Rolling until a 5 pays out or a 6 busts

A fair die is thrown over and over. The game ends the moment a $5$ or a $6$ appears. If the game ends on a $5$, you are paid the total of every throw made before it (the $5$ itself does not count); if it ends on a $6$, you get nothing. You cannot stop early. Find the expected payout.

$5/2$. The game ends on a $5$ or a $6$ with equal probability. The number of throws before the end is geometric with mean $2$, and each of those throws is uniform on $\{1,2,3,4\}$ with mean $2.5$, independently of how the game ends. So the expected payout is $\tfrac12\cdot 2\cdot 2.5=\tfrac52$.