Probabalistic formulation of the identity $4^{-n} \sum_{k=0}^n \binom{2k}{k}\binom{2n-2k}{n-k} = 1$?

Question

I was wondering if there was a probabilistic way to interpret the combinatorial identity $$\sum_{k=0}^n \binom{2k}{k}\binom{2n-2k}{n-k} = 4^n.$$ There are a number of ways to prove this result without probability (using generating functions, for example), however I was hoping there might be a more elegant probabilistic interpretation. By rewriting $$\sum_{k=0}^n \binom{2k}{k}\binom{2n-2k}{n-k} = \binom{2n}{n}\sum_{k=0}^n \frac{\binom{n}{k}^2}{\binom{2n}{2k}},$$ it appears that it might be possible to interpret this as a sort of "broken parts problem": given $2n$ lightbulbs, half of which are defective with probability $1/2$, you sample $2k$ of them. Then $\binom{n}{k}^2/\binom{2n}{2k}$ is the probability half of your sample is defective and the other half works. However, it is hard to see how summing these probabilities cancels the factor of $4^{-n}\binom{2n}{n}$.