Round of Eight tournament pairing combinations I notice recently that in a Starcraft tournament in the round of eight, there were 3 Zerg players, 3 Protoss players and 2 Terran players, yet none of the four pairings had a Terran versus Zerg matchup. What are the chances of that?

I computed an answer to this problem by writing a computer program that enumerated all the possibilities and counted those pairings in which a TvZ occurred, but is there a mathematical way to compute the odds without enumerating every possibility manually?

For example, lets say the players are Z1, Z2, Z3, P1, P2, P3, T1 and T2. One possible pairing might be:

Z1 vs P1
Z2 vs P2
Z3 vs P3
T1 vs T2


In this set of pairings there are no TvZ matches. What is the odds of this occurring?
I don't know the game, so will assume that anyone can be matched with anyone else. I will also assume all pairings are equally likely. Let the Terrans be A and B. Choose A's opponent first. Maybe it is B, in which case no Terran is matched with a Zerg. This has probability $\frac{1}{7}$.

But maybe A is matched with someone else. The probability this is a non-Zerg is $\frac{3}{7}$. And if that has happened, the probability B is matched with a non-Zerg is $\frac{2}{5}$. So the required probability is
$$\frac{1}{7}+\frac{3}{7}\cdot \frac{2}{5}.$$
This turns out to be $\frac{11}{35}$. In the language of odds, which is little used in probability theory, the odds are $11$ to $24$.
Total ways of pairing from a pool of $n = (n-1)\cdot(n-3). . .\cdot3\cdot1 = (n-1)!!$

Total ways of pairing from two groups with $m$ and $n = mn$



Pairings when both T's together = $5!!$

Pairings between T's and P's and between remaining pool of $4 = 2\cdot3\cdot3!!$

$$Pr = \frac{5!! + 2\cdot3\cdot3!!}{7!!} = \frac{33}{105} = \frac{11}{35}$$

Odds are usually used for betting as odds against

Odds against = $24:11$