Probability of a number rolled of a 20 sided dice being greater than the sum of the numbers rolled on 3 six sided die.
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 $P(Y=X)=\frac1{20}.$
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.
Your best bet is to keep track of two things for both sides: how likely this particular result is, and how likely anything less than this result is. Given these we can multiply them together relatively easily.
$$\begin{array}{r|rr|rr|r} x & 3\text{d}6 = x & 3\text{d}6 < x & 1\text{d}20 = x & 1\text{d}20 < x & 1\text{d}20 < x \cap 3\text{d}6 = x\\ \hline 1 & 0 & 0 & 1 & 0 & 0 \\ 2 & 0 & 0 & 1 & 1 & 0 \\ 3 & 1 & 0 & 1 & 2 & 2 \\ 4 & 3 & 1 & 1 & 3 & 9 \\ 5 & 6 & 4 & 1 & 4 & 24 \\ 6 & 10 & 10 & 1 & 5 & 50 \\ 7 & 15 & 20 & 1 & 6 & 90 \\ 8 & 21 & 35 & 1 & 7 & 147 \\ 9 & 25 & 56 & 1 & 8 & 200 \\ 10 & 27 & 81 & 1 & 9 & 243 \\ 11 & 27 & 108 & 1 & 10 & 270 \\ 12 & 25 & 135 & 1 & 11 & 275 \\ 13 & 21 & 160 & 1 & 12 & 252 \\ 14 & 15 & 181 & 1 & 13 & 195 \\ 15 & 10 & 196 & 1 & 14 & 140 \\ 16 & 6 & 206 & 1 & 15 & 90 \\ 17 & 3 & 212 & 1 & 16 & 48 \\ 18 & 1 & 215 & 1 & 17 & 17 \\ 19 & 0 & 216 & 1 & 18 & 0 \\ 20 & 0 & 216 & 1 & 19 & 0 \\ \hline \text{total} & 216 & & 20 & & 2052 \end{array}$$
Total up everything in the rightmost column, divide by the totals for the $3\text{d}6 = x$ and $1\text{d}20 = x$ columns, and you win: Bob wins $\frac{2052}{4320} = \frac{19}{40} = 0.475$ of the time.
To get ties, or where Bill wins, change what pairs you multiply: both $=$ ones for ties, and $3\text{d}6 < x$ and $1\text{d}20 = x$ for Bill's wins.
This particular one is interesting: because both distributions are symmetrical, and they have the same mean, Bill and Bob will both win the same proportion of the time.
It occurs to me that there's another part to this question: how do we efficiently calculate result probabilities for combinations of dice?
The answer to that is an operation called convolution, which I'll present here in discrete form.
Given two functions $f(x)$ and $g(x)$, the convolution is $$(f * g)(x) = \sum_{k = -\infty}^\infty f(k)g(x-k)$$
This can be interpreted in probability theory as the following: we have two random variables $f$ and $g$, with probability functions $f(x)$ and $g(x)$. the probability function for $f + g$ -- adding the two results together -- is equal to $(f * g)(x)$.
Obviously with those infinities in there, we have to fiddle with it a little to actually get anything done. In our case, because we're dealing with dice, our functions have what's called limited support: they're only non-zero in a small area, so we only need to cover that small area.
Let's do a specific example. Say I want to calculate the probability that I'll get a $7$ on $3\text{d}6$. $3\text{d}6$ is the same as $1\text{d}6$ + $2\text{d}6$, so I can convolve these two. I'll call their functions $f$ and $g$ respectively.
$f$ here has limited support: the only values it is non-zero for are $1$ through $6$. This allows us to change the limits of our summation to the bounds of $f$'s support.
$$\begin{align} (f * g)(7) &= \sum_{k=1}^6 f(k)g(7 - k)\\ &= f(1)g(6) + f(2)g(5) + f(3)g(4) + f(4)g(3) + f(5)g(2) + f(6)g(1) \\ &= \frac{1}{6}\cdot\frac{5}{36} + \frac{1}{6}\cdot\frac{4}{36} + \frac{1}{6}\cdot\frac{3}{36} + \frac{1}{6}\cdot\frac{2}{36} + \frac{1}{6}\cdot\frac{1}{36} + \frac{1}{6}\cdot\frac{0}{36} \\ &= \frac{15}{216} \end{align}$$
Using convolution, then, we can calculate the probabilities of the summed results of multiple dice, without necessarily considering every single simple event: to calculate, say, the probability distribution of $5\text{d}6$, we can take the distribution of $4\text{d}6$ and the distribution of $1\text{d}6$ and convolve them. And to get $4\text{d}6$'s distribution we can convolve $3\text{d}6$ and $1\text{d}6$, etc. So instead of counting out $7776$ possibilities, we instead handle $21\cdot6 + 16\cdot6 + 11\cdot6 + 6\cdot6 = 324$ total multiplications and a similar number of additions.