This question is slightly more complex than the Die Roll II. We now have to set up two simultaneous equations, due to the fact that we now have an intermediate transition in the Markov chain - if we roll a 1, and then another 1, we remain at the intermediate step where we only have to roll a 2 to end the game. $$ $$ So, we set up the Markov chain equations $$ E[12] = 1 + \frac{1}{6}\text{E}[12|1] + \frac{5}{6}\text{E}[12], $$ $$ E[12|1] = 1 + \frac{1}{6}\text{E}[12|1] + \frac{4}{6}\text{E}[12]. $$ Simplifying these equations gives $$ \text{E}[12] = 6 + \text{E}[12|1], $$ $$ 5\text{E}[12|1] = 6 + 4\text{E}[12]. $$ Solving these two equations for $\text{E}[12]$ gives $$ \text{E}[12] = 36. $$ We see that the expected number of rolls is lower than what we saw in Die Roll II, since in this situation we can remain at the intermediate state by rolling two consecutive $1$s.