Die game paying a dollar per low roll until it ends

You roll a die repeatedly. A $1$, $2$ or $3$ earns you $1 and you roll again. A $4$ or $5$ ends the game and you keep everything earned. A $6$ ends the game and you forfeit everything. What is the most you should pay to play?

$2/3$ of a dollar. You leave with $k$ dollars only if you roll $k$ low numbers then a $4$ or $5$, probability $\left(\tfrac12\right)^k\tfrac13$. So the value is $\tfrac13\sum_k k\,2^{-k}=\tfrac13\cdot2=\tfrac23$.