Probability of a number rolled of a 20 sided dice being greater than the sum of the numbers rolled on 3 six sided die. Bob rolls $3$ six-sided die and sums the numbers facing up. Bill rolls a single $20$ sided dice and records the number. What is the probability that Bob's number is greater than Bill's number.
I started the problem trying to come up with an equation, but that didn't work, so I resorted to creating $6$ six by six tables with all of the possible sums for Bob's die. Then, I counted the number of each number and created a chart and calculated the probability of each of those numbers occurring. Then, I multiplied the probability of each number occurring by the probability that Bill's number is greater than that number. Finally, I added them all up.
Obviously, this was very tedious and time-consuming. Is there a more elegant/less tedious way to do this problem. 
PS: Is there away to do a $n$ die vs $m$ dice problem without listing them all out? Is there a general formula? Up until now, that's what I've been doing.
 A: Let the total on the three six-sided dice be $X,$ $3\leq X\leq18.$
Let the number on the twenty-sided die be $Y,$ $1\leq Y\leq20.$
Given any particular value of $X,$ the probability
that the twenty-sided die will roll higher is 
$$P(Y>X \mid X) = \frac{20-X}{20} = 1 - \frac1{20}X.$$
The overall probability that the twenty-sided die will roll higher than
the total on the other three dice is
\begin{align}
P(Y>X) &= \sum_{n=3}^{18} P(Y > X \mid X)P(X=n) \\
&= \sum_{n=3}^{18} \left(1 - \frac1{20}X\right)P(X=n).
\end{align}
The last line of that set of equations is just the expected value of
$1 - \frac1{20}X.$
That is,
\begin{align}
P(Y>X) &= \mathbb E\left[1 - \frac1{20}X\right] \\
&= 1 - \frac1{20} \mathbb E[X] \\
&= 1 - \frac1{20} \left(\frac{21}{2}\right) \\
&= \frac{19}{40}.
\end{align}

If the question is actually the one posed in the original question body rather than in the original title,
namely the probability that $X > Y,$  then we simply observe that
for any given value of $X,$
$$ P(Y < X \mid X) = \frac{X-1}{20} = \frac{1}{20}X - \frac{1}{20}.$$
The rest of the calculation builds on this the same way the first calculation in this answer built on $P(Y > X \mid X).$
We find that
\begin{align}
P(Y<X) &= \mathbb E\left[\frac1{20}X - \frac1{20}\right] \\
&= \frac1{20} \mathbb E[X] - \frac1{20}\\
&= \frac1{20} \left(\frac{21}{2}\right) - \frac1{20} \\
&= \frac{19}{40}.
\end{align}
This should not be surprising, because it also follows from
$P(Y>X)=\frac{19}{40}$ and the "obvious" fact that $P(Y=X)=\frac1{20}.$
A: Whatever Bob rolls with the $6$-sided dice has a $1$ in $20$ chance of being matched, for a tie, by Bill's roll of the $20$-sided die.  Whatever sum, $S=a+b+c$, Bob rolls, if you turn his dice over, the sum is $(7-a)+(7-b)+(7-c)=21-S$.  Similarly, whatever number $T$ Bill rolls, if you turn the $20$-sided die over, the number is $21-T$.  Thus for each outcome in which Bob wins, there is an equally likely outcome in which Bill wins, and vice versa.  Hence the probability of winning for each of them is the same, namely
$${1\over2}\left(1-{1\over20}\right)={19\over40}$$
Remark:  It's not literally necessary that the "complementary" number for each side of a die be the opposite face, just that there be a complemenary number somewhere.  For $6$-sided dice, having opposite faces sum to $7$ is fairly standard; I believe it's also standard for $20$-sided dice to have opposite faces sum to $21$.
Also, I'd like to credit David K's answer with motivating this one.  When I saw from his analysis that the two probabilities were equal, I decided there ought to be simple reason why.  As luck would have it, I found one.
A: Call the probability of scoring a $k$ with the D20 $q_k=q=1/20$ and scoring a $k$ with the 3 D6'S $p_k$ then we are looking for the probability that the D20 has the greater score
$$p_3\sum_{k=4}^{20}q_k+p_4\sum_{k=5}^{20}q_k+\ldots +p_{18}\sum_{k=19}^{20}q_k$$
Or simply
$$q(17p_3+16p_4+\cdots +2p_{18})$$
but by symmetry for every way to score $k$ with 3 D6s there is a way to score $21-k$, simply by subtracting the score on each die from $7$. The probability of rolling a total of $18$ with $3$ D6s is therefore the same as rolling a total of $3$ ($p_3=p_{18}$) and rolling a total of $17$ is the same probability as rolling $4$ ($p_4=p_{17}$) etc. So
$$\text{required probability}=19q(p_3+p_4+\cdots +p_{10})=19q(p_{11}+p_{12}+\cdots +p_{18})$$
But of course
$$p_3+p_4+\cdots +p_{18}=1$$
So
$$2(p_3+p_4+\cdots +p_{10})=1=2(p_{11}+p_{12}+\cdots +p_{18})$$
$$\implies p_3+p_4+\cdots +p_{10}=p_{11}+p_{12}+\cdots +p_{18}=\frac{1}{2}$$
Giving
$$\text{required probability}=19\cdot\frac{1}{20}\cdot\frac{1}{2}=\frac{19}{40}\tag{Answer}$$
Note that, since the probability of the players getting the same score is $(1/20)\cdot 1=1/20$, then the probability that the D20 loses is the same as it winning i.e. $19/40$.
