the 10 teams are divided into 2 divisions of 5. So each team plays the other four teams in their division once each season. How many years would it take to a Team to play each of the other 9 teams in the league, the same number of times? i.e. play everyone 3 times, or 4 times. Thanks you.
asked Jul 30, 2013 in Statistics Answers by golfing sam

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Let the teams be denoted by 1A, 1B, 1C, 1D, 1E and 2A, 2B, 2C, 2D, 2E, where the numbers are for division 1 or 2, and the teams in the division are A to E.

The games for division 1 would be: 1A v 1B, 1A v 1C, 1A v 1D, 1A v 1E, 1B v 1C, 1B v 1D, 1B v 1E, 1C v 1D, 1C v 1E, 1D v 1E, each team playing 4 matches. That’s 10 matches in each season. Similarly for division 2. The league matches mean that each team in division 1 also has to play each team in division 2. So we have 1A v 2A, 1A v 2B, etc., and 1A will have 5 matches to play against the teams in division 2. Similarly for the other teams in division 1, making 25 matches between the divisions.

If it takes one season to play 10 games, in which each team plays each of the other teams in the same division once, and that is the maximum number of games in the season, then it would take another 25/10=2.5 years for all the teams in one division to play all the teams in the other division just once. So that’s one year (season) for the division games plus 2.5 for the league games=3.5 years. If 4 matches between the same pair of teams are to take place, it would take 14 years (4 times 3.5) for each team to play each of the other 9 teams; or 7 years to play each twice.

answered Jul 13 by Rod Top Rated User (570,500 points)

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