A group of $2n$ boys is to be divided into two groups of $n$ boys . What is the probability that the two tallest boys are in different groups ?
This is how I attempted it:
The probability that the two boys are in same group can be obtained as follows:
First we separate those two particular boys leaving us with $2n-2$ boys. We then form a group of $n$ boys not containing the two particular boys giving us a group $n-2$ boys in which the two boys can be accommodated. The probability of forming such groups is $\frac{\binom{2n-2}{n}}{\binom{2n}{n}}$. Thus the actual probability of forming the groups with the two boys in different groups is $1- \frac{\binom{2n-2}{n}}{\binom{2n}{n}}$ . However there seems a problem with this. Could you please point out where I was wrong ?
Also , please note that I already know the correct solution to this problem. I just wanted to correct my mistake.
Thanks for your help !
Best Answer
Let us assume that the tallest boys are Andrew and Bruce. The configurations of this kind $$ (A\text{ together with }n-1\text{ other people })\quad (B\text{ together with }n-1\text{ other people }) $$ are $\binom{2n-2}{n-1}$ (it is enough to select Andrew's mates), while the configurations of this kind $$ (A,B\text{ together with }n-2\text{ other people })\quad (n\text{ other people }) $$ are $\binom{2n-2}{n-2}=\binom{2n-2}{n}$ (it is enough to select Andrew and Bruce's mates). The wanted probability is so
$$ \frac{\binom{2n-2}{n-1}}{\binom{2n-2}{n-1}+\binom{2n-2}{n}}=\frac{\binom{2n-2}{n-1}}{\binom{2n-1}{n-1}}=\color{red}{\frac{n}{2n-1}}. $$