[EM] Multiple Same Choices
seppley at alumni.caltech.edu
Fri Mar 1 13:50:53 PST 1996
Lucien Saumur wrote:
>Steve Eppley writes:
> Or NOTR for None of the Rest.
Or NOTB for None of these Bums...
I don't care which term is used.
>>>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?
> I mean that ballots would be produced indicating:
> NO CANDIDATE IS RANKED
> NO CANDIDATE IS ACCEPTABLE
That's what is meant by simply voting: 1=NOTB
> 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.
Like I said, you can produce the same result by treating NOTB as just
another candidate in the matrix, simplifying the algorithm. You
don't need a separate data structure for approval counts. This is
just an implementation detail, though; what matters is what the
>>>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.
I meant a "Condorcet circular tie", in which no candidate beat all
the rest in the pairwise matchups. We're not yet in a coin-flipping
The tie-breaking should probably look at how badly each candidate is
beaten overall. If a voter ranks two choices equal, does that mean
each is unbeat or half-beat? Using .5 means you think each is
half-beat. (There was a message several days ago, from Rob or Mike
I think, on this question. I'll have to go back and reread it.)
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