I don't think it is possible to have a balanced schedule and avoid having one player who is always in the foursome. For example I have found the following schedule :
( 8 4 6 .) ( 7 11 3 .) (12 2 10 .) ( 5 1 13 9)
( 5 7 2 .) ( 4 13 3 .) (12 1 8 .) (10 11 9 6)
( 8 5 11 .) ( 2 13 6 .) ( 3 10 1 .) ( 9 4 12 7)
{ . . . .} ( 1 6 7 .) ( 4 5 10 .) ( 2 3 8 9)
by starting with an all foursome schedule for 16 players and then cutting it back. Player 9 is always in the foursome, the other players all have one game in a foursome. The schedule should have no pair of players who play together twice. Hope that helps.
Ian