Probability of a given 3 teams playing each other in 1st round There are 10 total teams in a company volleyball tournament and 3 of them are from the same department, let's just call it the finance department. What is the probability that two teams from the finance department will face each other in the first round? 
This is a seemingly simple probability problem that was posed to me. I found a solution with a spreadsheet simulation that I believe is correct, but I haven't been able to replicate it with an elegant analytical solution that makes sense to me. 
 A: It is 1 minus the probability of none of the teams playing each other. Now, I assume that there are 5 games in the first round where the 10 teams are randomly paired to play against each other. So, given that there are 5 games, after the 'first' of the three teams is given a slot, there is an 8 in 9 chance that the 'second' team does not play the 'first', and once that has happened, there is a 6 in 8 chance that the 'third' team does not play either the first or second.
So, there is a $\frac{8}{9} \cdot \frac{6}{8}=\frac{2}{3}$ chance none of them meet, and hence a $\frac{1}{3}$ chance that some of them will meet.
A: $\frac79$ ways for the second team to be chosen if the first is set.  $\frac67$ ways to avoid a meeting once the second team is set.  $\frac79×\frac67=\frac23$.  Take the complement :  $1-\frac23=\frac13$.
A: The total number of possible pairings, calculated by pairing the first team with any of the remaining nine, then the next with the remaining seven, etc., is $(10-1)!! = 9\cdot7\cdot5\cdot3\cdot1$.
Similarly, the number of pairings in which the three finance teams do not play each other is $7\cdot6\cdot5\cdot3$ (because the first finance team can be paired with any of seven non-finance teams, and similarly for the next finance teams, and finally there are three ways to pair off the last four non-finance teams).
Therefore the probability that no two finance teams play each other is $$\frac{7\cdot6\cdot5\cdot3}{9\cdot7\cdot5\cdot3} = \frac23,$$
so the probability that two of the finance teams do play each other is $\frac13$.
A: How I did it.
Lets call the three teams from finance $F_1,F_2, F_3$
There are 9 teams that team F_1 could match with two of which are in finance.
$F_1$ matches with another finance team $\frac 29$
$F_1$ does not match with another finance team $\frac 79$
If $F_1$ does not match with a finance team what is the chance $F_2$ matches with $F_3?$
There are $7$ teams $F_2$ could be matched with.  One of which is $F_3$
$\frac 29 + \frac 79\cdot\frac 17 = \frac 13$
A: $\boxed.\boxed.\quad\boxed.\boxed.\quad\boxed.\boxed.\quad\boxed.\boxed.\quad\boxed.\boxed.\quad$
If the three are to avoid each other, they can be placed in $10\cdot8\cdot6$ ways,
so P(they don't meet) = $\dfrac{10\cdot8\cdot6}{10\cdot9\cdot8} = \dfrac23$
and P(they meet) $=\dfrac13$ 
