1. **Problem statement:** Alex rolls an 8-sided die repeatedly and stops once he rolls two consecutive 8s. We want to find the expected number of rolls until he stops. 2. **Define states:** - Let $E_0$ be the expected rolls starting with no previous 8. - Let $E_1$ be the expected rolls starting with exactly one 8 rolled previously (waiting for the second 8). 3. **Set up equations:** - From state $E_0$, on the next roll: - With probability $\frac{1}{8}$, roll an 8 and move to state $E_1$. - With probability $\frac{7}{8}$, roll not 8 and stay in $E_0$. So, $$E_0 = 1 + \frac{1}{8}E_1 + \frac{7}{8}E_0$$ - From state $E_1$, on the next roll: - With probability $\frac{1}{8}$, roll an 8 and stop (0 more rolls). - With probability $\frac{7}{8}$, roll not 8 and return to $E_0$. So, $$E_1 = 1 + \frac{1}{8} \times 0 + \frac{7}{8}E_0 = 1 + \frac{7}{8}E_0$$ 4. **Solve the system:** From $E_0$ equation: $$E_0 - \frac{7}{8}E_0 = 1 + \frac{1}{8}E_1$$ $$\frac{1}{8}E_0 = 1 + \frac{1}{8}E_1$$ $$E_0 = 8 + E_1$$ Substitute $E_1$: $$E_0 = 8 + 1 + \frac{7}{8}E_0 = 9 + \frac{7}{8}E_0$$ Bring terms together: $$E_0 - \frac{7}{8}E_0 = 9$$ $$\frac{1}{8}E_0 = 9$$ $$E_0 = 72$$ 5. **Interpretation:** The expected number of rolls until two consecutive 8s appear is $\boxed{72}$. **Answer:** Option B) 72