[EM] Proof Borda Count best in the case of fully ranked preference ballots

[Steve Barney]

Forest:

No, only the Borda Count completely cancels such symmetric profiles and
produces a complete tie. For example, the plurality outcome for the reversal
profile:

	A>B>C
	C>B>A

is A~C>B. For the circular profile (what Saari calls a "Condorcet term":

	A>B>C
	B>C>A
	C>A>B

the pairwise vote outcome (Condorcet's method) is a cycle, which is not the
same as a tie. As Rob pointed out, the addition of such a circular profile can
change the outcome; that is, pairwise voting does not completely cancel such a
term and produce a complete tie. That's the difference between a cycle and a
tie.

SB

--- In election-methods-list at y..., Forest Simmons <fsimmons at p...> wrote:
> On Tue, 19 Feb 2002, Steve Barney wrote:
> 
> > Forest:
> > 
> > What do you think of circular triplets, such as:
> > 
> > 	A>B>C
> > 	B>C>A
> > 	C>A>B,
> > 
> > and reversals, such as:
> > 
> > 	A>B>C
> > 	C>B>A.
> > 
> > If that is all the information that we have to go on (when ordinal
preference
> > ballots are used, it is ), shouldn't either of these profiles cancel out
> > completely and yield a tie? The Borda Count is the only method which always
> > does that, according to Saari's analysis. From that simple fact, argues
Saari,
> > come voting paradoxes such as non-monotonicity, etc.
> 
> The two examples that you give yield ties in every serious method of which
> I am aware.
> 
> Forest


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