[sweetjazz]

[ QUOTE ]
I probably explained this poorly.

Suppose I have a team game that takes 3 players at a time. I want to schedule a series of games that has each player be in a game with every other player at least once, but no more than twice.

For example:
Game 1 Game 2
ABC DEF
ADE BCF
ABF CDE

In this case, a plays everyone once, and B twice. But D plays E 3 times, and doesn't play B ever.

There has to be some math-y name for this type of thing.

For 6 players in 3-man games, I think you'd have to have 4 matches. But I'm not even sure I can satisfy the play once but not twice criteria.

[/ QUOTE ]

Yeah, you can take your example and generalize it into an argument for impossibility.

Pick one team (call it 'A'). There must be some team they play twice (call them 'B'). In those two games, A plays B plus two other teams (call them 'C' and 'F'). Let 'D' and 'E' be the remaining teams. We know they play each other the two times that A plays B. But both D and E must play A once, and there is only game left for A, so that is the third time that D and E play.

I wonder if there is a quicker "slicker" way to prove this.