Die game that banks on four or five, busts on six

You roll a fair die repeatedly. Each roll of $1$, $2$ or $3$ adds one point to your score and you roll again. The first time a $4$, $5$ or $6$ appears the game ends: on a $4$ or $5$ you are paid your accumulated score in dollars, and on a $6$ you get nothing. What is the expected payout?

$2/3$ of a dollar. The number of low rolls before stopping is geometric with mean $1$ (each roll is low with probability $\tfrac12$). The stopping face is independent of that count and is a $4$ or $5$ with probability $\tfrac23$, so the expected payout is $\tfrac23\cdot1$.