Kendo match schedules are fair

Some time ago I posed the question wat the probability distribution was of X, the number of groups with more than one fellow dojo member, when m members of our dojo are randomly assigned to any one of g groups, each consisting of n group competitors.

The reason was that in the last competition eight members of our dojo competed. The first round was a group competition. In total there were 29 groups, all but one with three competitors. To our surprise two of our dojo members were in the same group. All the rest were divided over different groups, so the rest of us didn't have to fight friends from the dojo.

The reason why it surprised us was that until today, we were much less fortunate. Some began even to suspect that there was some evil scheming at work.
Finding the answer to the probability distribution of the number of groups with more than one fellow dojo member seemed quite a simple task. So, even though I have been quite busy, I wanted a generalisation of the question (see Kendo match schedules fair? (rev.)). However, this made the question much harder to answer and not at all clearer.

So, after a while I decided to post the answer to the original question.
Let X be the number of groups with more than one fellow dojo member; m is the number of members of our dojo; they are randomly assigned to any one of g groups, each consisting of n group competitors. In total the number of competitors is about N = n·g.

Then X is binomially distributed with the number of (Bernoulli) trials is equal to g (the number groups in the competition) and p the probability of more than one member of our dojo in the group:

Here:

This holds for all dojo in the competition, so it is much handier to calculate the probability distribution for each dojo separately; especially since we are only interested in proving or disproving that there is an evil scheme against our own dojo.

In the example of the last tournaments, this results in the following probability distribution*:




Now, X = 1 seems a perfectly reasonable outcome for the tournament. This makes me curious what is the probability of a (fictional, but close-enough to typical) other outcome. Typically our dojo competes with six members (m = 6), there are twelve groups (g = 12) of four competitors (n = 4; N = 48). We get two groups with multiple members of our dojo (X = 2). How unlikely is this typical situation?

P(X = 3 | g = 12; p = 0,0708) ≈ 0,159.

Now, I cannot be convinced that there are other forces at work than mere chance.