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In a game I was playing, you roll two six-sided dice. If either of them rolls a one, your points for the turn become zero and your turn is ended. If you roll greater than one on both dice, the sum of the rolls add to your points and you can choose to roll again or choose to end your turn and add your points from the turn to your total score.

I was wondering how many rolls per turn would give you the best possible score on average, so I programmed a simple simulation that tested rolling once, twice, three times, and four times in a turn. It turned out that the best score comes from rolling a pair three times, but I can't explain why. Can anyone explain this through a proof or some equations?

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    You can do better by letting your decision about the next roll depend not only on the number of rolls so far, but also the total you have so far (and are risking at the next roll). – 2017-02-07

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If you must decide in advance how many rolls to take, then the first roll leaves you with an average of $\frac{25}{36}\cdot 8$ points. But if you have survived to $n$ rolls, you will have an average of $8n$ points. On the next roll you will risk $8n$ (with a loss probability of $\frac{11}{36}$ to gain $\frac{25}{36}\cdot 8$ so to find the region where rolling again is advantageous we must solve for $$ \frac{25}{36}\cdot 8 > \frac{11}{36}\cdot 8n\\ n < \frac{25}{11} $$ So after two rolls you are still going to want to roll another; after 3 rolls you want to stop.

In reality, you want to stop if before the next roll you have more than $\left\lfloor\frac{200}{11}\right\rfloor = 18$ points.

By the way, it is somewhat harder to calculate the expected value of the game if your strategy is to stop whenever you have more than $18$ points. The expected value is $$ \frac{30599639}{3779136}\approx8.09079 $$ Although that looks rather small (after all, any time you get anything you get somewhere between $19$ and $30$ points), it becomes easier to understand when you see that you lose completely (with a score of zero) about 64% of the time.