What is the theoretical expected value of the sum of the highest 3 values of 4 fair dice?

Question

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.

Answer 1

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 $$\mu=1+(5/6)^4+(4/6)^4+\cdots +(1/6)^4.$$

Answer 2

Just to give a more computational approach than Andre, notice that what we want to calculate is $$ E=\sum_{i=1}^6 \sum_{j=1}^6 \sum_{k=1}^6 \sum_{l=1}^6 \left(i+j+k+l - \min \{i,j,k,l\} \right). $$ We can find this readily with various tools, such as PARI/GP, like this:

ss=0; for(i=1,6,for(j=1,6,for(k=1,6,for(l=1,6, ss=ss+i+j+k+l-min(min(min(i,j),k),l))))); print(ss/6^4)