Question

Person A rolls the dice 2 times while person B trolls the dice 3 times. The one with the highest individual score wins. ex. A: 2, 1 and B: 1,5, 1. In this case, B wins. If they have the same highest number ex. A: 5, 4 and B: 2, 2, 5, then they both win. Obviously, person B will have a higher probability. But what's a formula that proves that?

Answer 1

\[1/N^{n}\]
N = number in the dice
n = number of dice

Answer 2

The number of cases that A is win is as in the following.
\[\sum_{k=1}^{6}k^3(k^2-(k-1)^2)\]
\[=6^3(6^2-5^2)+5^3(5^2-4^2)+4^3(4^2-3^2)+3^3(3^2-2^2)+2^3(2^2-1^2)+1^3(1^2-0^2)\]
\[=4109\]
Similarly, the number of cases that B is win is the following.
\[\sum_{k=1}^{6}k^3(k^2-(k-1)^2)\]
\[=6^2(6^3-5^3)+5^2(5^3-4^3)+4^2(4^3-3^3)+3^2(3^3-2^3)+2^2(2^3-1^3)+1^2(1^3-0^3)\]
\[=5593\]