First HH completed exactly on toss ten

A fair coin is tossed until heads appears twice in a row. What is the probability that the game ends on toss number $10$?

$17/512$. Tosses $9$ and $10$ must be heads, toss $8$ must be a tail, and tosses $1$–$7$ must avoid HH. Strings of length $m$ with no HH are counted by Fibonacci numbers ($2,3,5,8,\dots$), giving $34$ for $m=7$, so the probability is $34/2^{10}=\tfrac{17}{512}$.