How to find out the probability that the tallest person in a group of people is a man? Assume we have a population of N men and women such that exactly $N/2$ people are men (set $M$) and $N/2$ people are women (set $W$).
Assume that the standard deviations for height between both groups is the same $\sigma$ however their averages are different with $\mu_M > \mu_W$.
We pick $n$ people at random from the population such that $n/2$ people are men and $n/2$ people are women.
Knowing $n$ what's the probability that the tallest person in the selected group is a man? 
Note: this is not homework, I came up with this question on my own.
Note 2: all distributions are normal distributions.
EDIT:
My current reasoning is that if $Z = M-W$:
$E(Z) = E(M-W) = E(M) - E(W)$
And 
$V(Z) = V(M-W) = V(M) + V(W) = \sqrt2\sigma$
Thus the probability that one man is taller than one woman is:
$P(Z>0)$
And so the probability that all women are taller than all men in the sample is: $(1 - P(Z>0))^{n/2}$:
However this result has lead me to some very odd conclusions. So I suspect I am wrong.

Let
\begin{align*}
M_1, \cdots, M_{N/2} &\overset{\text{iid}}{\sim} N(\mu_M, \sigma^2) \\
W_2, \cdots, W_{N/2} &\overset{\text{iid}}{\sim} N(\mu_W, \sigma^2) \\
S_M, S_W &\overset{\text{iid}}{\sim} \text{SRSWOR}(\{1, \cdots, N/2\}) \\
\end{align*}
where SRSWOR meaning simple random sample without replacement. Define
\begin{align*}
\textbf{A}_{S} = \{A_i\}_{i \in S}
\end{align*}
So we are looking at computing
\begin{align*}
P(\max(\textbf{M}_{S_M}) > \max(\textbf{W}_{S_W}))
\end{align*}
This is equal to
\begin{align*}
\sum_{s_M, s_W \in \binom{\{1, \cdots, N/2\}}{n/2}}P(\max(\textbf{M}_{S_M}) > \max(\textbf{W}_{S_W})|S_M=s_M, S_W=s_W)P(S_M=s_M)P(S_W=s_W)
\end{align*}
But since 
\begin{align*}
P(S_M=s_M) = P(S_W=s_W) = 1/\binom{N/2}{n/2}
\end{align*} due to SRSWOR and 
\begin{align*}
P(\max(\textbf{M}_{S_M}) > \max(\textbf{W}_{S_W})|S_M=s_M, S_W=s_W) = P(\max(\textbf{M}_{S_M}) > \max(\textbf{W}_{S_W}))
\end{align*}
due to i.i.d, we could just consider computing
\begin{align*}
P(\max(M_1, \cdots, M_{n/2}) > \max(W_1, \cdots, W_{n/2}))
\end{align*}
Working directly with the distribution of the maximum of normal random variables would be straightforward to compute numerically, although I don't think a closed-form would exist.
