



 
Tyler, It is not possible to have every pair together in the way you describe. What you are looking for corresponds to a mathematical object known as a resolvable transversal design with 4 groups of size 6. This is known not to exist. (Because it is equivalent to the existence of 3 mutually orthogonal Latin squares of side 6, which were shown not to exist by the Gaston Tarry in around 1900.) The problem arises mainly because of the number 6. If, for example, there were 7 teams, then you could do it. That opens the door to an approximate solution that is not perfect, but might not be too bad. Imagine you have 7 teams in round 1. (Team 7 is a fake.) In round 2, shift the leads by 1 (the lead from 7 wraps around to team 1), the seconds by 2 (wrapping around similarly, 6> 1, 7 > 2) and the thirds by 3 (again, wrapping around similarly. Now take the lead from team 7 and use it to replace lead #7 wherever that player occurs. Do the same thing with the second and third from team 7. (Hope this makes sense.) Now delete the seventh team in each round, and all of round 2. Good luck.  


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