Competition problem — Combinatorics

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Combinatorics Difficulty 5.1 AIME, harder Prove it United States

Problem:

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

Solution:

Answer: $\frac{19}{40}$

Let Brian's roll be $d$ and let George's rolls be $x, y, z$ . By pairing the situation $d, x, y, z$ with $21-d, 7-x, 7-y, 7-z$ , 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 $\frac{1-\frac{1}{20}}{2}=\frac{19}{40}$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.