Probably of winning with 2 dice (maximum of them) against another one could some of you help me to find out what is the probability of A) obtain with two dice a greather number than another die? B) and if the dice are 3 how can I do? Not the sum of the 2 dice, but the greatest value of those 2 against another die
 A: When rolling $1$ die, the probability of throwing "N or less" is $P(X \le N) =\frac{N}{6}$. When rolling $n$ dice, the probability of throwing "N or less" is $P_n(X\le N) = (\frac{N}{6})^n$. 
So with $n$ dice, the probability of throwing "at least one N and nothing higher than N" is  $P_n(N) = (\frac{N}{6})^n-(\frac{N-1}{6})^n$. This means the probability of someone with $n$ dice rolling a higher number than someone with $1$ die would be: $$P = P_n(6)\cdot P(X \le 5) + P_n(5)\cdot P(X \le 4)+ \ldots + P_n(2)\cdot P(X \le 1)$$
or
$$P =\left((\frac{6}{6})^n-(\frac{5}{6})^n\right)\frac{5}{6}+\left((\frac{5}{6})^n-(\frac{4}{6})^n\right)\frac{4}{6}+ \ldots$$
or $$P =\frac{1}{6^{n+1}} \left(6^n\cdot 5 - 5^n-4^n-3^n-2^n-1^n \right)$$ or
$$P= \frac{5}{6} - \frac{1}{6^{n+1}} \sum_{i=1}^5 i^n$$
For $n=1$, i.e two people with one die each, we get $P= \frac{15}{36}$, a known result. For $n=2$ we get $P= \frac{125}{216}$ and for $n=3$ we get $P= \frac{855}{1296}$.
EDIT
It makes sense that the probability tops out at $P=\frac{5}{6}$ as there is always $\frac{1}{6}$ probability that the person with $1$ die throws a $6$, in which case it doesn't matter how many dice the other person has as they will never beat the $6$. 
