[EM] Interesting use of Borda count

[Steve Barney]

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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