About the probability in rolling dice If A and B each roll two dice, how do we calculate the probability that the maximum number of A is greater than the maximum of B and that the minimum number of A is also greater than the minimum of B?

Hint: Start by computing the probability mass function of the maximum for player $A$, i.e., compute $Pr(\max=x)$ for each $x$.
By the way, if you are able to solve the problem for the maximum, it will be easy to get the solution for the minimum.

Edit: Since it's solved, I will have some fun adding the formulas for the cases:
Let $m$ be the maximum of the roll of two dices $d_1$ and $d_2$. Assuming that the dice rolls are independent, we can compute the probability of the maximum being $x \in \{1,2,3,4,5,6\}$ by
$$
Pr(m=x) = \\
\ \ \ 
  Pr(d_1=x \wedge d_2<x) \\
+ Pr(d_1<x \wedge d_2=x) \\
+ Pr(d_1=x \wedge d_2=x)
$$
Which means that either the first dice $d_1$ is the maximum, or the second dice $d_2$ is the maximum or both are the maximum.
This gives:
$$
Pr(M=x) = \frac{1}{6} \cdot \frac{x-1}{6} 
+ \frac{x-1}{6} \cdot \frac{1}{6} 
+ \frac{1}{6} \cdot \frac{1}{6} 
= \frac{2x-1}{36} \ .
$$
Alternatively, you could compute 
$$
Pr(m=x) = Pr(d_1 \leq x \wedge d_2 \leq x) 
- Pr(d_1 \leq x-1 \wedge d_2 \leq x-1)
$$
Which means the probability of at least one of the dice being less than or equal to $x$ but not both.
This gives:
$$
Pr(M=x) = \frac{x}{6} \cdot \frac{x}{6} 
- \frac{x-1}{6} \cdot \frac{x-1}{6} 
= \frac{x^2 - (x-1)^2}{36} \ .
$$
And simplifying we get the same result as before.
The computation for the minimum $m$ is similar, with the inequality signs switched. The end result will be symmetrical, i.e.:
$$
Pr(m=x) = Pr(M=7-x) = \frac{13-2x}{36}\ .
$$
If you make the computations for each $x$ you'll get the results presented by HJ_beginner.
While we are at it, we may take the opportunity to compute the cumulative probability distribution of the maximum $M$ for each $x \in \{1,2,3,4,5,6\}$:
$$
F(M=x) = Pr(M \leq x) = \sum_{j=1}^x Pr(M = x) = \frac{x^2}{36} \ .
$$
Similarly, for the minimum $m$:
$$
F(m=x) = Pr(m \leq x) = \sum_{j=1}^x Pr(m = x) = \frac{12x-x^2}{36} \ .
$$

Now, for the question, if we want to know the probability of player $A$ having his/her maximum above player $B$, i.e., $M_A > M_B$, we will have:
$$
Pr(M_A > M_B) = \sum_{j=2}^6 Pr(M_A = j \wedge M_B < j)
$$
Assuming that the rolls of $A$ and $B$ are independent, we get
$$
Pr(M_A > M_B) = \sum_{j=2}^6 Pr(M_A = j) \times Pr(M_B < j)
$$
Noting that $Pr(M_B < j) = Pr(M_B \leq j-1)$, we get
$$
Pr(M_A > M_B) = \sum_{j=2}^6 Pr(M_A = j) \times Pr(M_B \leq j-1)
$$
Using the formulas for each probability we have:
$$
Pr(M_A > M_B) = \sum_{j=2}^6 \frac{2j-1}{36} \times \frac{(j-1)^2}{36}
$$
$$
Pr(M_A > M_B) = \sum_{j=1}^5 \frac{(2j+1)j^2}{36}
$$
Using Faulhaber's formulas we obtain:
$$
Pr(M_A > M_B) = \frac{2(5^4 + 2 \times 5^3 + 5^2)/4 + (2 \times 5^3 + 3 \times 5^2 + 5)/6}{6^4}
$$
$$
Pr(M_A > M_B) = \frac{505}{6^4} = 0.38966
$$
Of course, we could also use the symmetry argument by HJ_beginner to get the same result:
$$
Pr(M_A > M_B) = \frac{1}{2} \times Pr(M_A \neq M_B) = \frac{1}{2} \times (1 - Pr(M_A = M_B))
$$
And $Pr(M_A = M_B)$ is:
$$
Pr(M_A = M_B) = \sum_{j=1}^6 Pr(M_A = j) \times Pr(M_B = j)
$$
$$
Pr(M_A = M_B) = \sum_{j=1}^6 \frac{2j-1}{36} \times \frac{2j-1}{36} = \frac{1}{6^4} \times \sum_{j=1}^6 4j^2-4j+1 
$$
$$
Pr(M_A = M_B) = \frac{1}{6^4} \times 
\left( 
4 \times \frac{\color{red}{6} \times (6+1) \times (2 \times 6 + 1)}{\color{red}{6}} 
- 4 \times \frac{6 \times (6+1)}{2} + 6 \right)
$$
$$
Pr(M_A = M_B) = \frac{286}{6^4} = 0.22068
$$
Hence
$$
Pr(M_A > M_B) = \frac{1}{2} \times \left(1 - \frac{286}{6^4} \right) = \frac{505}{6^4}
$$

As for the minimum, $Pr(m_A > m_B)$ the result is the same by symmetry.
