The rich party problem (was Re: Briefly why I prefer Reg-Champ t

[Steve Eppley]

Bruce A wrote:
-snip-
>Personally, I would feel quite comfortable defending any
>Regular-Champion (or Consensus-Champion, Complete-Champion, or
>Qualified-Champion) winner before a group of intelligent but
>hostile critics.
-snip-

I hope the following is sufficiently "hostile" to merit a reply.  :-)

I looked in Bruce's "Brief Descriptions of Voting Methods" for his
definition of Regular-Champion, but he must have defined it
somewhere else.  I'll assume in the following analysis that if
there's no beats-all-others candidate, it picks the candidate(s)
which maximizes (number of pairwins minus number of pairlosses).

Suppose there are three political parties Red, Green, and Blue.  
The Reds and Blues are rich, so they can afford to field more
candidates than the Greens: the Reds and Blues each field two
candidates and the Greens field one.

Suppose also there's a circular tie, with the voters preferring
Blue to Red, Green to Blue, and Red to Green:
   G > B1&B2 > R1&R2 > G

Not yet counting the intraparty pairings, the Blue candidates have a
clear advantage because each Blue loses only one interparty pairing
(to the lone Green candidate) but each Red and Green candidate loses
two interparty pairings.  Expect a Blue to win, since this is an
advantage in voting methods which count the *number* of pair losses
and ignore the *size* of the losses.  But this win has nothing to do
with the popularity of Blueness and is strictly a matter of money. 
For me, this disqualifies such methods. 

Here's an example:
  48:    G   > B1=B2 > R1=R2
  24:  B1=B2 > R1=R2 >   G
  28:  R1=R2 >   G   > B1=B2
 ---
 100

The ten pairings:
  B1 loses to G,  24 to 76.
  B2 loses to G,  24 to 76.
  R1 loses to B1, 28 to 72.
  R1 loses to B2, 28 to 72.
  R2 loses to B1, 28 to 72.
  R2 loses to B2, 28 to 72.
  G  loses to R1, 48 to 52.
  G  loses to R2, 48 to 52.
  B1 ties B2 unanimously.
  R1 ties R2 unanimously.

Candidate    Wins - Losses
---------    -------------
B1              2 - 1 = 1
B2              2 - 1 = 1
R1              2 - 2 = 0
R2              2 - 2 = 0
G               2 - 2 = 0

So a Blue wins, by maximizing (wins-losses).  But this measure looks
irrelevant to me here, being related more to the party financial
assets and not to the relative popularity of the candidates.  Perhaps 
I could state this better as an "Independence from Twins" criterion, 
if I'm clever enough.

It appears that the Green candidate would be a better winner here,
since there's a far smaller majority against G than against the
others. 

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