Optimal stopping with up to three die rolls

You may throw a fair six-sided die up to three times. After each of the first two throws you can either stop and be paid the face showing, or discard it and throw again; if you reach the third throw you must accept it. What is the game worth under optimal play, and what rule achieves it?

$14/3\approx4.67$. Work backwards: the last throw is worth $3.5$, so on the second throw stop at $4$ or more, worth $\tfrac{4+5+6}{6}+\tfrac12(3.5)=4.25$. On the first throw stop only at $5$ or $6$, worth $\tfrac{11}{6}+\tfrac46(4.25)=\tfrac{14}{3}$.