3 fair but unequal dice thrown, 2 greatest values summed 3 fair dice with number of sides d1, d2, and d3 and with face values each of 1, 2, ...(d). The 2 greatest values thrown are summed, while the least-valued die is discarded. What is the expected value?
 A: Let $X$ be the sum of all $3$. It is easy to calculate $E(X)$.
We change notation slightly. Let the three dice have $a$ sides, $b$ sides, and $c$ sides respectively. We interpret the problem as specifying that the sides of the $a$-sided die are labelled $1$ to $a$, with all sides equally likely. Make similar assumptions about the other two dice.
Without loss of generality, we may arrange things so that $a\le b\le c$.
The smallest throw $Y$ is one of $1$ to $a$. 
We have that $\Pr(Y=1)$  is $1$ minus the probability there is no $1$ thrown. This is $1-\frac{(a-1)(b-1)(c-1)}{abc}$.
We can get similar expressions for $\Pr(Y=2)$, $\Pr(Y=3)$, and so on up to $\Pr(Y=a)$. Once we have these probabilities we can compute $E(Y)$ is the usual way.
We do not do the details, because we will use a shortcut. Note that
$$E(Y)=\Pr(Y\ge 1)+\Pr(Y\ge 2)+\cdots +\Pr(Y\ge a).$$
We have $\Pr(Y\ge 1)=1$. In general, for $i\le a$, the probability that $Y\ge i$ is the probability that all the dice show a number $\ge i$. This probability is
$$\frac{(a-i+1)(b-i+1)(c-i+1)}{abc}.$$
To calculate $E(Y)$, add up, from $i=1$ to $i=a$.
Finally, note that the answer to the question of the post is $E(X)-E(Y)$.
Remark: We can use basically the same idea if we have "weird" dice, with unequal probabilities for the various numbers on the faces. Things do get a bit messy notationally. 
