[EM] Interesting use of Borda count

[Forest Simmons]

Bart,

this discussion reminds me of the time Tom Ruen was toying with the idea
of modifying Approval by requiring all of the approved candidates on one
ballot to share one vote equally, i.e. if you approve three candidates
they each get one third of your vote, a kind of constrained cumulative
voting where all of the non-zero votes on your ballot have equal value.

The same averaging argument you use below shows that both Tom's idea and
ordinary cumulative voting are strategically equivalent to ordinary
plurality. No wonder we couldn't find a violation of monotonicity in his
examples!

Also you have used the same argument to show that Approval is
strategically equivalent to Cardinal Ratings when the latter is scored
according to the sum or average of rates.

It seems to be a useful device.

Forest


On Fri, 4 Jan 2002, Bart Ingles wrote:

> 
> I don't recall using the term "average ranking".  My focus was on
> average (or total) point counts (i.e. Borda scores), as a way of showing
> the practical and strategic equivalence among the Borda variations
> mentioned.
> 
> 
> 
> Steve Barney wrote:
> > 
> > Bart:
> > 
> > OK, I get it now. When I see the term "average ranking" I think of something
> > other than what you describe. I think you get a more intuitive, and perhaps
> > more descriptive sense of "average ranking" if you do as follows. You average
> > the RANKINGS for each candidate by dividing the sum of the rankings of each
> > candidate by the number of voters. For example, if there are three candidates
> > and two voters, one voting for A>B>C and the other voting for A>C>B, it goes
> > like this:
> > 
> >            1 1
> >            A A 1st
> >            B C 2nd
> >            C B 3rd
> > 
> >            A: (1+1)/2=1
> >            B: (2+3)/2=2.5
> >            C: (3+2)/2=2.5
> > 
> > That makes more sense to me, on an intuitive level, than averaging the total
> > point scores. Don't you agree?
> > 
> > Steve Barney
> > 
> > PS: Thanks for the reference. That will help my education along.
> > 
> > --- In election-methods-list at y..., Bart Ingles <bartman at n...> wrote:
> > >
> > > Steve:
> > >
> > > I agree with your Saari results, if the two voters are ignorant enough
> > > to actually bullet vote (even though this may accurately represent their
> > > preferences).
> > >
> > > One of the ways to defeat Saari's variation is for the two voters to
> > > collaborate:  One voter agrees to rank A, B, and C in order, while the
> > > other ranks A, C, B.  So the individual ballots are worth (2, 1, 0) and
> > > (2, 0, 1).  The combined total is (4, 1, 1), hence the per-ballot
> > > average (2, .5, .5) I claimed below.
> > >
> > > Thus the suggestion that Saari's variation could function as a sort of
> > > voter intelligence screen, since a potential bullet voter who doesn't
> > > understand the above strategy has his voting power reduced by one-third.
> > >
> > > Samuel Merrill III (Making Multicandidate Elections More Democratic,
> > > 1988) includes the following citation:
> > >   Black, D. (1958) *The Theory of Committees and Elections*, Cambridge
> > > University Press.
> > >
> > > I haven't read Black's work.  But the issue seems moot to me, since with
> > > the voter strategy above (or the equivalent coin-toss strategy), the
> > > three variations we have discussed (Borda, Black, and Saari) are all
> > > equivalent.
> > >
> > > Bart
> > 
> > =====
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