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 <[email protected]>
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 <[email protected]>
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: "[email protected]" <[email protected]>
>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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