Multiple Same Choices

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Multiple Same Choices

Craig Carey-2
Lucien Saumur wrote:
>Steve Eppley writes:
>>Lucien wrote:
>          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
voters see.

>>>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
situation.  

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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Multiple Same Choices

Craig Carey-2
When a ballot ranks 2 or more candidates at the same rank position,
the voter isn't expressing any preference between them. If you rank
A & B equally, then you aren't one of the voters ranking A over B,
& you aren't one of the voters ranking B over A. So it wouldn't
be right to count you as saying A is 1/2 better than B, & B is 1/2
better than A. That isn't what you're saying.

Purely on principle, then, the voter should be able to decline to
express a preference between A & B, and should be able to not be
counted as expressing preference between them.

***

Aside from that, there's the results to consider. As Steve said,
it matters how circular ties are solved. There are circular tie
solutions that don't care whether we count that shared rank position
as no preference, or 2 equal & opposite preferences. Copeland's method
is one such. Another is Lucien's suggestion to solve circular ties
randomly.

But Condorcet's method _does_ care which way we count shared rankings.
Condorcet counts votes-against. I've talked about why that's desirable,
in order to achieve what we want from a single-winner method. Do
you wanted to be counted against A & B both, because you didn't
vote for either over the other?

***

To summarize:

1. If you don't express a prefrence between A & B that means that
you should be counted as voting neither over the other. Counting
you as voting each over the other is both self-contradictory &
contrary to your expressed wishes.

2. Condorcet's method requires that voters not be counted as voting
a preference that they didn't vote. Its valuable properties depend
on that.

***

Mike


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