New pairwise tie-breaker (was Re: Various Condorcet Tie

[Steve Eppley]

Mike O wrote:
>Even if it's a fairly good size margin for Clinton > Dole, maybe
>Dole voters' truncation makes Clinton beaten by a huger margin, and
>Dole voters outnumber Nader voters well enough that (majority-
>rejected) Dole's margin of defeat, though not small, is still 
>the smallest. 

Mike may be right, and maybe this new method flops on majority rule
and other important criteria.  But I always like to see an example
instead of depending on word outlines.  Mike, have you got one which
illustrates your point?

By the way, I've looked at Bruce's "brief descriptions" article and
found that it's Dodgson, not Young, which this new method most
closely resembles:

  > Dodgson: selects the candidate(s) that minimize the number of
  > (contiguous) pair reversals in voters rankings needed to make 
  > that candidate not lose any pairwise matchup.

If the Dole voters truncate so they (un)rank Clinton=Nader, then this
neo-Dodgson method will modify their Clinton=Nader rankings before it
attempts to modify anyone's '>' or '<' rankings.  This means that it
will make a beats-all winner out of Clinton or Nader, and not Dole.
It won't matter that Dole's margin of defeat to Clinton is smaller
than Clinton's margin of defeat to Nader--it's still the Clinton vs
Nader pairing that will be modified first, to benefit Clinton or
Nader.  Isn't this a significant difference from what ordinary
Dodgson would do?  It seems to protect against tactical truncation
better than Dodgson, because it's not simply a matter of who has the
smallest margin (or combined margins, if the circular tie happens to
include more than three candidates) of defeat.

Because of this, I think Mike's example (if he has time to provide 
one) will probably not use truncation as the tactic.  Maybe 
order-reversal?  Or maybe no tactic at all, but just a violation of 
some important criterion when everyone votes sincerely.

Ok, here's an example showing how neo-Dodgson can elect Dole if the
Dole voters order-reverse:

  Sincere preferences:
    46 D>C>N
    10 C>N>D
    10 C>D>N
    34 N>C>D
  The pairings:
    N<D 44 to 56
    N<C 34 to 66
    D<C 46 to 54
  Clinton is the sincere beats-all winner.

  The 46 order-reverse:
    46 D>N>C
    10 C>N>D
    10 C>D>N
    34 N>C>D
  The pairings:
  N<D 44 to 56 (margin=12)
  C<N 20 to 80 (margin=60)
  D<C 46 to 54 (margin= 8)

Since there weren't any equal rankings, neo-Dodgson behaves
identically to ordinary Dodgson here and rewards the order-reversers.
The 10 C>D>N voters, if they knew about the reversals, could vote
C>N>D instead to elect N, and the 34 N voters could vote C>N instead
to elect C.  This is a strategic mess compared to Condorcet.

I think neo-Dodgson (and Dodgson) may also have the rich party problem.
Since it's the *combined* margins of defeats which must be overcome,
if there are many candidates in the circular tie and some of these
are near-clones, these near-clones would "gang up" against the
candidates they each defeat pairwise, multiplying the combined
margin of defeat that needs to be surmounted.  I'm not sure though:
the method isn't making tweaks to pairvotes, it's making tweaks to 
ballots.  One tweak of a ballot ranking can affect many pairings 
because of the transitive ripple-throughs:
  Nine voters:
  C1 > C2 > C3 > A > B1 > B2
  C2 > C3 > C1 > A > B2 > B1
  C3 > C1 > C2 > A > B1 > B2
  B2 > B1 > C1 > C2 > C3 > A
  B1 > B2 > C2 > C3 > C1 > A
  B2 > B1 > C3 > C1 > C2 > A
  A > B1 > B2 > C1 > C2 > C3
  A > B2 > B1 > C2 > C3 > C1
  A > B1 > B2 > C3 > C1 > C2
  Fifteen pairings:
  B1<A   3 to 6
  B2<A   3 to 6
  C1<B1  3 to 6
  C2<B1  3 to 6
  C3<B1  3 to 6
  C1<B2  3 to 6
  C2<B2  3 to 6
  C3<B2  3 to 6
  A<C1   3 to 6
  A<C2   3 to 6
  A<C3   3 to 6
  B2<B1  4 to 5
  (The three Cx vs Cy pairings are ties.)
  Dodgson and neo-Dodgson will elect B1 here, I think, tweaking only 
  three ballots.  B1 wins simply because there aren't many A's running,
  another instance of the "candidate-counting" fallacy shared by 
  Copeland and Regular-Champion.

* *

I didn't do a good job of describing what neo-Dodgson would do if
merely altering '=' rankings isn't enough to make some candidate a
beats-all.  I'm not sure I have the words at the moment to make it
clear, but the principle of treating all the changes of '=' to '>' or
'<' as smaller than changes of '>' or '<' to '=' would persist.  

Even changing a billion '=' to '>' would be treated as smaller than
changing one '>' to '='.  Here's why: if you remember your basic
calculus, then for any small delta one can find an epsilon much
smaller than delta.  Even an epsilon a billion times smaller than
delta.  So if one sets delta equal to the smallest nonzero difference
between any two candidates' ratings by any voter, and epsilon less
than (delta / (nVoters * nCandidates)), then modifying *all* the
voters' equal ratings by epsilon would still be a smaller change than
modifying one voter's '>' in one pairing. 

---Steve     (Steve Eppley    seppley at alumni.caltech.edu)