John,
Thanks for the interesting question. Certainly I can make a 16 round schedule where there are 12 games played each. For example:
( 1 3 11 v 10 14 13) ( 7 9 16 v 5 4 15) ( 2 12 6 8)
( 2 4 12 v 11 15 14) ( 8 10 1 v 6 5 16) ( 3 13 7 9)
( 3 5 13 v 12 16 15) ( 9 11 2 v 7 6 1) ( 4 14 8 10)
( 4 6 14 v 13 1 16) (10 12 3 v 8 7 2) ( 5 15 9 11)
( 5 7 15 v 14 2 1) (11 13 4 v 9 8 3) ( 6 16 10 12)
( 6 8 16 v 15 3 2) (12 14 5 v 10 9 4) ( 7 1 11 13)
( 7 9 1 v 16 4 3) (13 15 6 v 11 10 5) ( 8 2 12 14)
( 8 10 2 v 1 5 4) (14 16 7 v 12 11 6) ( 9 3 13 15)
( 9 11 3 v 2 6 5) (15 1 8 v 13 12 7) (10 4 14 16)
(10 12 4 v 3 7 6) (16 2 9 v 14 13 8) (11 5 15 1)
(11 13 5 v 4 8 7) ( 1 3 10 v 15 14 9) (12 6 16 2)
(12 14 6 v 5 9 8) ( 2 4 11 v 16 15 10) (13 7 1 3)
(13 15 7 v 6 10 9) ( 3 5 12 v 1 16 11) (14 8 2 4)
(14 16 8 v 7 11 10) ( 4 6 13 v 2 1 12) (15 9 3 5)
(15 1 9 v 8 12 11) ( 5 7 14 v 3 2 13) (16 10 4 6)
(16 2 10 v 9 13 12) ( 6 8 15 v 4 3 14) ( 1 11 5 7)
Above all pairs of players are together in the same group of 6 exactly 4 times, pairs that partner only once oppose 3 times, while pairs that partner twice, oppose twice. So this should be optimal for 16 rounds.
It is theoretically possible to have a perfect 20 round schedule where there are 15 games played each - all partners twice and all opponents three times. Unfortunately I have not been able to find this yet, but if I do, I will post it here.
