What is the theoretical expected value of the sum of the highest 3 values of 4 fair dice? Many have estimated the expected value of the common "4d6, drop lowest" roll or averaged a complete enumeration of the possible outcomes. I would like to know the theoretical expression of this value, but can't wrap my head around the significance of dropping the lowest value.

The mean of the sum of the $4$ is $(4)(7/2)=14$. Let $\mu$ be the mean value of the smallest number. Then the mean of the three largest is $14-\mu$.
So we only need to find $\mu$. Let random variable $X$ be the smallest number obtained. We will know $\mu=E(X)$ once we know the distribution of $X$.
So let us find $\Pr(X=1)$, $\Pr(X=2)$, and so on up to $\Pr(X=6)$. 
It is simplest to calculate the probabilities backwards. So we find first the probability that $X=6$. This is the probability all throws are $6$, which is $(1/6)^4$.
Next we calculate $\Pr(X=5)$. This is the probability that all throws are $\ge 5$, minus the probability they are all $\ge 6$. So $\Pr(X=5)=(2/6)^4-(1/6)^4$.
Next we calculate $\Pr(X=4)$. This is the probability all throws are $\ge 4$, minus the probability they are all $\ge 5$. So $\Pr(X=4)=(3/6)^4-(2/6)^4$.  
Continue. At the end we find $\Pr(X=1)=(6/6)^4-(5/6)^4$.
Now use the usual formula for expectation once we know the distribution.
Remark: There is another way of doing it that introduces a useful trick. 
 Note that 
$$E(X)=1\cdot\Pr(X=1)+2\cdot\Pr(X=2)+3\cdot\Pr(X=3)+\cdots+6\cdot\Pr(X=6).$$
Rearrange this sum. We get 
$$\left[\Pr(X=1)+\Pr(X=2)+\cdots+\Pr(X=6)\right]+\left[\Pr(X=2)+\cdots+\Pr(X=6)\right]+\left[\Pr(X=3)+\cdots +\Pr(X=6)\right]+\cdots+\left[\Pr(X=6)\right] $$
The first sum in square brackets is $1$. The second is the probability that the minimum is $\ge 2$, which is $(5/6)^4$. (All tosses must be $\ge 2$). Similarly, the third term in square brackets is $(4/6)^4$, and so on up to $(1/6)^4$. It follows that
$$\mu=1+(5/6)^4+(4/6)^4+\cdots +(1/6)^4.$$
