Two rolls two apart

Idea. The state of the game is the last face rolled. From an “outer” face (1, 2, 5, 6) one face out of six ends the game; from a “middle” face (3, 4) two do. Two equations, by symmetry.

Step 1: the unknowns. Let $a$ be the expected number of further rolls when the last face is outer, and $b$ when it is middle.

Step 2: outer face. Take face 1. The next roll ends the game with a 3. Otherwise it is 1, 2, 5 or 6 (outer) with probability $\frac{4}{6}$, or 4 (middle) with probability $\frac{1}{6}$: $$a=1+\frac{4}{6}a+\frac{1}{6}b.$$ The other outer faces give the same equation: for face 2, the stopping face is 4 and the other middle face is 3.

Step 3: middle face. Take face 3. The game ends with a 1 or a 5. Otherwise it is 2 or 6 (outer) or 3 or 4 (middle), each pair with probability $\frac{2}{6}$: $$b=1+\frac{2}{6}a+\frac{2}{6}b.$$

Step 4: solve. The first equation gives $2a=6+b$, the second $4b=6+2a$. So $4b=12+b$, $b=4$, and $a=5$.

Step 5: from the start. The first roll is outer with probability $\frac{4}{6}$ and middle with probability $\frac{2}{6}$: $$1+\frac{4}{6}\cdot 5+\frac{2}{6}\cdot 4=1+\frac{20}{6}+\frac{8}{6}=\frac{17}{3}\approx \mathrm{5.67.}$$