What is the average number of rolling 4 six-sided die and choosing the 2 higher rolls( which will be added)?

Question

As an example, if you roll 6,5,3,3 you will have 11 as the sum of 6 + 5.

The average is obviously higher than 7 but i have no idea how to calculate it other than by brute force.

Answer 1

Let $X_{(4)}, X_{(3)}$ be the highest and second highest roll. That means the other two rolls are no greater than $X_{(3)}$ so:

$$\mathsf P(X_{(4)}=y, X_{(3)}=x) = \frac 1 {6^4} \begin{cases} \binom{4}{2}(y-1)^2+\binom{4}{3}(y-1)+1 & : 1\leq x=y\leq 6\\ 4\Big(\binom 3 1 (x-1)^2+\binom 3 2(x-1)+1\Big) & : 1\leq x< y\leq 6 \\ 0 & :\text{otherwise}\end{cases}$$ Thus: $$\begin{align} \mathsf E(X_{(4)}+X_{(3)}) & = \frac 1{6^4}\sum_{y=1}^6(2y)\left(6(y-1)^2+(y-1)+1\right) \\[0ex] & +\frac 4{6^4}\sum_{y=2}^6\sum_{x=1}^{y-1}(x+y)\left(3(x-1)^2+3(x-1)+1\right)\end{align}$$

$$\begin{align}\mathsf E(X_{(4)}+X_{(3)}) & =\frac{6055}{648} \\[2ex] & \approx 9.3{\small 4}{\tiny 4135802469135802469135802469135802469135802469135802469135802469135802469135802469135802469135802}\end{align}$$