Expected value of highest die roll Daniel will roll a fair, six-sided die until he gets a 4. What is the expected value of the highest number he
rolls through this process?

It seems the expected number of rolls is 6. Can we reword this question to what is the expected value of the maximum of 6 die rolls?


Let $P(6)$ denote the probability of getting a $(6)$ before the first $(4)$.
Let $P(5)$ denote the probability of getting a $(5)$ before the first $(4)$ and simultaneously, not getting a $(6)$ before the first $(4)$.
Then, the computation for the expected value of the highest roll is
$$4 +~ \left[ ~2 \times P(6) ~\right] ~+~ \left[ ~1 \times  ~P(5) ~\right]. \tag1 $$

$$P(6) = \frac{1}{6} + \left[\frac{4}{6}P(6)\right] \implies \frac{2}{6}P(6) = \frac{1}{6} \implies P(6) = \frac{1}{2}.$$

$$P(5) = \frac{1}{6} + \left[\frac{3}{6}P(5)\right] \implies \frac{3}{6}P(5) = \frac{1}{6} \implies P(5) = \frac{1}{3}.$$
Thanks to lulu, for her comment, immediately following this posting, which indicated a flaw in the above analysis, pertaining specifically to the computation of $P(5).$
$\color{red}{\text{The above analysis is wrong because:}}$
The above equation assumes that if the first roll is a $(5)$, that it is game over, and that the high roll will be a $(5)$.  This overlooks the fact that the die rolls do not stop, if the first roll is a $(5)$.
Instead, the die rolls continue, and by the already computed $P(6)$, (1/2) the (subsequent) time, there will be a $(6)$ before the $(4)$.
Therefore, the correct equation for $P(5)$ is
$$P(5) = \left[ ~\frac{1}{6} \times \left( ~1 - P(6) \right) ~\right] + \left[\frac{3}{6}P(5)\right] \implies $$
$$\frac{3}{6}P(5) = \left[\frac{1}{6} \times \left( ~1 - \frac{1}{2} ~\right) ~\right] \implies P(5) = \frac{1}{6}.$$
Note:
As indicated in the comment of lulu, an alternative (less convoluted) approach is to:

*

*Let $P(4)$ denote the probability that the high roll will be $(4)$.


*Recognize that $P(4) + P(5) + P(6) = 1$.


*Recognize that 
$\displaystyle P(4) = \frac{1}{6} + \left[\frac{3}{6}P(4)\right] \implies \frac{3}{6}P(4) = \frac{1}{6} \implies P(4) = \frac{1}{3}.$

Plugging these $\color{red}{\text{now corrected}}$ results back into (1) above, the computation of expected value becomes
$$4 +~ \left[ ~2 \times \frac{1}{2} ~\right] ~+~ \left[ ~1 \times  \frac{1}{6} ~\right] = \frac{31}{6}. $$
