Speaky,
Have a look at this message where I explain about a problem that arises when you assign 12 players to foursomes. This works against you when trying to arrange that everyone plays everyone else equally often. If your aim is to have no pair of players play together more than twice, then I think you must have at least 6 pairs of players who never get placed together in a foursome. Such a schedule would be,
(C F I K) (J L E A) (G B D H)
(E H A F) (B K G J) (D C L I)
(A D B I) (L J K H) (F G C E)
(D G F L) (A J C B) (K I H E)
(I E J G) (H B L C) (F A K D)
where AG BE BF DE DJ FJ are the pairs who never play together.
There is an alternative you could ask the players to form buddy pairs and then use the following:
(A G B H) (C I D J) (E K F L)
(A G C I) (B H F L) (D J E K)
(A G D J) (B H E K) (C I F L)
(A G E K) (B H C I) (D J F L)
(A G F L) (B H D J) (C I E K)
where the pairs AG BH CI DJ EK and FL always play together. It has the advantage that all pairs do occur at least once.
If you are considering more than 5 rounds then see this topic or this one.
Ian.