I did this calculation this morning. There are 16 teams in the AJHL. We currently have players on 7. If all play on the same date which happens often (e.g., this evening) what is the chance that no two of these teams are playing each other? IIRC, that happened last Saturday.
I will assume that the schedule is balanced, that is there are no teams that play each other a different number of times. One can place the first team (whichever you want) in any of 16 positions (home and away in each game) with equal likelihood, i.e. 1 = 16/16. Then consider the 2nd of the 7 teams. It can be placed in any of 15 remaining positions with 14 resulting in it not playing the one already seated, i.e., 14/15. Then the third team can be placed in any of 14 positions with 12 not resulting in them playing one of the two already seated teams, i.e., 12/14. Then follows 10/13, 8/12, 6/11, and 4/10. Multiplying one gets (16*14*12*10*8*6*4) / (16*15*14*13*12*11*10). The numerator = 2^7 * 8!. The denominator = 16! / 9!. The result is 2^7 * 8! * 9! / 16! which is about .08951... Thus more then 10 of 11 times we should have AJHL recruits playing each other.