[EM] Multiple Same Choices

Lucien Saumur aa447 at freenet.carleton.ca
Fri Mar 1 08:48:05 PST 1996

In an article, seppley at alumni.caltech.edu ("Steve Eppley") writes:

>Lucien Saumur wrote:
>This is what I called Condorcet+NOTB.  The "dividing line" is 
>inserted by ranking None Of The Below in the appropriate position.

          Or NOTR for None of the Rest.

>>A ballot may indicate that no candidate is acceptable.
>The voter ranks NOTB highest.
>>It may also indicate that no candidate is ranked.
>I'm unclear on this.  Do you mean some are equally approved and the 
>rest are equally disapproved?
>  Some approved:  1=A,B,C   2=NOTB   3=D,E,F,G
>  All approved:   1=A,B,C,D,E,F,G
>  None approved:  1=NOTB

          I mean that ballots would be produced indicating:

                    NO CANDIDATE IS RANKED
                    NO CANDIDATE IS ACCEPTABLE

          Such ballots would be tallied by adding .5 vote
to every count of the matrix and by adding zero to the
acceptability count of every candidate.

>> My system is also designed to tally the ballots.
>>When candidates are not ranked (indicating equal
>>preference) the related counts for both candidates are
>>updated with half (1/2) a vote.
>I haven't thought about this, but my guess is that whether you should
>score equally ranked choices as .5 and .5, or 1 and 1, or 0 and 0,
>depends on the choice of tie-breaking algorithm you plan on using. 
>It may not matter--is it possible to specify a sensible tie-breaking
>method that doesn't care how the equal pairs are counted?

          A tie means a tie and it may be broken most
fairly by flipping a coin. It cannot be broken by looking
at some arbitrary aspect the election data without
introducing a reason for strategic voting in the voting

          aa447 at FreeNet.Carleton.CA

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