The highest roll before a 4

1. Step 1 · The highest is at least 4

   $\displaystyle \text{the last roll is always } a 4$

   So the highest is 4, 5 or 6. You only need the chance of each.

2. Step 2 · The highest stays at 4

   $\displaystyle a 4 \text{ comes before any } 5 \text{ or } 6\colon 1 \div 3 = \frac{1}{3}$

   Ignore 1s, 2s and 3s. Of 4, 5 and 6, each is equally likely to show up first, and the 4 must be first.

3. Step 3 · The highest is 5 or less

   $\displaystyle a 4 \text{ comes before any } 6\colon 1 \div 2 = \frac{1}{2}$

   Now only 4 and 6 matter: whichever shows up first decides it.

4. Step 4 · Chances of a highest of 4, 5 and 6

   $\displaystyle \frac{1}{3},\, \frac{1}{2} - \frac{1}{3},\, 1 - \frac{1}{2} = \frac{1}{3},\, \frac{1}{6},\, \frac{1}{2}$

   The chance the highest is exactly 5 is the chance it’s at most 5, minus the chance it’s at most 4.

5. Step 5 · Average highest

   $\displaystyle 4 \times \frac{1}{3} + 5 \times \frac{1}{6} + 6 \times \frac{1}{2} = \frac{31}{6}$

   Multiply each highest by its chance and add them up.

Answer$\displaystyle \frac{31}{6} \approx 5.167$