# [EM] Multiple Same Choices

Steve Eppley seppley at alumni.caltech.edu
Fri Mar 1 13:50:53 PST 1996

```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.
>>
>>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.)

```