Cycle Combinations tie breaker

[DEMOREP1 at aol.com]

Assume that there is a Condorcet cycle with 4 candidates such as A>B>C>D>A.

There are 24 combinations of the candidates (1 of the 4 for first, 1 of the
remaining 3 for second, 1 of the remaining 2 for third and the last remaining
candidate-- or 4 x 3 x 2 x 1 or 4 factorial (4 ! ) ).

Should the tie breaker be the one combination (if any) that the most voters
agree with ?

The votes in each X versus Y pairing in each of the 24 combinations would
have to be added together.

Example-
Assume the combination D>B>A>C has the highest sum (that is, the votes for D
in the D-B, D-A and D-C pairings plus the votes for B in the B-A and B-C
pairings plus the votes for A in the A-C pairing). 
Note 1- no voter may have actually voted for such final one combination.
Note 2- all of the 12 head to head pairings from the head to head (Condorcet)
matrix are used.
Note 3- If a majority coalition exists (such as D and B), then such coalition
should defeat the minority candidates (A or C).
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With the standard 3 candidate cycle (A>B>C>A), there are 3 x 2 x 1 or 6
combinations. 
Do any folks who have done 3 candidate examples want to do the 6 combinations
math in their examples ???