[EM] An ABE solution

[Jameson Quinn]

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

>
> 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,
>

I count 9 and 2, for a total of 20.


> B has 8 top-ratings and 1 middle-rating to make a score of  17,
>

I count 8 and 2, for a total of 18.


> and A has 5 top-ratings and 2 middle-ratings
> to make a score of 12.
>

I count 5 and 5, for a total of 15.

My counts differed from yours in the other cases; in particular, in the
second "failed strategy" case, I found a tie between A and B, which is why
I said you need an extra half-voter plumping B for the scenario to work as
you claim.

Again, this is mostly irrelevant, because I do agree that with this minor
modification your scenario proves what you say it does.

Jameson


> 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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