Race between a second five and seeing both four and six

A fair die is rolled repeatedly and the results are recorded. Rolling stops at the first moment when either the value $5$ has appeared twice in total (not necessarily in a row), or both a $4$ and a $6$ have appeared at least once each. What is the probability that the stopping reason is the pair of $5$s?

$\tfrac7{18}$. Ignore 1s, 2s and 3s: each relevant roll is a 4, 5 or 6 with probability $\tfrac13$, and repeating an already-seen 4 or 6 changes nothing. If the first relevant roll is a 5, the fives win with probability $\tfrac13+\tfrac23\cdot\tfrac12=\tfrac23$; if it is a 4 or 6, they win with probability $\tfrac12\cdot\tfrac12=\tfrac14$. Hence $\tfrac13\cdot\tfrac23+\tfrac23\cdot\tfrac14=\tfrac7{18}$.