If 2n boys are divided into two equal subgroups randomly, find the probability that the two tallest boys are in the same subgroup.
Here is what I though:
There are $$\frac{2n!}{n!\cdot n!\cdot2!}$$ ways to divide 2n boys into two subgroups.
The tallest two will be in one group, thus the rest of boys have $$\begin{pmatrix}2n-2 \\n-2\end{pmatrix}$$ possibilities.
So my answer is $$\frac{\begin{pmatrix}2n-2 \\n-2\end{pmatrix}}{\frac{2n!}{n!\cdot n!\cdot2!}}$$
I'm not really sure how to simplify this, and it seems too complicated to be the right answer…
Is there anywhere I made a mistake or any other easier way to do this problem?
Best Answer
It's not that hard. Say the tallest boys are A and B, and the groups are picked by lining them all up and dividing the first $n$ from the second $n$. Then WLOG A is in the first $n$. Then it is equally likely for B to be in any of the remaining $2n-1$ spots, of which $n-1$ would place him in the same group as A. So the probability is $\boxed{\frac{n-1}{2n-1}}$.