Probabilities of the second highest roll in a dice pool. I've been trying to figure out the probabilities of rolling a specific number in a pool of 4 20-sided dice, assuming I discard the highest and two lowest rolls. My instinct was to compound the probabilities of two dice being lower or equal to a target n, one being higher or equal, and one being the specific roll I want, like so $P(n) = \frac{(n^2)(21-n)}{20^4}$. 
However, I noticed that the sum of the probabilities for each numbers was only ~10% instead of a flat 1. My first instinct was that this was the probability of a specific dice being the second highest, so not all outcomes were accounted for, but factoring in permutations brought me to ~242%. Still not close. Is this a case where the sum of the individual probabilities doesn't need to equal 1, or is my math wrong and where?.
 A: The probability for the highest die to show $k$ and all other dice to show at most $n\lt k$ is $4\cdot\frac{n^3}{20^4}$, so the probability for the highest die to show $k$ and the next highest die to show $n\lt k$ is $4\cdot\frac{n^3-(n-1)^3}{20^4}$. So the probability for the second-highest die to be lower than the highest and have the value $n$ is
$$
4\sum_{k=n+1}^{20}\frac{n^3-(n-1)^3}{20^4}=4(20-n)\frac{n^3-(n-1)^3}{20^4}\;,
$$
whereas the probability for the second-highest die to be equal to the highest and have the value $n$ is
$$\frac{n^4-(n-1)^4-4(n-1)^3}{20^4}\;.$$
The probability you seek is the sum of those two probabilities,
$$
\frac{4(20-n)\left(n^3-(n-1)^3\right)+n^4-(n-1)^4-4(n-1)^3}{20^4}=\frac{83-252n+258n^2-12n^3}{20^4}\;.
$$
Here's a table of the numerators for $n=1,\ldots,20$:
\begin{array}{r|r}
n&\\\hline
1&77\\
2&515\\
3&1325\\
4&2435\\
5&3773\\
6&5267\\
7&6845\\
8&8435\\
9&9965\\
10&11363\\
11&12557\\
12&13475\\
13&14045\\
14&14195\\
15&13853\\
16&12947\\
17&11405\\
18&9155\\
19&6125\\
20&2243
\end{array}
