re: Compromise for an 8 game schedule: a 3-5-5 format

So instead of playing all the other teams 10 times in 20 years, it’d be reduced to playing them 8.33 times in 20 years.

Total number of games against Team A = 5 games per year x 20 years = 100 games

Number of teams excluding Team A and its 3 permanent rivals = 12 teams

Again, each team plays Team A an equal number of times, so we divide the total number of games against Team A by the number of remaining teams:

100 games / 12 teams = 8.33 games
8 teams play Team A 8 times in 20 years
4 teams play Team A 9 times in 20 years