I came across these numbers in my work some time ago. This type of expressions do not exist in closed form (not to confuse with Vandermonde convolution), I already know that. To simplify I denote $$P(r;s;k)=\binom{\frac{n}{2}+r}{k} \binom{\frac{n}{2}-r}{k+s}$$
Here is why I think these numbers are interesting: anyone can see combinatorial structures in algorithms or elsewhere, but these expressions converge.
Experiment 1, $r=0, s=0:$ There are $n$ fair coins, of them half are heads $H$, half are tails $T$. Each coin is flipped w.p. $\frac{1}{n}$, i.e. proportional to size of the sample. The question is, after the experiment is over, what is the probability that the ratio of $H$/$T$ remains the same, i.e. half and half. The tricky thing is, in order to preserve the ratio, we need to flip an $\mathit{even}$ number of coins, or, more specifically, an equal number of $H$ and $T$ (obviously two $H$ or two $T$ won't do): either 0 $H$ and 0 $T$, or 1 $H$ and 1 $T$ and so on till $\frac{n}{2} \ H, \frac{n}{2} \ T$. The probability to preserve the ratio (and the first such number) is therefore $$ h_{0,0}= \lim _{n \to \infty} \sum_{k=0}^{\frac{n}{2}} P(0;0;k) \frac{1}{n^{2k}} \bigg(1-\frac{1}{n}\bigg)^{n-2k} \approx 0.465267 $$
Experiment 2, $r=0,s=1$: same setup as before, but now I need to obtain $\textit{exactly}$ one more $H$ coin as a result of the experiment (or $T$, it does not matter by symmetry). To do this, I need to flip $exactly$ one more $T$ than $H$, hence we need to flip an $odd$ number of coins:
$$ h_{0,1}= \lim _{n \to \infty} \sum_{k=0}^{\frac{n}{2}-1} P(0;1;k) \frac{1}{n^{2k+1}} \bigg(1-\frac{1}{n}\bigg)^{n-2k-1} \approx 0.208912 $$
Experiment 3, $r=1,s=0$. This comes directly after experiment 2: we have $\frac{n}{2}+1 \ H$ and $\frac{n}{2}-1 \ T$. We want the same result as in Experiment 1: maintain the current proportion of $H$ and $T$ (i.e. flip an even number of coins), but clearly this time around the upper bound on the summation is $\frac{n}{2}-1$.
$$ h_{1,0}= \lim _{n \to \infty} \sum_{k=0}^{\frac{n}{2}-1} P(1;0;k) \frac{1}{n^{2k}} \bigg(1-\frac{1}{n}\bigg)^{n-2k} \approx 0.465225 $$
and so on for $h_{r,s}, 0 \leq r,s \leq n$. So here is what I'd like to know:
1) Have these numbers arisen in some other context (I never ended up publishing a paper)
2) Is there some rigorous way to prove these numbers are irrational? I've shown the $\lim$ part simply by taking large $n$, so I guess this not rigorous enough.
3) If there is something I've missed, I'd like to know too.
Thanks for the suggestions.