Running Total of a Die

Two approaches. Monte Carlo gives 13 at roughly 28%. The analytic approach uses integer partitions: the number of ways to reach each total with parts $\leq 6$ and penultimate sum $\leq 12$. Comparing $p(13,6)$ to $\sum_{i=8}^{12} p(i,6)$ shows $\Pr[13] \approx 28\%$ exceeds $\Pr[14] \approx 24\%$. Full code in the PDF.