Pattern recurrence

High-level idea

Model the process as a Markov chain with two transient states tracking the current "run" of consecutive 1s. Set up first-step equations and solve the $2\times 2$ linear system.

States and equations

Let $p = \tfrac{1}{6}$ be the probability of rolling a 1. Define:

• $S_0$: start, or the last roll was not a 1.
• $S_1$: the last roll was a 1 (one consecutive 1 in progress).

Let $\mu_0,\mu_1$ be the expected number of additional rolls needed to reach absorption (two consecutive 1s) from each state.

From $S_0$ — one roll is used; with probability $\tfrac{1}{6}$ we move to $S_1$, with probability $\tfrac{5}{6}$ we remain in $S_0$:

From $S_1$ — one roll is used; with probability $\tfrac{1}{6}$ we finish (0 more rolls), with probability $\tfrac{5}{6}$ we return to $S_0$:

Solving the system

Rearranging the first equation:

Substituting $\mu_1 = 1 + \tfrac{5}{6}\mu_0$:

Check: $\mu_1 = 1 + \tfrac{5}{6}(42) = 36$, and $6 + 36 = 42$. ✓

Final answer

The expected number of rolls is $\boxed{42}$.