Tournament probability question If we have 8 teams in the quarter final
if 3 teams are from the same country
what is the probability that 2 of the 3 will be competing against each other in the quarter final?
From my understanding, if we only had 2 team, probability is basically 1/7 because we are drawing 1 from the remaining 7 teams to pair up.
But what about 3 teams, 4 teams? If we have 5 teams then definitely the probability will be over 1 because we are guaranteed to have a pair that are from same country.
I want to know how to calculate this
 A: Number the $3$ teams and let $E_i$ denote the event that team $i$ does not have to play against a team of the same country.
Then the probability equals $$1-P(E_1\cap E_2\cap E_3)=1-P(E_1)P(E_2\mid E_1)P(E_3\mid E_1\cap E_2)=1-\frac57\frac45\frac33=\frac37$$
A: It's the number of $4$ match sets containing a pairing of $2$ of the $3$ teams from the same country divided by the total number of $4$ match sets.
We can number teams $1$ to $8$ with $1$ to $3$ being from the same country. For the number of $4$ match sets containing a pairing of $2$ of the $3$ teams from the same country, there are $3$ possible pairings $(1,2)(1,3)(2,3)$ and for every one of those, from the remaining $6$ teams there are $5$ ways to match the next team, and from the remaining $4$ there are $3$ ways to match the next team leaving a matched pair. This number is therefore $3\cdot 5\cdot 3 = 45$
For the total, there are $7$ ways to match the first team and from the remaining $6$, there are $5$ ways to match the second team and from the remaining $4$ there are $3$ ways to match the third team leaving a matched pair. The total is therefore $7\cdot 5\cdot 3 = 105$
$$P = \frac{3\cdot 5\cdot 3}{7\cdot 5\cdot 3} = \frac{3}{7}$$
