Three fair twenty-sided dice are rolled, and then arranged in decreasing order. The expected value of the largest die can be written in the form $p/q$ where $p$ and $q$ are relatively prime positive integers. Find $p+q$.
I try to find the exoected value and got $31.5$ and representing that as $p/q$, I got $63/2$ so $p + q = 65$. But the quiz master said The answer is $4$ digit. Am I missing something from the $q$ or have a wrong approach?
On
One approach is to use the theorem that for a random variable $X$ that takes on only non-negative values, $$E(X) = \sum_{n=0}^{\infty} P(X>n)$$ If $X$ is the maximum of the three dice, then $$P(X>n) = 1 - P(X \le n) = 1 - (n/20)^3$$ for $0 \le n \le 20$. So by the theorem, $$\begin{align} E(X) &= \sum_{n=0}^{20} ( 1 - (n/20)^3 ) \\ &= 21 - (1/20)^3 \sum_{n=0}^{20} n^3 \\ &= 21 - (1/20)^3 \cdot (1/4) \cdot20^2 \cdot 21^2 \\ &= \frac{1239}{80} \end{align}$$ where we used the formula for the sum of the first $n$ cubes to simplify computation.
The probability that the maximum of three fair dices is $k$ is $\tfrac{k^3-(k-1)^3}{20^3}$ for each $k\in \{1,\ldots,20\}$ (each dice should show at most $k$, but not all of them show at most $k-1$.)
Thus, the maximum is expected to be $\sum\limits_{k=1}^{20} k\tfrac{k^3-(k-1)^3}{20^3}=\tfrac{1239}{80}$.