[EM] An ABE solution

Thu Nov 24 19:41:47 PST 2011

Jameson,

"Your range scores are a little bit wrong,.."

I've re-checked them and I don't see how. I gave each candidate 2 points for a top-rating, 1 for a middle-rating
and zero for a bottom rating (or truncation).

So in the initial "sincere" scenario for example C has 9 top-ratings and 1 middle-rating to make a score of  19,
B has 8 top-ratings and 1 middle-rating to make a score of  17, and A has 5 top-ratings and 2 middle-ratings
to make a score of 12.

Chris  Benham

________________________________
From: Jameson Quinn <jameson.quinn at gmail.com>
Sent: Friday, 25 November 2011 5:39 AM
Subject: Re: An ABE solution

Chris: 

Your range scores are a little bit wrong, so you have to add half a B vote for the example to work (or double all factions and add one B vote if you discriminate against fractional people), but yes, this is at heart a valid example where the method fails FBC. 

Note that in my tendentious terminology this is only a "defensive" failure, that is, it starts from a position of a sincere condorcet cycle, which I believe will be rare enough in real elections to be discountable. In particular, this failure does not result in a stable two-party-lesser-evil-strategy self-reinforcing equilibrium. 

Jameson

2011/11/24 Chris Benham <cbenhamau at yahoo.com.au>

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=A or B or B>C or B=C 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
>
>
>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  
>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.
>
>
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