Dice game: throw a 6-dice, until a 6 is shown, in that case you lose everything, or until you stop, winning the total $S_n$ of the face values seen so far.
I want to compute the average time to termination of the game (by winning or losing) under an optimal strategy.
By value function/Markov Chain arguments plus MC simulation, I know that I should stop when $S_n$ exceeds 14, that the expected value of $S_n$ is 6.15, that the chance of actually winning is 37.7%, and that the average time to termination is 3.73 throws. One just needs to take account of the fact that under the optimal strategy the game is a Markov chain with states going from 0 to 19 (roll a 14 and then roll again: all or nothing, for sure stop).
I wanted to compute again the time to termination via martingale arguments, so I state that the process $$M_n:=\left(\frac65\right)^n\left(S_n-\frac{25}{12}\frac n{n+1}\right)$$ is a martingale.
i) I am not sure this actually is, I came up with it doing some patchwork, plus I am not sure also of the way I verified the martingale property.
ii) I would like to compute the average time to termination as done for standard random walks: $$ 0=M_0=\mathbb E[M_n|\text{win before rolling a 6}]\mathbb P[\text{win before rolling a 6}]+\mathbb E[M_n|\text{instead}](1-\mathbb P[\text{win before rolling a 6}])$$ knowing that $\mathbb P[\text{win before rolling a 6}]=37.7\%$. It should be doable, since the value of $S_n$ conditioned to winning can be computed, and in case of defeat is just null, so the only random variable whose expectation appears is $n$. However, I find myself carrying an $\left(\frac65\right)^n$ in expectation, that is defeating me.
How would you proceed?
