When to stop rolling a die in a game where 6 loses everything You play a game using a standard six-sided die. You start with 0 points. Before every roll, you decide whether you want to continue the game or end it and keep your points. After each roll, if you rolled 6, then you lose everything and the game ends. Otherwise, add the score from the die to your total points and continue/stop the game.
When should one stop playing this game? Obviously, one wants to maximize total score.
As I was asked to show my preliminary results on this one, here they are:
If we simplify the game to getting 0 on 6 and 3 otherwise, we get the following: 
$$\begin{align}
EV &= \frac{5}{6}3+\frac{25}{36}6+\frac{125}{216}9+\ldots\\[5pt]
&= \sum_{n=1}^{\infty}\left(\frac{5}{6}\right)^n3n
\end{align}$$ 
which is divergent, so it would make sense to play forever, which makes this similar to the St. Petersburg paradox. Yet I can sense that I'm wrong somewhere!

The question is missing the concept of utility, a function that specifies how much you value each possible outcome. In the game, the utility of ending the game with a certain score would be the value you place on that outcome. Although you could certainly argue it is implied in the question that the utility of a score is simply the score, I would like to add an answer that takes a non-trivial utility into account. If each point translated to 1000 USD, for example, you might have a utility that looks more like $U(x) = \log(x)$ than $U(x) = x$.
Let's say that $U(x)$ is the utility of score $x$ for $x \ge 0$ and assume that $U$ is monotonically non-decreasing. Then we might say that the optimal strategy is that which maximizes $E[U(X)]$, where $X$ is a random variable representing the final score if one plays with the policy where you roll the die if and only if your current score is less than $t \in \mathbb{Z}_{\ge 0}$. (It is clear that the optimal policy must have this form because the utility is non-decreasing.)
Let $Z$ denote current score. Suppose we are at a point in the game where our current score is $z \ge 0$. Then
$$E[U(X)|Z = z] = \frac{1}{6} \left( U(0) + \sum_{i=1}^5  E[U(X)|Z=z+i] \right) \text{ if } z < t$$
$$E[U(X)|Z = z] = U(z) \text{ if } z \ge t$$
Note that for many choices of $U(x)$ the recurrence relation is very difficult to simplify, and that, in the case of choosing to roll the die, we must consider the expected change in utility from that roll and all future rolls. The figures below are examples of what the above recurrence relation gives for $U(x) = x$ and $U(x) = \log_2(x + 1)$. The expression $E[U(X)]$ means $E[U(X)|Z=0]$, because at the start we have $0$ points. The horizontal axis corresponds to different policies, and the vertical axis corresponds to expected utility under each policy.

Gist with Python code
