The objective of a dice rolling game is to roll the highest possible value. The player is allowed $2$ consecutive rolls of a fair $6$ sided dice. The player is given $2$ distinct choices: stop after one roll and keep that value, or continue to roll the second time. If the player chooses to roll the second time, the value of the second roll becomes the player’s final value, no matter what was rolled the first time. What is the expected value of this game, assuming the optimal strategy?
I know that the expected value of roll $1$ is $3.5$ and that if the $1$st roll is greater than $3.5$, i.e. $4, 5,$ or $6$ then don't roll again. If roll $1$ is less than $3.5,$ i.e. $1,2 3,$ then roll again, which increases the odds of winning. So assuming the probability for second roll would be $\frac{3}{6} \times 3.5 + \frac{3}{6} \times 5$, which yields an expected value of $4.25$, is this assumption correct? I'm assuming expected value of $5$ for second roll - average of $4,5,6 = 5$, but is this the correct assumption as there is still $\frac{1}{6}$ probability for each number?
