Let $X$ be a random variable following the hypergeometric distribution with parameters $N,K,n$, where \begin{equation} Pr(X=k) = \frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}. \end{equation} To make things easier, we bring symmetry and avoid the boundry issues by letting $K=\frac{N}{2}$ and $n<K$. For an odd $n$, we know by symmetry \begin{equation} Pr(X~even)=Pr(X~odd). \end{equation} For an even $n$, my conjecture is that \begin{equation} \vert Pr(X~even)-Pr(X~odd) \vert \leq \frac{\binom{N}{n/2}}{\binom{N}{n}}. \end{equation} I ran some simulations and the result shows that it should be true, but I have not been able to prove it so far. Could someone help me with that?

If this question is answered, a subsequent question is the case where $K$ is roughly $\frac{N}{2}$ but not exact. A corresponding condition is $\vert K-\frac{N}{2}\vert \leq c N$, for a constant $0<c<0.5$. We still assume $n<min (K, N-K)$ to avoid boundary issues. The conjecture is still the same in this case.

I try not to be greedy and will be much satisfied if only the first question is answered. Many thanks in advance!