Tony,
With 12 players you can play 11 rounds where everybody plays each other three times.
( 1 2 4 8) ( 3 5 10 11) (12 6 7 9)
( 2 3 5 9) ( 4 6 11 1) (12 7 8 10)
( 3 4 6 10) ( 5 7 1 2) (12 8 9 11)
( 4 5 7 11) ( 6 8 2 3) (12 9 10 1)
( 5 6 8 1) ( 7 9 3 4) (12 10 11 2)
( 6 7 9 2) ( 8 10 4 5) (12 11 1 3)
( 7 8 10 3) ( 9 11 5 6) (12 1 2 4)
( 8 9 11 4) (10 1 6 7) (12 2 3 5)
( 9 10 1 5) (11 2 7 8) (12 3 4 6)
(10 11 2 6) ( 1 3 8 9) (12 4 5 7)
(11 1 3 7) ( 2 4 9 10) (12 5 6 8)
With 14 players the problem is harder. Counting pairs, it is easy to see that you will need to play 91 games at which point all players will have played each other six times each. This can be arranged as 30 rounds with 2 byes and one additional game.
(A B C G) (D L M N) (K E F H)
(B C D H) (E M N A) (L F G I)
(C D E I) (F N A B) (M G H J)
(D E F J) (G A B C) (N H I K)
(E F G K) (H B C D) (A I J L)
(F G H L) (I C D E) (B J K M)
(G H I M) (J D E F) (C K L N)
(H I J N) (K E F G) (D L M A)
(I J K A) (L F G H) (E M N B)
(J K L B) (M G H I) (F N A C)
(K L M C) (N H I J) (G A B D)
(L M N D) (A I J K) (H B C E)
(M N A E) (B J K L) (I C D F)
(N A B F) (C K L M) (J D E G)
(A F I K) (H J N C) (B D E L)
(B G J L) (I K A D) (C E F M)
(C H K M) (J L B E) (D F G N)
(D I L N) (K M C F) (E G H A)
(E J M A) (L N D G) (F H I B)
(F K N B) (M A E H) (G I J C)
(G L A C) (N B F I) (H J K D)
(H M B D) (A C G J) (I K L E)
(I N C E) (B D H K) (J L M F)
(J A D F) (C E I L) (K M N G)
(K B E G) (D F J M) (L N A H)
(L C F H) (E G K N) (M A B I)
(M D G I) (F H L A) (N B C J)
(N E H J) (G I M B) (A C D K)
(B F I M) (C G J N) (D H K A)
(E I L B) (F J M C) (G K N D)
(H L A E) (- - - -) (- - - -)
Finally try searching the message board for the phrase '10 players', I think you may find something that you can adapt.
Ian.
