Cash or Reroll

Before solving, let's name the one idea behind this game: backward induction for an optimal stopping decision. When you face a "keep it or try again" choice, you compare what you already have against the expected value of the alternative, and you keep what you have only when it is at least as good.

What is a fresh roll worth?
If you discard the first roll, you are paid whatever the second roll shows, with no further choices. The second roll is a fair eight-sided die, so its expected value is the average of the faces: \begin{equation} E[\text{reroll}] = \frac{1 + 2 + 3 + 4 + 5 + 6 + 7 + 8}{8} = \frac{36}{8} = 4.5 \end{equation} So choosing to re-roll is worth $4.5$ on average, no matter what the first roll was.

The decision rule
After seeing the first roll $v$, you choose the better of two options: keep $v$, worth $v$, or re-roll, worth $4.5$. You therefore keep the first roll exactly when \begin{equation} v \geq 4.5 \end{equation} that is, when $v \in \{5, 6, 7, 8\}$, and you re-roll when $v \in \{1, 2, 3, 4\}$.

Computing the game's value
Each first-roll value $v$ occurs with probability $\frac{1}{8}$. For the four low values you re-roll and collect $4.5$; for the four high values you keep $v$. So the fair value is \begin{equation} E[\text{game}] = \frac{1}{8}\Big(\underbrace{4.5 + 4.5 + 4.5 + 4.5}_{v = 1,2,3,4 \text{, re-roll}} + \underbrace{5 + 6 + 7 + 8}_{v = 5,6,7,8 \text{, keep}}\Big) \end{equation} \begin{equation} E[\text{game}] = \frac{1}{8}\big(18 + 26\big) = \frac{44}{8} = 5.5 \end{equation} Notice the value 5.5 sits comfortably above the 4.5 you would get from a single roll with no choice: the option to re-roll the bad half of the die is worth exactly one extra point.

So, the fair value of this game is 5.5