# 3 Choices again, 4 July 1998

DEMOREP1 at aol.com DEMOREP1 at aol.com
Sat Jul 4 14:02:43 PDT 1998

```Back to basics (again)
The 3 choice case--
N1 ABC
N2 ACB
N3 BAC
N4 BCA
N5 CAB
N6 CBA

(N1+N2)+N5    A/B  (N3+N4)+N6
(N3+N4)+N1    B/C  (N5+N6)+N2
(N5+N6)+N4    C/A  (N1+N2)+N3
(first choices)

Assuming A>B>C>A and no number ties (as is most likely in a mass public
election), then there are limited possibilities.

One of the majority sums in the left column is highest or lowest and one of
the first choice sums is lowest. Which sum should be the tiebreaker (either
directly or as an eliminator of a choice) ?

Condorcet[EM] claims that the lowest majority sum in the left column should be
the tiebreaker (which means that the highest minority sum in the right column
is the winner).  Does this comply with *common sense* ?

Bucklin has the highest majority sum of the 1st and 2nd choices being the
winner.
A  (N1+N2)+N3+N5
B  (N3+N4)+N1+N6
C  (N5+N6)+N2+N4

Reverse Bucklin has the highest reverse sum (of the 2nd and 3rd choices) being