False votes in Condorcet ties ?
DEMOREP1 at aol.com
DEMOREP1 at aol.com
Thu Oct 31 21:07:49 PST 1996
Assume three candidates are in a standard Condorcet Circular tie with the
votes for each candidate such that--
A > B
B > C
C > A
which means
A1 + C1A2 > B1 + C1B2
B1 + A1B2 > C1 + A1C2
C1 + B1C2 > A1 + B1A2
1= first choice vote, 2= second choice vote
By splitting each of the first choice votes into the votes with no second
choice or the second choices for one of the other candidates, then it is
possible to get some idea of the possibly false votes which produced the
circular tie in the first place.
(A1N2+A1B2*+A1C2) + C1A2 > (B1N2+B1A2*+B1C2) + C1B2
(B1N2+B1C2*+B1A2) + A1B2 > (C1N2+C1B2*+C1A2) + A1C2
(C1N2+C1A2*+C1B2) + B1C2 > (A1N2+A1C2*+A1B2) + B1A2
N= no second choice vote
Note that each X1N2 vote (such as C1N2) appears twice and that each X1Y2 vote
(such as A1C2) appears on all three lines. Thus, only the nine (3 x 3) first
choice subvotes are unique.
Are the head to head votes with an asterisk false (such as A1B2*) ? If false,
then does it matter ?
Are such same votes (such as A1B2) on the other two lines true ?
If the fewest votes against is to be the tie breaker (i.e. the lowest total
amount on the left), then I will say again it will be the strategy for the
candidate with the plurality of first choice votes (as shown in the polls) to
especially try to maximize the votes against the other two candidates so that
such plurality candidate would win.
With 4 or more candidates in a tie, the number of first choice subvotes
multiplies in pyramid fashion, especially due to truncated votes (such as
first choice X- no second choice; first choice X- second choice Y- no third
choice, etc.).
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