Christopher,

My program that searches for "most balanced" schedules found the one pasted below. The computation is too big for me to verify that this is best possible in less than a week. All I can say is that this one is likely close to best.

The 12 players correspond to the positions in the sequences, and the numbers indicate the foursomes. In this schedule not very pair of players get to play in the same group (e.g. 1 and 6, or 1 and 12). I suspect that isn't possible.

Day 0 : (0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2)
Day 1 : (0, 0, 1, 2, 0, 1, 1, 2, 0, 1, 2, 2)
Day 2 : (0, 1, 0, 2, 0, 1, 2, 1, 2, 1, 0, 2)
Day 3 : (0, 1, 1, 2, 1, 2, 0, 0, 2, 0, 2, 1)

--Victoria

In May of 2016 we were asked to clarify Victoria's schedule above.

Let the 12 players have names A, B, C and so on to L and put the names of the players on the table.

Each day there are 3 foursomes labeled foursome 0, foursome 1 and foursome 2. Each player looks down the column of digits below their name to see which foursome they are in. For example player E is in foursome 1 on day 0, foursome 0 on day 1, foursome 0 on day 2 and foursome 1 on day 3.

Reading across the rows you can see the foursomes on each day. For example on Day 2 the foursomes are

Foursome 0: A, C, E, K
Foursome 1: B, F, H, J
Foursome 2: D, G, I, L