Expected vegetables eaten in a reroll die game

You roll a fair six-sided die and eat as many Brussels sprouts as the number shown. If the roll was $3$ or higher, the game stops there. If it was a $1$ or $2$, you eat that many sprouts and must roll again, repeating until a roll of $3$ or more ends the game. What is the expected total number of sprouts you eat?

$21/4$. Condition on the first roll: with probability $\tfrac13$ it is $1$ or $2$ (average $1.5$) and the game restarts; with probability $\tfrac23$ it is $3$–$6$ (average $4.5$) and stops. So $E=\tfrac13(1.5+E)+\tfrac23(4.5)$, which solves to $E=\tfrac{21}{4}$.