[EM] An ABE solution

[Chris Benham]

Forest,

In reference to your new Condorcet method suggestion (pasted at the bottom), which elects an
uncovered candidate and if there is none one-at-time disqualifies the Range loser until a remaining
candidate X covers all the other remaining candidates and then elects X, you wrote:

"Indeed, the three slot case does appear to satisfy the FBC..".

No. Here is my example, based on that Kevin Venke proof you didn't like.
 
Say sincere is
 
3: B>A
3: A=C
3: B=C
2: A>C
2: B>A
2: C>B
1: C
 
Range (0,1,2) scores: C19,   B17,   A12.

C>B 8-5,   B>A 10-5,   A>C  7-6.
 
C wins.
 
Now we focus on the 3 B>A preferrers. Suppose (believing the method meets the FBC)
they vote B=A.
 3: B=A  (sincere is B>A)
3: A=C
3: B=C
2: A>C
2: B>A
2: C>B
1: C
 Range (0,1,2) scores: C19,   B17,   A15.


C>B 8-5,   B>A 7-5,   A>C  7-6.
 
C still wins.
 
Now suppose they instead rate their sincere favourite Middle:
 
3: A>B  (sincere is B>A)
 
3: A=C
3: B=C
2: A>C
2: B>A
2: C>B
1: C
 Range (0,1,2) scores: C19,   A15,   B12.

A>B  8-7,   A>C  7-6,    C>B  8-5
 
Now those 3 voters get a result they prefer, the election of their compromise
candidate A. Since it is clear they couldn't have got a result for themselves as
good or better by voting B>A or  B>C or B this is a failure of the FBC.
 
 
Chris Benham 


________________________________

From: "fsimmons at pcc.edu" <fsimmons at pcc.edu>
Sent: Wednesday, 23 November 2011 9:01 AM
Subject: Re: An ABE solution

voters to avoid the middle slot.  Then the method reduces to Approval, which does satisfy the FBC."
  
The FBC doesn't stipulate that all the voters use "optimal  strategy", so that isn't relavent.
 
http://wiki.electorama.com/wiki/FBC
  
http://nodesiege.tripod.com/elections/#critfbc
 
Chris  Benham

Forest Simmons wrote (17 Nov 2011):

Here’s my current favorite deterministic proposal: Ballots are Range Style, say three slot for simplicity.

When the ballots are collected, the pairwise win/loss/tie relations are
determined among the candidates.

The covering relations are also determined.  Candidate X covers candidate Y if X
beats Y as well as every candidate that Y beats.  In other words row X of the
win/loss/tie matrix dominates row Y.

Then starting with the candidates with the lowest Range scores, they are
disqualified one by one until one of the remaining candidates X covers any other
candidates that might remain.  Elect X.


You are right that although the method is defined for any number of slots, I suggested three slots as 
most practical.

So my example of two slots was only to disprove the statement the assertion that the method cannot be 
FBC compliant, since it is obviously compliant in that case.  

Furthermore something must be wrong with the quoted proof (of the incompatibility of the FBC and the 
CC) because the winner of the two slot case can be found entirely on the basis of the pairwise matrix.  
The other escape hatch is to say that two slots are not enough to satisfy anything but the voted ballots 
version of the Condorcet Criterion.  But this applies equally well to the three slot case.

Either way the cited "therorem" is not good enough to rule out compliance with the FBC by this new 
method.

Indeed, the three slot case does appear to satisfy the FBC as well.  It is an open question.  I did not 
assert that it does.  But I did say that "IF" it is strategically equivalent to Approval (as Range is, for 
example) then for "practical purposes" it satisfies the FBC.  Perhaps not the letter of the law, but the 
spirit of the law.  Indeed, in a non-stratetgical environment nobody worries about the FBC, i.e. only 
strategic voters will betray their favorite. If optimal strategy is approval strategy, and approval strategy 
requires you to top rate your favorite, then why would you do otherwise?

Forest

----- Original Message -----
From: Chris Benham 

Forest,
 
"When the range ballots have only two slots, the method is  simply Approval, which does satisfy the 
FBC."
  
When you introduced the method you suggested that 3-slot ballots be used "for simplicity".
 I thought you might be open to say 4-6 slots, but a complicated algorithm on 2-slot ballots
 that is equivalent to Approval ??
  
"Now consider the case of range ballots with three slots: and  suppose that optimal strategy requires the  
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