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
$$4 +~ \left[ ~2 \times \frac{1}{2} ~\right] ~+~ \left[ ~1 \times \frac{1}{6} ~\right] = \frac{31}{6}. $$