<br><br><div class="gmail_quote">2011/11/24 Chris Benham <span dir="ltr"><<a href="mailto:cbenhamau@yahoo.com.au">cbenhamau@yahoo.com.au</a>></span><br><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex;">
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<div>Jameson,</div>
<div> </div>
<div>"Your range scores are a little bit wrong,.."<br></div>
<div>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</div>
<div>and zero for a bottom rating (or truncation).</div>
<div> </div>
<div>So in the initial "sincere" scenario for example C has 9 top-ratings and 1 middle-rating to make a score of 19,</div></div></div></blockquote><div><br></div><div>I count 9 and 2, for a total of 20.</div><div>
</div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex;"><div><div style="color:#000;background-color:#fff;font-family:times new roman,new york,times,serif;font-size:12pt">
<div>B has 8 top-ratings and 1 middle-rating to make a score of 17, </div></div></div></blockquote><div><br></div><div>I count 8 and 2, for a total of 18.</div><div> </div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex;">
<div><div style="color:#000;background-color:#fff;font-family:times new roman,new york,times,serif;font-size:12pt"><div>and A has 5 top-ratings and 2 middle-ratings</div>
<div>to make a score of 12.</div></div></div></blockquote><div><br></div><div>I count 5 and 5, for a total of 15.</div><div><br></div><div>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.</div>
<div><br></div><div>Again, this is mostly irrelevant, because I do agree that with this minor modification your scenario proves what you say it does.</div><div><br></div><div>Jameson</div><div><br></div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex;">
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<div><br>Chris Benham<var></var></div>
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</div><b><span style="FONT-WEIGHT:bold">From:</span></b> Jameson Quinn <<a href="mailto:jameson.quinn@gmail.com" target="_blank">jameson.quinn@gmail.com</a>><br><b><span style="FONT-WEIGHT:bold">Sent:</span></b> Friday, 25 November 2011 5:39 AM<div class="im">
<br><b><span style="FONT-WEIGHT:bold">Subject:</span></b> Re: An ABE solution<br></div></font><br>
<div><div><div class="h5">Chris:
<div><br></div>
<div>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. </div>
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</div></div><div><div><div class="h5">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.
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</div></div><div>Jameson<br><br>
<div>2011/11/24 Chris Benham <span dir="ltr"><<a href="mailto:cbenhamau@yahoo.com.au" rel="nofollow" target="_blank">cbenhamau@yahoo.com.au</a>></span><br>
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<div><span>Forest,<br></span><span></span></div>
<div><span>In reference to your new Condorcet method suggestion (pasted at the bottom), which elects an<br>uncovered candidate and if there is none one-at-time disqualifies the Range loser until a remaining</span></div>
<div><span>candidate X covers all the other remaining candidates and then elects X, you wrote:<br></span></div>
<div><span>"Indeed, the three slot case does appear to satisfy the FBC..".<br><br>No. Here is my example, based on that Kevin Venke proof you didn't like.</span></div>
<div><span></span> </div>
<div><span>Say sincere is</span></div>
<div><span></span> </div>
<div><span>3: B>A</span></div>
<div>
<div><span>3: A=C<br>3: B=C</span></div>
<div><span>2: A>C<br>2: B>A</span></div>
<div><span>2: C>B</span></div></div>
<div><span>1: C</span></div>
<div><span></span> </div>
<div><span>Range (0,1,2) scores: C19, B17, A12.<br></span></div>
<div><span>C>B 8-5, B>A 10-5, A>C 7-6.</span></div>
<div><span></span> </div>
<div><span>C wins.</span></div>
<div><span></span> </div>
<div><span>Now we focus on the 3 B>A preferrers. Suppose (believing the method meets the FBC)</span></div>
<div><span>they vote B=A.</span></div>
<div><span> </span></div></div></div><span><div><div class="h5">
<div><span>3: B=A (sincere is B>A)</span></div>
<div>
<div><span>3: A=C<br>3: B=C</span></div>
<div><span>2: A>C<br>2: B>A</span></div>
<div><span>2: C>B</span></div></div>
<div><span>1: C</span></div>
<div><span></span> </div></div></div><span><div><div class="h5">
<div><span>Range (0,1,2) scores: C19, B17, A15.<br><br></span></div>
<div><span>C>B 8-5, B>A 7-5, A>C 7-6.</span></div>
<div><span></span> </div>
<div><span>C still wins.</span></div>
<div><span></span> </div>
<div><span>Now suppose they instead rate their sincere favourite Middle:</span></div>
<div><span></span> </div>
<div><span><span><span>3: A>B (sincere is B>A)</span></span></span></div>
<div>
<div><span>3: A=C<br>3: B=C</span></div>
<div><span>2: A>C<br>2: B>A</span></div>
<div><span>2: C>B</span></div></div>
<div><span>1: C</span></div>
