TITLE: Probability of a number rolled of a 20 sided dice being greater than the sum of the numbers rolled on 3 six sided die.
QUESTION [14 upvotes]: 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.

REPLY [10 votes]: 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.