Krdytkyu

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

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 $P_n(Xle N) = (frac{N}{6})^n$.

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
$$P =left((frac{6}{6})^n-(frac{5}{6})^nright)frac{5}{6}+left((frac{5}{6})^n-(frac{4}{6})^nright)frac{4}{6}+ ldots$$
or $$P =frac{1}{6^{n+1}} left(6^ncdot 5 - 5^n-4^n-3^n-2^n-1^n right)$$ 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 $6$.