HMMT-Feb 2012 comb, Problem 2

Brian has a 20-sided die with faces numbered from 1 to 20 , and George has three 6 -sided dice with faces numbered from 1 to 6 . Brian and George simultaneously roll all their dice. What is the probability that the number on Brian's die is larger than the sum of the numbers on George's dice?

Solution: Let Brian's roll be $d$ and let George's rolls be $x, y, z$. By pairing the situation $d, x, y, z$ with $21-\overline{d, 7}-x, 7-y, 7-z$, we see that the probability that Brian rolls higher is the same as the probability that George rolls higher. Given any of George's rolls $x, y, z$, there is exactly one number Brian can roll which will make them tie, so the probability that they tie is $\frac{1}{20}$. So the probability that Brian wins is $\frac{1-\frac{1}{20}}{2}=\frac{19}{40}$. $\fbox{\frac{19}{40}}$.

Answer: \frac{19}{40}