What is the average of rolling a die twice, with the option of opting out the second rolling?

Question

I saw the question What is the average of rolling two dice and only taking the value of the higher dice roll?.

What about the case that instead of rolling two dice simultaneously, rolling of the dice in the game is this:

1. The player rolls the dice.

2. The player is asked whether he/she want to roll the dice again.

3. If the player want to roll the dice again, he/she will roll the dice again and the points obtained will be the final score; if the player does not want to roll the dice again, he/ she will have the dice roll points in 1. as the final score

The purpose of the game is to get the as high score as possible.

Am I correct that the expected value of the game score is also 4.47 (as in the question mentioned at the beginning of this question)? Am I correct that the optimal strategy is that if a player get a dice roll point smaller than 5, he/ she should roll it again?

If yes, does that mean this game is equivalent mathematically to the game mentioned at the beginning of this question?

Is such result extensible to more than 2 dice (e.g. 3 dice and rolling a die for 3 times)?

Answer 1

The two games are not equivalent. Roll two and take the highest will give you $6$ from a roll of $5,6$. Roll one and decide will stop at $5$ and give you $5$.

In fact, roll one and decide should stop at $4$ as well, because the average roll is $3.5$ and you are above that. The average roll for roll one and decide is $\frac 12 \cdot 5 + \frac 12\cdot 3.5=4.25$ because the average roll you keep is $5$, which you do with chance $\frac 12$ and the rest of the time you average $3.5$.

Answer 2

The expected score of a single die role is $3\frac 12$. Hence if I roll a die and decide to retry whenever the first roll was $n$ or lower, the expected value of the score is $$\begin{align} \underbrace{\frac 16\cdot 3\frac12+\ldots \frac 16\cdot 3\frac12}_n+\frac 16\cdot (n+1)+\frac 16\cdot (n+2)+\ldots+\frac16\cdot 6&=\frac{7n}{12}+\frac{42-n(n+1)}{12}\\&=\frac{42+6n-n^2}{12}\\&=\frac{51-(n-3)^2}{12}\\&\le \frac{17}{4}=4\frac 14\end{align}$$ with equality iff $n=3$.