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.

First off, for a normal distribution to be applicable, $N$ needs to be sufficiently large, so let's assume it is and we can then ignore $N$ in our analysis.  For small $N$, we would need additional information to solve this problem, preferably the exact discrete distribution of heights.
What you need to solve this is "the distribution of the nth order statistic".  Or as Wolfram language calls it: the OrderDistribution
Let Dm be the distribution of the tallest male and Dw likewise for females
Dm = OrderDistribution[{NormalDistribution[\[Mu]m, \[Sigma]], n/2}, n/2]
Dw = OrderDistribution[{NormalDistribution[\[Mu]w, \[Sigma]], n/2}, n/2]

You want to know the probability that x>y given that x is distributed as Dm and y is distributed as Dw.
Mathematica can do the math for us as follows.
Probability[x > y, {x \[Distributed] Dm, y \[Distributed] Dw}, Method -> "Trace"]

yields:
$\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\frac{2^{-n-1} n^2 \text{erfc}\left(\frac{\text{$\mu $1}-x}{\sqrt{2} \sigma }\right)^{\frac{n-2}{2}} \text{erfc}\left(\frac{\text{$\mu $2}-y}{\sqrt{2} \sigma }\right)^{\frac{n-2}{2}} e^{-\frac{\text{$\mu $1}^2+\text{$\mu $2}^2+x^2-2 \text{$\mu $1} x+y^2-2 \text{$\mu $2} y}{2 \sigma ^2}} \text{Boole}[x>y]}{\pi  \sigma ^2}dydx$
which is the answer as an improper double integral.  It will only evaluate symbolically for special cases like these where 1) our random selection has just one man and woman each or 2) men and women have the same average height:

Now WolframAlpha will tell us that the average man's height is 5'6" and 5'2" for a woman.  Standard deviation of height yields 7.9".
With these numbers we can plot the answer corresponding to the real world:
ListPlot[Table[{2^nn,NProbability[x>y,{x\[Distributed]Dm,y\[Distributed]Dw}/.{n->2^nn,\[Mu]m->5+6/12,\[Mu]w->5 +2/12,\[Sigma]->7.9/12}]},{nn,1,8}],Joined->True,PlotMarkers->Automatic,AxesLabel->{"n","p"}]

Note that I used NProbability to essentially compute the value of the above improper double integral numerically.  I just chose to plot powers of 2 for n.
Obviously, this becomes asymptotic to 1 but quite a bit more slowly than I would have guessed.

