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