@matthewconroy

Thank you for this fun decision problem!

To start getting a picture I examined the much simpler case of a "two-sided die" or a coin with faces 1 and 2. Same rules otherwise.

Below is a sketch of a truncated decision tree for this case, when the initial roll is "1". Decision nodes are blue squares with decisions as solid blue lines; the decisions are "K" for "keep" and "R" for "roll". Inference/chance nodes are red circles with outcomes as dashed red lines; outcomes are "1" and "2"; it's understood that each has 50% probability.

The expected utility of each decision is in grey at the end of the corresponding line. The fold-back value at each decision branch is in grey above the corresponding decision node.

If the first roll is "2" then it's clearly best to keep it, as the subsequent roll would yield a divisor and a 0$ outcome.

If the first roll is "1", the best decision apparently is to roll once more, and then, if the outcome is "2", keep the result.

Assume that the player stops in any case with "Keep" at the Nth decision, say N = 5. Then the expected utility of the previous "Roll" decision is 9$/2, which is less that the utility of the "Keep" decision, $7. So the previous decision should be "Keep". Folding backwards this situation remains up to the very first decision, for which the expected utility of "Roll" is 3$/2, whereas that for "Keep" is 1$.

It *seems* that the reasoning above would still apply if the player could continue indefinitely, since at each decision the utility of "Keep" is (1+2N)$, whereas that of "Roll" is (1.5 + N)$. But I'm not completely sure about this, there may be some logical gap.

Unfortunately this simplest case is too special, owing to its binary nature, to say something about your score-bound-strategy assumption. It'll be cool to check a 3-sided die :)

#decisiontheory