[EM] Paul: All preferences counted equally

[Dave Ketchum]

Tempting to charge all with vagueness.

In a 5-way race there are 10 pairs of candidates, for which standard
Condorcet scoring counts:
       A>B
       B>A (separate count)
       A=B gets no count - general agreement among most EM members

Meaning 20 counters in  the 5x5 array for 5 candidates.

Agreed that voters could consider 4th and 5th almost even but:
       COULD consider 1st and 2nd almost even as acceptables.
       COULD consider 5th much worse than 4th for 5th being seriously
objectable for some reason.

On Mon, 24 Oct 2005 08:02:34 +0200 Jobst Heitzig wrote:

 > Dear Mike!
 >
 > What Paul meant by "10 counters" is the pairwise matrix, which in case
 > of 5 candidates and complete ballots without equal ranks contains only
 > 10 significant numbers (one for each of the pairwise races). His point
 > seems to be that basing the method only on those numbers ignores
 > valuable information.
 >
 > Yours, Jobst
 >
 >
 >
 > MIKE OSSIPOFF wrote:
 >
 >
 >>Paul--
 >>
 >>You said:
 >>
 >>My "discomfort" (can't quite call it a criticism) with any method that
 >>counts votes using the pairwise matrix is that my "A (1st) > B (5th)" vote
 >>in a 4-way race is negated by some other voter's "B (fourth) > A (fifth)"
 >>vote. Neither of us particularly want B, but by the time the other voter is
 >>ranking fourth and fifth she's in the "who cares?" part of her ballot.

The voters could have ranked:
       A>C>D>E>B
       C>D>E>B>A
 >>
 >>I reply:
 >>
 >>Counting all voted pairwise preferences equally does a lot to get rid of
 >>strategy problems. But James GA has proposed a method that weights
 >>pairwise preference votes with ratings differences, and it works fine,
 >>for an FBC-failing method.  But it adds too much complication for an
 >>initial proposal.
 >>
 >>Anyway, you can't have everything. If you want minimum strategy need,
 >>and simplicity, then you've got to count all voted pairwise preferences
 >>equally.
 >>
 >>You continued:
 >>
 >>In a 5-way race there are 120 unique sets of preferences (151 if equal
 >>rankings are allowed) and any method thhat only uses 10 counters to
 >>determine a winner is going to mess things up somehow.

Mess how?  With ranking we only know A>B or A=B or A<B, but have no 
measure of how much difference.
 >>
 >>I reply:
 >>
 >>What is a counter? Do you refer to the hiring of only 10 people to
 >>handcount the ballots? Nowadays a computer program would probably be
 >>used. But of course a computer shouldn't be used unless there are paper
 >>ballots too, and unless the sourcecode is available to everyone, and
 >>checkable by anyone. I don't know the details of how computer-counting
 >>could be made really secure, but most feel that it can.
 >>
 >>Lacking secure computer counts, we could hire lots of people for the
 >>pairwise counting. It needn't take longer than IRV. In fact it could be
 >>quicker, with one team for each pairwise comparison. Quicker but more
 >>expensive.
 >>
 >>But why would it mess things up if only 10 ballot-counters were hired? A
 >>national presidential count with lots of candidates would take a while
 >>then, of course.
 >>
 >>Mike Ossipoff
-- 
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   Dave Ketchum   108 Halstead Ave, Owego, NY  13827-1708   607-687-5026
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