Four times as long to the first head

MediumProbability Random Variables~6m

Two people flip fair coins independently, each until their first head. Let $G$ and $T$ be the number of flips each needs. Find $\mathbb{P}[\,T \ge 4G\,]$ - the probability the second person needs at least four times as many flips as the first. Writing it as $\tfrac{p}{q}$ in lowest terms, report $p + q$ .

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Both are Geometric $(1/2)$ : $\mathbb{P}[G=g]=(1/2)^g$ , $\mathbb{P}[T\ge m]=(1/2)^{m-1}$ .
Condition on $G=g$ , then sum $\sum_g (1/2)^g (1/2)^{4g-1}$ as a geometric series.

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Four times as long to the first head

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