TITLE: Identity for convolution of central binomial coefficients: $\sum\limits_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k}=2^{2n}$
QUESTION [89 upvotes]: It's not difficult to show that
$$(1-z^2)^{-1/2}=\sum_{n=0}^\infty \binom{2n}{n}2^{-2n}z^{2n}$$
On the other hand, we have $(1-z^2)^{-1}=\sum z^{2n}$.  Squaring the first power series and comparing terms gives us
$$\sum_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k}2^{-2n}=1$$
that is,
$$\sum_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k}=2^{2n}$$
My question: is there a more direct, combinatorial proof of this identity?  I've been racking my brains trying to come up with one but I'm not having much success.

REPLY [11 votes]: There is also a probabilistic proof of this identity.
Start with an urn containing one red marble and one blue marble. Make a series of $n$ draws from the urn; for each draw, remove a random ball in the urn, then put it back, along with two extra balls of the same color. We then ask, what is the probability that exactly $k$ of the draws were red?
The probability that the first $k$ draws are red and the last $n-k$ are blue is
$$
\frac12\cdot\frac{3}4\cdot\frac5{6}\cdots\cdot\frac{2k-1}{2k}\cdot\frac1{2k+2}\cdot\frac{3}{2k+4}\cdots\frac{2(n-k)-1}{2n}=\frac{(2k-1)!!(2(n-k)-1)!!}{(2n)!!}
$$
where $n!!=\prod_{k=0}^{\lfloor n/2\rfloor-1}(n-2k)=n(n-2)(n-4)\cdots$.
It is not hard to see that every sequence of $k$ red and $n-k$ blue draws has this same probability; rearranging the order of draws just changes the order of the factors in the numerator. Therefore, the probability of $k$ red draws is, using the identities $(2n)!!=2^nn!$ and $(2k-1)!!=\frac{(2k)!}{(2k)!!}=\frac{(2k)!}{2^kk!}$,
$$
\binom{n}k\frac{(2k-1)!!(2(n-k)-1)!!}{(2n)!!}=\frac{\binom{2k}k\binom{2(n-k)}{n-k}}{2^{2n}}
$$
Since these probabilities must sum to $1$, the desired identity follows!