A probability game involving dice Consider two players A and B.
Player A rolls a fair, six-sided die $m$ times and notes the highest number on the upper face out of all of the rolls.
Player B rolls the same die $n$ times and notes the highest number on the upper face out of all of the rolls. (It's clear that as $m$ and $n$ increase, the score of 6 becomes more likely).
If $m>n$, what is the probability player A will win in terms of $m$ and $n$? Note that if the numbers are equal, player B wins.
(I know the total number of outcomes is $6^{m+n}$, but I need help counting the winning games).
 A: 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.