<div><span></span> </div></div></div><span><div><div class="h5">
<div><span>Range (0,1,2) scores: C19, A15, B12.<br><br>A>B 8-7, A>C 7-6, C>B 8-5</span></div>
<div><span></span> </div>
<div><span>Now those 3 voters get a result they prefer, the election of their compromise</span></div>
<div><span>candidate A. Since it is clear they couldn't have got a result for themselves as</span></div>
</div></div><div><span>good or better by voting B>A or B=A or B or B>C or B=C this is a failure</span></div><div><div class="h5">
<div><span>of the FBC.</span></div>
<div><span></span> </div>
<div><span><var></var>Chris Benham</span></div>
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<div><span></span> </div>
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<div style="FONT-FAMILY:times new roman,new york,times,serif;FONT-SIZE:12pt"><b><span style="FONT-WEIGHT:bold">From:</span></b> "<a href="mailto:fsimmons@pcc.edu" rel="nofollow" target="_blank">fsimmons@pcc.edu</a>" <<a href="mailto:fsimmons@pcc.edu" rel="nofollow" target="_blank">fsimmons@pcc.edu</a>><br>
<b><span style="FONT-WEIGHT:bold">Sent:</span></b> Wednesday, 23 November 2011 9:01 AM
<div><br><b><span style="FONT-WEIGHT:bold">Subject:</span></b> Re: An ABE solution<br></div></div></font>
<div>
<div><br>You are right that although the method is defined for any number of slots, I suggested three slots as <br>most practical.<br><br>So my example of two slots was only to disprove the statement the assertion that the method cannot be <br>
FBC compliant, since it is obviously compliant in that case. <br><br>Furthermore something must be wrong with the quoted proof (of the incompatibility of the FBC and the <br>CC) because the winner of the two slot case can be found entirely on the basis of the pairwise matrix. <br>
The other escape hatch is to say that two slots are not enough to satisfy anything but the voted ballots <br>version of the Condorcet Criterion. But this applies equally well to the three slot case.<br><br>Either way the cited "therorem" is not good enough to rule out compliance with the FBC by this new <br>
method.<br><br>Indeed, the three slot case does appear to satisfy the FBC as well. It is
an open question. I did not <br>assert that it does. But I did say that "IF" it is strategically equivalent to Approval (as Range is, for <br>example) then for "practical purposes" it satisfies the FBC. Perhaps not the letter of the law, but the <br>
spirit of the law. Indeed, in a non-stratetgical environment nobody worries about the FBC, i.e. only <br>strategic voters will betray their favorite. If optimal strategy is approval strategy, and approval strategy <br>requires you to top rate your favorite, then why would you do otherwise?<br>
<br>Forest<br><br>----- Original Message -----<br>From: Chris Benham <br><br>Forest,<br> <br>"When the range ballots have only two slots, the method is simply Approval, which does satisfy the <br>FBC."<br> <br>
When you introduced the method you suggested that 3-slot ballots be used "for simplicity".<br> I thought you might be open to say 4-6 slots, but a
complicated algorithm on 2-slot ballots<br> that is equivalent to Approval ??<br> <br>"Now consider the case of range ballots with three slots: and suppose that optimal strategy requires the </div></div></div>
<div>
<div>
<div style="FONT-FAMILY:times new roman,new york,times,serif;FONT-SIZE:12pt">voters to avoid the middle slot. Then the method reduces to Approval, which does satisfy the FBC."<br> <br>The FBC doesn't stipulate that all the voters use "optimal strategy", so that isn't relavent.<br>
<br><a href="http://wiki.electorama.com/wiki/FBC" rel="nofollow" target="_blank">http://wiki.electorama.com/wiki/FBC</a><br> <br><a href="http://nodesiege.tripod.com/elections/#critfbc" rel="nofollow" target="_blank">http://nodesiege.tripod.com/elections/#critfbc</a><br>
<br>Chris Benham<br></div></div></div>
<div style="FONT-FAMILY:times new roman,new york,times,serif;FONT-SIZE:12pt">Forest Simmons wrote (17 Nov 2011):</div>
<div>
<div style="FONT-FAMILY:times new roman,new york,times,serif;FONT-SIZE:12pt"> </div>
<div style="FONT-FAMILY:times new roman,new york,times,serif;FONT-SIZE:12pt">Here’s my current favorite deterministic proposal: Ballots are Range Style, say three slot for simplicity.<br><br>When the ballots are collected, the pairwise win/loss/tie relations are<br>
determined among the candidates.<br><br>The covering relations are also determined. Candidate X covers candidate Y if X<br>beats Y as well as every candidate that Y beats. In other words row X of the<br>win/loss/tie matrix dominates row Y.<br>
<br>Then starting with the candidates with the lowest Range scores, they are<br>disqualified one by one until one of the remaining candidates X covers any other<br>candidates that might remain. Elect X.<br><br></div></div>
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