Lowest Number of First Choice votes tie breaker, Ver.2

DEMOREP1 at aol.com DEMOREP1 at aol.com
Thu Oct 17 17:06:45 PDT 1996


The original had an obvious error. Expanded Version 2----

Assume 3 Condorcet candidates are in a circular tie.

Example (left side is greater than right side on each line,1=first choice
votes, 2 = second choice votes)----

(A1 + C1A2) > (B1 + C1B2) (A beats B)
(B1 + A1B2) > (C1 + A1C2) (B beats C)
(C1 + B1C2) > (A1 + B1A2) (C beats A)

If the fewest-votes-against-in-worst-defeat (FVA) method is used as a tie
breaker, then some of the voters by voting sincerely may defeat their first
choice.

Example- 
40 + 15 (= 55) > 35 + 10  (= 45) (A beats B)
35 + 19 (= 54) > 25 + 21  (= 46) (B beats C) (C wins using FVA)
25 + 31 (= 56) > 40 + 4    (= 44) (C beats A)

The 21 A1C2 voters on the right side by voting sincerely for C have caused C
to win.

However, if 2 or more of such 21 A1C2 voters vote insincerely for B, then B
will win using FVA (B will beat C by at least 56 to 44, B wins by having only
55 votes against B in the A beats B pairing). 

40 + 15 (= 55) > 35 + 10  (= 45) (A beats B) (B wins using FVA)
35 + 21 (= 56) > 25 + 19  (= 44) (B beats C)
25 + 31 (= 56) > 40 + 4    (= 44) (C beats A)

Should the tie breaker results be determined whether or not there are sincere
or insincere voters ?

If FVA is the tie breaker, then all candidates will play games and try to get
their first choice voters to vote insincerely to maximize the number of votes
against in all pairings (i.e. to cause the loser to get only the votes from
his/her first choice supporters- which means that the candidate with the most
first choice votes will win- who would be the ordinary plurality winner). In
the above example-
40 + 25 (= 65) > 35 (A beats B)
35 + 40 (= 75) > 25 (B beats C)
25 + 35 (= 60) > 40 (C beats A) (A wins using FVA)

To avoid such game playing, I suggest the candidate with the lowest number of
first choice (LNFC) votes should lose. Thus in the example, C loses and the
C1 voters can vote sincerely for A or B (A wins).

With 4 or more candidates in a tie and after each loser is eliminated the
head to head pairings among the remaining tied candidates would be looked at
again to see if there is a Condorcet winner. 

Eliminating just one candidate may produce a Condorcet winner among the
remaining candidates.

Note- LNFC is not direct instant run off but only a tie breaker after doing
the standard Condorcet pairings. LNFC is also easy to explain to the public
if there are ties.

I maintain that since rankings give only an indication of "relative" support,
that there should first be an absolute yes/no vote on each candidate with
only those candidates who have an absolute yes majority go to the Condorcet
head to head test. 

If only 2 candidates can get majority support, then obviously a head to head
test will determine the winner.

Only if there are 3 or more majority support candidates is there even a
possibility of a tie (the above examples presume that A, B and C each have a
majority of all voters support).




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