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re: Compromise for an 8 game schedule: a 3-5-5 format
Posted on 6/1/23 at 5:58 pm to mckibaj
Posted on 6/1/23 at 5:58 pm to mckibaj
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
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
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