I read this site all the time for solutions to problems at work, but I cant find anything that quite applies to this problem
There is currently an african snooker championship taking place in Egypt (6 reds) and there were 3 saffas (south africans like me) who made it through to the last 16. So for the purposes of this calculation we must also allow for seedings, which 1 saffa made from what i understand..
In brief: 3 of 16 players ended up in the same quadrant(i.e. 4 places playing to a top 8 spot in the knockout stages) of the draw, with one player being a seeded saffa (assuming there are 8 seeds). What is the probability and statistical likelihood (with all the bells and deviations you can throw at me) of this happening in a random draw for all 16 places please?
I had no idea where to start with this.. but using some game theory i tried $n(n-1)(n-2)/10000$ - so I used the $25%$ for a quadrant and came up with 1.38% .. yes yes this is black magic math to you bright guys probably, but I was lost ok, so I am here now..
I know that in other major comps the saffas could not be drawn against their seeded fellow countryman, but that is a different fight and post altogether.
The question has many terms which I do not understand (some of which was asked in the comments), however here is my interpretation:
There are four quadrants, and the seeded player must land in one of them. There are three other places in that quadrant, and twelve places not in that quadrant. We now place the two other saffa players in these 15 places. There are ${3\choose 2}=3$ ways to place them in that same quadrant with the seeded player, and ${15\choose 2}=105$ ways to place them among $15$ places. Hence, if the nonseeded players are placed at random, there is a probability of $$\frac{3}{105}\approx 2.9\%$$ that they will both be placed together with the seeded player.