[EM] Voting by selecting a published ordering

[David Cary]

--- "Simmons, Forest" <simmonfo at up.edu> wrote:

> David Cary asked ...
> 
> Does geometric consistency mean here that the candidates and voters
> can be represented by points in some space so that a voter prefers
> candidate A over Candidate B iff the voter is "closer" to A than to
> B?
> 
> Basically, yes, any metric on the candidates would do, though the
> triangle inequality is not needed.
>  

So using the standard metric in the Cartesian plane, with three
candidates positioned:
  A at (0,0)
  B at (4,4)
  C at (8,0)
(A and C are furthest apart)

A voter at the following positions would have the preferences:
  V1 at (1,1)  prefers A>B>C
  V2 at (3,-3) prefers A>C>B
  V3 at (3,3)  prefers B>A>C
  V4 at (5,3)  prefers B>C>A
  V5 at (5,-3) prefers C>A>B
  V6 at (7,1)  prefers C>B>A
  V7 at (2,-2) prefers A>(B=C)
  V8 at (4,3)  prefers B>(A=C)
  V9 at (6,-2) prefers C>(A=B)

With such a full range of preferences possible, it would seem that
just about any ordinally ranked ballot is possible, and with an
appropriate  distribution of voters, just about any election result
with 3 candidates is geometrically consistent.  That includes a
result with A beating B beating C beating A in pair-wise elections
that does not produce a Condorcet winner.  For example 2 voters at
V1, 3 voters at V4, and 4 voters at V5.

This example seems to contradict what I understood to be Forest's
earlier claim that a geometrically consistent election with 3
candidates always produced a Condorcet winner.  Am I missing something?

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