, for the above value of $K$, the hypergeometric probability will be withing $\frac{n-1}{N-1}$ of $\frac12$, which is very close as long as $N$ is large compared to $n$.