Sets and Plurality losers

Mon Jul 6 16:32:16 PDT 1998

Another circular tiebreaker involves sets.

I again note that both YES/NO and number votes (1, 2, etc.) would be used.
Minority YES choices would lose.  The remaining (lesser of evils) choices on
the ballots of the voters who voted for such losers would move up.   Thus, the
early numbered choices would be a mixture of desired and compromise choices of
most voters.

The sets idea comes about due to having divided majorities that create the
circular ties (a divided majority may be composed of all but one of the
51 A
49 Z

B is close to A.

26 A > B
25 B > A
49 Z

Both the majority and the minority might obviously be more divided.

This leads to--
TC = tied choices in a circular tie. 

TC   sets (combinations of the tied choices)
[2  1 vs. 1]  not in tie
3   2 vs. 1
4   3 vs. 1, 2 vs. 2
5   4 vs. 1, 3 vs. 2
6   5 vs. 1, 4 vs. 2, 3 vs. 3
7   6 vs. 1, 5 vs. 2, 4 vs. 3

Note the even-odd pairings of tied choices.

If there are 2 or more choices in the set, then it means the various
combinations (such as AB and BA).

Example- Question to each voter--
Which combination of 2 is your choice for first and second place ?
N1   AB   (C)
N2   BA   (C)

N3   AC   (B)
N4   CA   (B)

N5   BC   (A)
N6   CB   (A)  

One of sets of votes (N1+N2, N3+N4, N5+N6) will have the highest majority. 

The choice in parenthesis would lose (i.e. the plurality choice in last place)
--- which is what also happens in 1 vs. 1.

With the 4 or more choice ties, the head to math would be looked at again.
The cycle would be repeated, if necessary.

Once again I note that the 2 vs. 1 set is a special limit set (which does not
immediately indicate the *best* tiebreaker).

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