This is a typical problem that is solved via recursion. The problem can also be described as a Markov Chain. Here, we will solve with recursion. $$ $$ The first case to consider is that we immediately roll a 1 and then another 1, and the game ends in 2 rolls. The second case is that our first roll is not a 1, in which case we add 1 to the expected value. The final case is that we roll a 1 and then a number which is not a 1, and we add 2 to the expected value. As a result, the equation to solve is $$ \text{E}[11] = 2 P(11) + P(\text{not } 1)\left( \text{E}[11] + 1 \right) + P(1, \text{not } 1) \left( \text{E}[11] + 2 \right). $$ It is straightforward to determine the probabilities in our equation $$ P(11) = P(1)P(1) = \frac{1}{36} $$ $$ P(\text{not } 1) = \frac{5}{6} $$ $$ P(1, \text{not } 1) = P(1)P(\text{not } 1) = \frac{5}{36} $$ and solving the equation gives us $$ \text{E}[11] = 42. $$