First of three alternating 50 percent shooters to hit

Three marksmen $X$, $Y$ and $Z$ fire at a target in the repeating order $X,Y,Z,X,Y,Z,\dots$, stopping as soon as someone hits. Every shot hits with probability $\tfrac12$, independently. What is the probability that $X$ is the one who hits first?

$\tfrac47$. The first hit is on shot $k$ with probability $2^{-k}$, and $X$ fires shots $1,4,7,\dots$. Summing, $\tfrac12\left(1+\tfrac18+\tfrac1{64}+\cdots\right)=\tfrac12\cdot\tfrac87=\tfrac47$. Equivalently, the three win chances are in ratio $1:\tfrac12:\tfrac14$.