I am trying to generalize the following well-known 3 roll die problem:
"We roll a single die no more than 3 times. We can stop immediately after the first roll, immediately after the second roll, or we can wait for the third. The gain is the same number of dollars as there are dots on the single upturned face of the last roll. What is the expected value of this game?"
Proceeding by backward reasoning, we can easily show that the expected value $E_3$ of the 3 roll problem is equal to $14/3$. This is my attempt to find the expected value $E_n$ of the $n$ roll problem. Let $n_0$ be the number of rolls such that $E_{n_0}$ reaches $5$ for the first time (in actuality, it turns out that $n_0=5$ because we easily find $E_4=4.94$ and $E_5=5.13$). For all $n>n_0$, we then have the relation:
$$E_n=\frac{5}{6}\times E_{n-1}+\frac{1}{6}\times 6=\left(\frac{5}{6}\right)^{n-n_0}\left(E_{n_0}-6\right)+6,$$
as we proceed to the $n$-th roll if we get $1,2,3,4$ or $5$ on roll $n-1$ and stop if we get $6$. As $n$ approaches infinity, we also see that:
$$E:=\lim_{n\rightarrow\infty}E_n=6.$$
Finally, as a sanity check, given that $E_{n_0}-6<0$, we have $E_n<6$ for all $n\geq n_0$.
I guess I don't really have a specific question, but just wondering if the general reasoning and arguments behind this generalization seemed correct? Any comments or ideas would be greatly appreciated.