Votes Against Tiebreaker

New Democracy donald at
Tue Oct 20 01:57:11 PDT 1998


     On Sat, 17 Oct 1998, MikeO wrote:  Here's the votes-against:

A>B 40
B>C 61
C>A 51

B wins by any of the VA methods, such as plain Condorcet(EM),
Smith//Condorcet(EM), Schulze, or SD.

Dear MikeO,

     B  also wins by Plurality. This tiebreaker solution is the same as

     You have three candidates, each with a majority win in one pairing,
but none of the candidates have won a majority of the pairs.
     When you pick the candidate with the "largest absolute Majority" that
is the same as picking the leading candidate in plurality.

     Does candidate B have a majority in his win?
     If so, - where are the numbers that prove the majority?

     Majority is defined as a number greater than one half of the total.
     61 is not greater than one half of 152 (40+61+51).
     If you say that 61 is a majority of 100, the number of votes, I will
say 51 is also a majority of 100. And, we cannot have two majorities - the
defination of majority rules out more than one majority.

     You talk about majorities as if you invented the word, but the bottom
line is that you are willing to accept less in order to force some result
out of a tie situation. You are not willing to accept the fact that it is
impossible to have a majority winner after Condorcet goes into a circular
     DEMOREP falls back on Approval Voting. You fall back on Plurality.
     Neither will give us a majority winner.

Donald Davison

     ///                 N E W    D E M O C R A C Y                ///
     \\\ Home of Citizen's Democracy \\\

More information about the Election-Methods mailing list