Top 3 of 4 Dice Rolls

We use the same idea as Ross Millikan. In order to bring out the structure, there will be as little calculation as possible.

Let the random variable $X$ denote the smallest roll, and let $Y$ denote the sum of the three larger rolls. Then $$E(X)+E(Y)=E(X+Y)=14.$$ We calculate $E(X)$. From this, $E(Y)$ is easily found.

Let $q_1$ be the probability that $X\ge 1$, let $q_2$ be the probability that $X\ge 2$, and so on. Then $P(X=1)=q_1-q_2$, $P(X=2)=q_2-q_3$, and so on until $P(X=6)=q_6$. Thus $$E(X)=1\cdot(q_1-q_2)+2\cdot(q_2-q_3)+3\cdot(q_3-q_4)+4\cdot (q_4-q_5)+5\cdot (q_5-q_6) +6q_6.$$ This simplifies to $$q_1+q_2+q_3+q_4+q_5+q_6.$$ But $$q_i=\frac{(7-i)^4}{6^4},\qquad\text{and therefore}\qquad E(X)=\frac{1^4+2^4+3^4+4^4+5^4+6^4}{6^4}.$$

Comment: Let $W$ be a random variable that only takes on integer values. For any positive integer $n$, let $q_n=P(W\ge n)$. Then $$E(W)=\sum_{n=1}^\infty q_n,$$ provided the sum is defined.

One approach is to find the number of rolls with each number as the lowest. There is only one roll with $6$ the minimum. There are $2^4-1=15$ with $5$ the minimum. This continues to the fact that there are $6^4-5^4=1296-625$ rolls with $1$ the minimum. The total of all the dice is $1296*4*3.5=18144$ The thrown out dice are $6*1+5*15+4*65+3*175+2*369+1*671=2275$ so the total score after throwouts is $15869$, giving an average of about $12.2446$