Exactly two fives before seeing both 4 and 6

Kathy throws a fair six-sided die again and again, stopping once both a $4$ and a $6$ have shown up (not necessarily consecutively). What is the probability that during this whole sequence of throws she saw the face $5$ exactly twice?

$19/108$. Faces $1,2,3$ are irrelevant. Before the first $4$ or $6$, the number of fives is $k$ with probability $(1/3)^k\tfrac23$; after it, waiting for the missing face, it is $j$ with probability $(1/2)^{j+1}$. Hence $P(\text{two fives})=\tfrac23\left(\tfrac18+\tfrac1{12}+\tfrac1{18}\right)=\tfrac{19}{108}$.