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.
Probability problem: what’s the chance that two strongest teams will play in the same subgroup.
combinatoricsprobability
Best Answer
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.