One geometric wait at least four times another

Two people each toss a fair coin until they see their first head; let $T$ and $G$ be the number of tosses for the first and second person. What is the probability that $T\ge4G$?

$2/31$. Condition on $G=g$ (probability $2^{-g}$). Then $T\ge4g$ means the first $4g-1$ tosses are tails, probability $2^{1-4g}$. Summing, $\sum_{g\ge1}2\cdot32^{-g}=\tfrac{2}{31}$.