Rolling two four-sided (tetrahedron) dice, what is the expected value of the higher and of the smaller of the two numbers shown?
Partial Solution: $$EX = 1 \cdot 1/16 + 2\cdot 3/16 +3\cdot 5/16 +4 \cdot7/16 =3.215$$ This will be the expectation of the higher number. But I don't know how to find the expectation of the smaller number.
Hint:
The expectation of the sum of higher and smaller number is $2\times\frac52$ (do you understand why?).
Also this expectation is the sum of their expectations by linearity of expectations.
So finding one of them is enough.
addendum:
Let $D_1$ and $D_2$ denote the two scores and let $H$ and $S$ denote the highest and smallest score respectively. Then: $$D_1+D_2=H+S$$ so that $$\mathbb ED_1+\mathbb ED_2=\mathbb E(D_1+D_2)=\mathbb E(H+S)=\mathbb EH+\mathbb ES$$
It is not difficult to find that $\mathbb ED_i=\frac52$ so we end up with $$\mathbb EH+\mathbb ES=5$$