Probability of inequalities between max values of samples from two different distributions Given samples from two empirical distributions (not necessarily passing tests of normality, but a solution for the normal case would definitely be useful) what's the probability that the maximum value from sample 1 will be greater than the maximum value from sample 2? 
I have methods for approximating the probability by brute force computation using Mathematica's built in empirical distribution / probability handling, but I would like a less opaque way to cross check my results. 
To give some context - imagine you are proposed with the following: 


*

*The average height of citizens in Country A is 72 inches with a standard deviation of 6 inches

*The average height of citizens in country B is 66 inches with a standard deviation of 3 inches

*A random sample of 5 citizens is drawn from country A

*A random sample of 25 citizens is drawn from country B


Question: What is the probability that the tallest member of sample A is taller than the tallest member of sample B?

I hope I got your question right and your question is mathematical in the sense that you want to get an answer in terms of the distribution of the sample.
Assume you have two samples $X_1,...,X_n\sim X$ and $Y_1,...Y_n\sim{Y}$. If the two samples are from the same distribution you can just take $X=Y$, this doesn't matter here.
Since you seem to be motivated by a practical application, I will assume that that both $X$ and $Y$ are absolutely continuous (i. e. have densities $p_x, p_y$). Let $M_x^n=\max_{i=1...n}X_i, M_y^m=\max_{i=1...m}Y_i$.
First, for any iid sample from $Y$ we have
$$
P(M_y^m<t)=\prod_{i=1}^m P(Y_i<t)=(F_y(t))^m
$$
and the corresponding density is
$$
p_{M_y^m}(t)=mF_y^{m-1}(t)p_y(t).
$$
Now,
$$
P(M_x^n<M_y^m)=\int P(M_x^n<t)p_{M_y^m}(t)dt=\int F_{M_x^n}(t)p_{M_y^m}(t)dt=\int m(F_x(t))^n (F_y(t))^{m-1}p_y(t)dt.
$$
