You roll a fair, 6-sided die repeatedly until you get a streak of three increasing rolls (three consecutive rolls with numbers in strictly increasing order, but not necessarily consecutive numbers). What is the expected number of rolls?
If we let $X$ be the number of rolls until we satisfy our required condition, let $E=\mathbb{E}[X]$ and $E_i = \mathbb{E}[X\space|\space i], \space i\in\{1,\dots, 6\}$. So $E_1$ is the expected number of rolls until we satisfy our required given that we have currently rolled a 1, meaning that, for instance $E_{31} = E_1$ where we define $E_{ij}$ similarly (rolled an $i$ and then a $j$).
Using this notation and considering the first roll, I found, $E = \frac{1}{6}(1 + E_1) + \frac{1}{6}(1+E_2) + \frac{1}{6}(1+E_3) + \frac{1}{6}(1+E_4) + \frac{1}{6}(1+E) + \frac{1}{6}(1+E)$
$\implies \frac{2}{3}E = 1 + \frac{1}{6}(E_1 + E_2 + E_3 + E_4)$
given that $E_5 = E_6 = E$.
Similarly, by considering the scenario where we have rolled a 4, I found $E_4 = \frac{1}{6}(1+E_1) + \frac{1}{6}(1+E_2) + \frac{1}{6}(1+E_3) + \frac{1}{6}(1+E_4) + \frac{1}{6}(1+E_{45}) + \frac{1}{6}(1+E)$
where $E_{45} = \frac{1}{6}(1+E_1)+\frac{1}{6}(1+E_2)+\frac{1}{6}(1+E_3)+\frac{1}{6}(1+E_4)+\frac{1}{6}(1+E)+\frac{1}{6}\cdot 1 $
$= 1 + \frac{1}{6}(E+E_1 + E_2 + E_3 + E_4)$
and so
$E_4 = \frac{7}{6} + \frac{7}{36}(E + E_1 + E_2 + E_3 + E_4)$
It seems that by continuing in this fashion I can find $E_1, E_2, E_3$ in terms of $E$ and the other $E_i$'s but will be become increasing tedious as we go from $E_3$ to $E_1$. However, after finding other linear equations involving $E$ and the $E_i$'s, we could find $E$ by inverting a matrix so the method seems to work, but is quite long and would not be very feasible to use when considering an $n$ sided dice.
I was wondering if there is a nicer approach to this problem which would also allow a generalisation to an $n$ sided dice?