Three geometric waits in strict increasing order

Three players each toss their own fair coin until they get their first head. Let $T$, $G$ and $P$ be the number of tosses the first, second and third player need, respectively. Compute $\mathbb{P}[T<G<P]$.

$1/21$. With $T,G,P$ i.i.d. $\text{Geom}(1/2)$, sum the joint pmf over $t<g<p$: the innermost sum gives $2^{-g}$, then summing $4^{-g}$ over $g>t$ and finally over $t$ yields $\tfrac{1}{21}$. (Equivalently, all three distinct has probability $\tfrac{6}{21}$ and each of the $6$ orders is equally likely.)