A dice is rolled and it's values are summed until a 6 occurs. You can stop at any time and take your profit (the sum of the values) with you, however, as soon as a 6 occurs, it'll be decreased to zero and the game ends.
Now consider strategy $H_k$, where you stop if you reach a value of $\ge k$ or as soon as a six occurs.
Find the value of $k$, for which the profit is maximal. In other words: For which $k$ is the gain per game not positive anymore?
If the expected value of the next roll is positive then roll again, otherwise stop.
Let $k$ be the current profit. Then the expected value of the next roll is
$$\frac{1+2+3+4+5-k}{6}$$
and it's clear that this is zero when $k=15$.
So, if $k\le14$, roll again; if $k=15$ it doesn't matter whether you roll again; if $k\ge16$, stop.
Strategy $H_{15}$ or $H_{16}$ will maximize profit in the long run. A simulation confirms this and suggests the maximal profit per game is about $6.15$.