We roll a fair 6 sided die and keep a running sum $S$ of the rolls so far. We stop as soon as $S$ is greater than 12. What is the expected value for $S$.
The simplest approach I can think of is using states: Let $E_i$ denote the expected end result given that the sum is currently $i$, we want $E_0$. Then:
$$E_i=\frac{1}{6}(E_{i+1}+\dots E_{i+6}), \forall 0\leq i\leq 12$$ and for $i>12$, we simply have $E_i=i$. This gives 12 independent linear equations with 12 unknowns $E_0,E_1,\dots, E_{12}$, so we could solve for $E_0$. Is there a clean way to do this? I've tried adding all the equations up and also the shifting indices trick, e.g. $E_{i+1}-E_i=\frac{1}{6}(E_{i+7}-E_{i+1})$, but I can't see how to get nice simplification. Perhaps there is not a nice simplification. But the symmetry in the system does make me think there could be. Any ideas?
Also if anyone has any other approaches besides this algebraic approach (e.g. a probability based approach with less computation), that would also be appreciated. Thanks!
Source: https://www.cut-the-knot.org/Probability/RollingDie.shtml (this has a solution, but it is incorrect)