At Least Six I

Work backwards. Let $E_i$ be the expected number of further rolls when the total is $i$. What is $E_5$?

From total $i < 6$, one roll takes you to $i + k$ with $k = 1, \dots, 6$ equally likely, so $E_i = 1 + \tfrac16\big(E_{i+1} + \dots + E_{i+6}\big)$, where any $E_j$ with $j \ge 6$ is $0$.

Set up. Let $E_i$ be the expected number of further rolls when the running total is $i$. Once the total is $6$ or more we stop, so $E_i = 0$ for $i \ge 6$.

From a total $i < 6$ we always spend one roll, and then land on $i + 1, \dots, i + 6$ with chance $\tfrac16$ each:

Only the terms that land below $6$ are nonzero.

Work backwards from $5$.

• $E_5 = 1$, since any roll finishes.
• $E_4 = 1 + \tfrac16 E_5 = \tfrac76$.
• $E_3 = 1 + \tfrac16(E_4 + E_5) = 1 + \tfrac16 \cdot \tfrac{13}{6} = \tfrac{49}{36}$.
• $E_2 = 1 + \tfrac16(E_3 + E_4 + E_5) = \tfrac{343}{216}$.
• $E_1 = 1 + \tfrac16(E_2 + E_3 + E_4 + E_5) = \tfrac{2401}{1296}$.
• $E_0 = 1 + \tfrac16(E_1 + E_2 + E_3 + E_4 + E_5) = \tfrac{16807}{7776}$.

Answer:

Notice the pattern: $E_4 = \tfrac76$, $E_3 = \left(\tfrac76\right)^2$, …, $E_0 = \left(\tfrac76\right)^5$. At Least Six II explains why.