Confusion about Banach Matchbox problem

Apart from doubling $p$ at the end, your answer is correct: your denominator is actually equal to $1$. It can be rewritten as

$$\frac1{2^{2n}}\sum_{i=0}^n\binom{2n-i}n2^i=\frac1{2^{2n}}\sum_{m=n}^{2n}\binom{m}n2^{2n-m}=\frac1{2^{2n}}\sum_{i=0}^n\binom{n+i}n2^{n-i}\;,$$

and

$$\begin{align*} \sum_{i=0}^n\binom{n+i}n2^{n-i}&=\sum_{i=0}^n\binom{n+i}n\sum_{k=0}^{n-i}\binom{n-i}k\\\\ &=\sum_{i=0}^n\sum_{k=0}^{n-i}\binom{n+i}i\binom{n-i}k\\\\ &=\sum_{k=0}^n\sum_{i=0}^{n-k}\binom{n+i}n\binom{n-i}k\\\\ &\overset{*}=\sum_{k=0}^n\binom{2n+1}{n+k+1}\\\\ &=\sum_{k=n+1}^{2n+1}\binom{2n+1}k\\\\ &=\frac12\sum_{k=0}^{2n+1}\binom{2n+1}k\\\\ &=2^{2n}\;, \end{align*}$$

where the starred step invokes identity $(5.26)$ of Graham, Knuth, & Patashnik, Concrete Mathematics. Thus, your result can be simplified to

$$p=\binom{2n-k}n\left(\frac12\right)^{2n-k}\;.$$

And you don’t want to multiply this by $2$: no matter which pocket empties first, this is the probability that the other pocket still contains $k$ matches.