TITLE: Closed form $\int_{0}^{\frac{r}{2}} {\binom{n}{p} \binom{n-p}{r-2p} 2^{r-2p}}{\binom{2n}{r}^{-1}} \ \text{d}p$
QUESTION [5 upvotes]: Note: This is exact copy of my Math.SE question, which I am reposting here, as despite bounty it did not receive any answers.

Let there be $n$ pairs of shoes in a box.
The the probability that from the $r \le n$ shoes I am taking out of the box there are exactly $p$ pairs is given by
\begin{equation*}
        \mathbb{P}_{n}^{(r)}(p)
        = \frac{\binom{n}{p} \binom{n-p}{r-2p} 2^{r-2p}}{\binom{2n}{r}}.
\end{equation*}
For $n = 15$ and $r \in \{6,8,10\}$. The function (assuming the continuous factorial equivalents) looks like this:


I am interested in finding the area under that curve, namely
  $$\int_{0}^{\frac{r}{2}} \frac{\binom{n}{p} \binom{n-p}{r-2p} 2^{r-2p}}{\binom{2n}{r}} \ \text{d}p$$

I consulted this question but could derive how that would help me.
I also thought about writing the first product of binomial coefficients as $$
\binom{n}{n - p}\binom{n - p}{r - 2p}
$$
which is similar to the form $\binom{f(x)}{f(y)} \binom{f(y)}{f(x)}$ mentioned in this question.

REPLY [3 votes]: I suspect some form of CLT for large $n, r$, which may possibly be proved by adopting Laplace's method of approximating the integrand by an appropriate gaussian kernel.
Interestingly enough, we can prove that:
$$ \int_{-\infty}^{\infty} \frac{\binom{n}{p}\binom{n-p}{r-2p}2^{r-2p}}{\binom{2n}{r}}\,\mathrm{d}p = 1. $$
To show this, we may apply the Legendre duplication formula to write
$$ \frac{\binom{n}{p}\binom{n-p}{r-2p}2^{r-2p}}{\binom{2n}{r}} = \frac{n!}{\binom{2n}{r}} \frac{\sqrt{\pi}}{\Gamma(1+p)\Gamma(n-r+1+p)\Gamma(\frac{r}{2}+\frac{1}{2}-p)\Gamma(\frac{r}{2}+1-p)}$$
and then apply the Ramanujan's beta integral (see the formula (5.3.14) of DLMF: 5.13).