Leon,

It is difficult to compute the best schedule as the number of possibilities to consider is about 10 to the power 30. I have a program that will give pretty good solutions (close to best) in cases like this. Here is what it found:

Day 1 : (0, 0, 1, 2, 0, 1, 1, 2, 0, 1, 2)
Day 2 : (0, 1, 0, 2, 0, 1, 2, 1, 2, 1, 0)
Day 3 : (0, 1, 1, 2, 2, 0, 1, 1, 0, 2, 0)
Day 4 : (0, 1, 2, 0, 1, 1, 0, 2, 2, 0, 1)
Day 5 : (0, 1, 2, 2, 0, 2, 1, 0, 1, 0, 1)
Day 6 : (0, 0, 1, 0, 1, 0, 2, 2, 1, 1, 2)
Day 7 : (0, 1, 1, 2, 2, 0, 2, 0, 0, 1, 1)
Day 8 : (0, 1, 2, 1, 2, 2, 0, 1, 1, 0, 0)
Day 9 : (0, 0, 0, 1, 1, 2, 0, 1, 2, 2, 1)
Day 10 : (0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2)

The 11 positions in each sequence represent the players, 1 through 11. The numbers represent the group that player is in. For example, on Day 1, the groups are 1, 2, 5, 9; 3, 6, 7, 10; 4, 8, 11.

Because of the way the possibilities are generates, a flaw arises when there are groups of unequal size. In the schedule above, players 1 and 2 are never in the threesome.

Hope this works for you.
--Victoria