To play this game with N dice (where N is greater than or equal to 3), each die must have 2N sides, numbered as follows: A: 1, 2, 3, ..., 2N B: N+1 of the faces should have value N. The remaining (N-1) of the faces should have value N+1 For all subsequent dice EXCEPT the last: start with the values on the previous die. SUBTRACT 1 from the value on the first half of the faces, and ADD 1 to the value on the second half. For the last (Nth) die: follow the same pattern, EXCEPT that for the final 2 faces, ADD 2 instead of 1. This will result in the first half of the faces having value "2", and the last 2 faces having value 2N. For example: 4 dice, 8 sides (N=4): A: 1, 2, 3, 4, 5, 6, 7, 8 B: 4, 4, 4, 4, 4, 5, 5, 5 C: 3, 3, 3, 3, 5, 6, 6, 6 D: 2, 2, 2, 2, 6, 7, 8, 8 5 dice, 10 sides (N=5): A: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 B: 5, 5, 5, 5, 5, 5, 6, 6, 6, 6 C: 4, 4, 4, 4, 4, 6, 7, 7, 7, 7 D: 3, 3, 3, 3, 3, 7, 8, 8, 8, 8 E: 2, 2, 2, 2, 2, 8, 9, 9, 10, 10 6 dice, 12 side (N=6): A: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 B: 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7 C: 5, 5, 5, 5, 5, 5, 7, 8, 8, 8, 8, 8 D: 4, 4, 4, 4, 4, 4, 8, 9, 9, 9, 9, 9 E: 3, 3, 3, 3, 3, 3, 9, 10, 10, 10, 10, 10 F: 2, 2, 2, 2, 2, 2, 10, 11, 11, 11, 12, 12 Following this pattern, you will get the following Win-Lose-Tie ratio for ALL N: For A against B, AND for the final Nth die against A: W: 2N^2 - N + 1 L : 2N^2 - N - 1 T: 2N (total outcomes 4N^2) For all other pairs (B against C, C against D, etc): W: 2N^2 L : 2N^2 - N + 1 T: N - 1