The following ingredients are enough to determine the probabilities.
$1.$ If you toss $n$ times, the probability that the highest number is $\le k$ is $\frac{k^n}{6^n}$.
$2.$ If you toss $n$ times, the probability that the highest number is equal to $k$ is $\frac{k^n -(k-1)^n}{6^n}$.
In ($1$) and ($2$), $k$ ranges from $1$ to $6$.
Added: We do the cooking, in case the list of ingredients was not enough. Assume that $m\ge 1$ and $n \ge 1$.
Player A wins if (i) her highest number is a $6$, and B's number is $\le 5$ or (ii) A's highest number is $5$, and B's is $\le 4$, or (iii) A's highest number is $4$, and B's is $\le 3$, or (iv) A's highest number is $3$, and B's is $\le 2$, or (v) A's highest number is $2$, and B's is $\le 1$.
The probability of (i) is $\frac{6^m-5^m}{6^m}\cdot \frac{5^n}{6^n}.$ The probability of (ii) is $\frac{5^m-4^m}{6^m}\cdot \frac{4^n}{6^n}.$ The probability of (iii) is $\frac{4^m-3^m}{6^m}\cdot \frac{3^n}{6^n}.$ The probability of (iv) is $\frac{3^m-2^m}{6^m}\cdot \frac{2^n}{6^n}.$ The probability of (v) is $\frac{2^m-1^m}{6^m}\cdot \frac{1^n}{6^n}.$ Add up.
There are various other ways to compute the probability that A wins. If we look at the probabilities that we are adding, we can see that there are opportunities to simplify the expression for the sum. If we were dealing with $d$-sided dice, then simplification might be worthwhile, but for the case $d=6$ it probably isn't.