# Probability of dividing boys in 2 groups

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.

• Now these two tallest could be in either group, is it not? – Satish Ramanathan Sep 30 '18 at 17:01
• Yes but I thought that each group is identical right ? Because the probability of selecting each boy is equally likely , I guess – Aditi Sep 30 '18 at 17:02
• Any group of $n-2$ boys that we form would have boys being equally likely to be selected – Aditi Sep 30 '18 at 17:04

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

$$\frac{\binom{2n-2}{n-1}}{\binom{2n-2}{n-1}+\binom{2n-2}{n}}=\frac{\binom{2n-2}{n-1}}{\binom{2n-1}{n-1}}=\color{red}{\frac{n}{2n-1}}.$$

• Thank you very much :) now I understood ! – Aditi Sep 30 '18 at 17:25

Given that $$2n$$ boys are randomly divided into two subgroups containing $$n$$ boys each.

So, we can choose $$n$$ boys from $$2n$$ boys in $$\dfrac{\dbinom{2n}{n}}{2!}=\dfrac{(2n)!}{2!\times(n!)^2}$$ ways.

We need to find the probability of two tallest boys in different groups.

So, if we seperate $$2$$ tallest boys from $$2n$$ boys we have $$2n-2$$ boys.

Now we can arrange the group of boys in $$\dfrac{\dbinom{2n-2}{n-1}}{2!}=\dfrac{(2n-2)!}{((n-1)!)^2\times 2!}$$ ways

Now we can separate these $$2$$ tallest boys into $$2$$ different groups in $$\dbinom{2}{1}$$ ways.

So, in total we have $$\dfrac{(2n-2)!}{((n-1)!)^2\times 2!}\times \dbinom{2}{1}$$ ways

The required probability is $$\dfrac{\dfrac{(2(n-1))!\times2}{((n-1))^2\times2!}}{\dfrac{(2n)!}{(n!)^2\times2!}}=\dfrac{n}{2n-1}$$

• Thank you for your help :) but could you please point out where I went wrong ? – Aditi Sep 30 '18 at 17:17
• @Aditi I think you made a mistake in forming the groups after separating the two boys. – Key Flex Sep 30 '18 at 17:21
• Ohh okay thank you ! – Aditi Sep 30 '18 at 17:26