Probability problem: what's the chance that two strongest teams will play in the same subgroup. The problem: To reduce the amount of games between teams, $2n$ teams were divided to 2 subgroups with $n$ teams in each group. What is the chance that the two strongest teams will play in the same subgroup? I have no idea how to solve this problem. There is $\dfrac{1}{2}$ chance that one group will play at a certain subgroup and also $\dfrac{1}{2}$ chance that the other group will play at a certain group. So would itthere be $\dfrac{1}{4}$ chance that two strongest teams will play at the same team? That would be too easy. 
 A: The naive guess would be $\frac 12$, not $\frac 14$:
To see this, suppose teams were assigned to subgroups $A,B$ independently with probability $\frac 12$ for each.  Then there are four equally probable ways to assign the two teams, namely $(A,A), (A,B), (B,A), (B,B)$. Each of these has probability $\frac 14$ and two of them have both teams in the same group.  And $\frac 14+\frac 14=\frac 12$.
This is only an approximation though, since the selections aren't independent.  Knowing, say, that one team has been assigned to $A$ reduces the probability that the second team is also assigned to $A$ (since one slot from $A$ is now occupied).  Of course, if $n$ is very large this effect can be negligible.
For a fixed $n$:  Assign the strongest team first.  Now there are $n-1$ slots left in whichever subgroup you picked and there are $2n-1$ teams left to fill them.  Each team is equally likely to occupy any given slot, so $\frac {n-1}{2n-1}$.
Note:  for large $n$ this approaches $\frac 12$ as it should.
A: The probability you are asking for is:
$\frac{n-1}{2n-1}$
Explanation:
Let's say the strongest team was put in Group A. There will be other $n-1$ out of the remaining $2n-1$ teams in Group A. So $P(both\_at\_the\_same\_group)=\frac{n-1}{2n-1}$
Your answer $1/4$ would be the answer to the question:
If not both groups are required to have $n$ teams (and each team goes to either group with probability $1/2$), what is the probability that the two strongest teams both end up in group A.
A: By direct counting we have


*

*number of division in two groups $\frac12\binom{2n}{n}$

*cases with the $2$ strongest team in the same group $\binom{2n-2}{n-2}$
then
$$p=\frac{\binom{2n-2}{n-2}}{\frac12\binom{2n}{n}}=\frac {n-1}{2n-1}$$
