Optional cash-out in a roll-until-5-or-6 game

You roll a fair die repeatedly, adding up the results of $1$–$4$. Rolling a $5$ ends the game and pays you the accumulated total (the $5$ itself is not added); rolling a $6$ ends the game with nothing. Before any roll you may also choose to stop and take your current total. With optimal play, what is the expected payout, to the nearest hundredth?

About $3.03$. With bank $k$, another roll is worth $\tfrac16k+\tfrac23\left(k+\tfrac52\right)$, which beats $k$ only when $k<10$, so stop once the total reaches $10$. Evaluating the game value backwards from the stopping totals $10$–$13$ down to $0$ gives about $3.03$.