<html><body><div style="color:#000; background-color:#fff; font-family:times new roman, new york, times, serif;font-size:12pt"><div style="RIGHT: auto"><SPAN style="RIGHT: auto">Forest,<BR><BR style="RIGHT: auto">"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."</SPAN></div>
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<div style="RIGHT: auto"><SPAN style="RIGHT: auto">The likely explanation for some odd remarks by you and Jameson has just occurred to me. Could it be<BR>that you and Jameson have mistaken a set of ballots for a "pairwise matrix" ??</SPAN></div>
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<div style="RIGHT: auto"><SPAN style="RIGHT: auto">Here is that quoted proof again, with the ballots represented in the more familiar EM notation:</SPAN></div>
<div style="RIGHT: auto"><SPAN style="RIGHT: auto"> <BR>Hello,<BR><BR>This is an attempt to demonstrate that Condorcet and FBC are incompatible.<BR>I modified Woodall's proof that Condorcet and LNHarm are incompatible.<BR>(Douglas R. Woodall, "Monotonicity of single-seat preferential election rules",<BR>Discrete Applied Mathematics 77 (1997), pages 86 and 87.)<BR><BR>I've suggested before that in order to satisfy FBC, it must be the case<BR>that increasing the votes for A over B in the pairwise matrix can never <BR>increase the probability that the winner comes from {a,b}; that is, it must<BR>not move the win from some other candidate C to A. This is necessary because<BR>if sometimes it were possible to move the win from C to A by increasing<BR>v[a,b], the voter with the preference order B>A>C would have incentive to<BR>reverse B and A in his ranking (and equal ranking would be inadequate).<BR><BR>I won't presently try to argue that this
requirement can't be avoided somehow.<BR>I'm sure it can't be avoided when the method's result is determined solely<BR>from the pairwise matrix.<BR><BR>Suppose a method satisfies this property, and also Condorcet. Consider this <BR>scenario:<BR><BR>3: A=B<BR>3: A=C<BR>3: B=C</SPAN></div>
<div style="RIGHT: auto"><SPAN style="RIGHT: auto">2: A>C</SPAN></div>
<div style="RIGHT: auto"><SPAN style="RIGHT: auto">2: B>A<BR>2: C>B</SPAN></div>
<div style="RIGHT: auto"><SPAN style="RIGHT: auto"><BR>There is an A>C>B>A cycle, and the scenario is "symmetrical," as based on<BR>the submitted rankings, the candidates can't be differentiated. This means<BR>that an anonymous and neutral method has to elect each candidate with 33.33%<BR>probability.<BR><BR>Now suppose the a=b voters change their vote to a>b (thereby increasing v[a,b]).<BR>This would turn A into the Condorcet winner, who would have to win with 100% <BR>probability due to Condorcet.<BR><BR>But the probability that the winner comes from {a,b} has increased from 66.67%<BR>to 100%, so the first property is violated.<BR><BR>Thus the first property and Condorcet are incompatible, and I contend that FBC<BR>requires the first property.<BR><BR>Thoughts?<BR><BR>Kevin Venzke</SPAN></div>
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<div style="RIGHT: auto"><SPAN style="RIGHT: auto"><A style="RIGHT: auto" href="http://lists.electorama.com/pipermail/election-methods-electorama.com/2005-June/016410.html">http://lists.electorama.com/pipermail/election-methods-electorama.com/2005-June/016410.html</A></SPAN></div>
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<div style="RIGHT: auto"><SPAN style="RIGHT: auto">It is certainly a clear proof of the incompatibilty of the Condorcet criterion and Kevin's later</SPAN></div>
<div style="RIGHT: auto"><SPAN style="RIGHT: auto">suggested "variation" of the FBC, "Sincere Favorite":</SPAN></div>
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<div><I>Suppose a subset of the ballots, all identical, rank every candidate in S (where S contains at least two candidates) equal to each other, and above every other candidate. Then, arbitrarily lowering some candidate X from S on these ballots must not increase the probability that the winner comes from S.</I></div>
<div>A simpler way to word this would be: <I>You should never be able to help your favorites by lowering one of them.</I></div></SPAN>
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<div style="RIGHT: auto"><SPAN style="RIGHT: auto"><A style="RIGHT: auto" href="http://nodesiege.tripod.com/elections/#critfbc">http://nodesiege.tripod.com/elections/#critfbc</A><BR></SPAN></div>
<div style="RIGHT: auto"><SPAN style="RIGHT: auto">I can't see any real difference between this and regular FBC, which probably partly explains</SPAN></div>
<div style="RIGHT: auto"><SPAN style="RIGHT: auto">why it didn't catch on.<VAR id=yui-ie-cursor></VAR></SPAN></div>
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<div style="RIGHT: auto"><SPAN style="RIGHT: auto">Chris Benham</div></SPAN>
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<DIV style="FONT-FAMILY: times new roman, new york, times, serif; FONT-SIZE: 12pt">
<DIV style="FONT-FAMILY: times new roman, new york, times, serif; FONT-SIZE: 12pt"><FONT size=2 face=Arial>
<DIV style="BORDER-BOTTOM: #ccc 1px solid; BORDER-LEFT: #ccc 1px solid; PADDING-BOTTOM: 0px; LINE-HEIGHT: 0; MARGIN: 5px 0px; PADDING-LEFT: 0px; PADDING-RIGHT: 0px; HEIGHT: 0px; FONT-SIZE: 0px; BORDER-TOP: #ccc 1px solid; BORDER-RIGHT: #ccc 1px solid; PADDING-TOP: 0px" class=hr contentEditable=false readonly="true"></DIV><B><SPAN style="FONT-WEIGHT: bold">From:</SPAN></B> "fsimmons@pcc.edu" <fsimmons@pcc.edu><BR><B><SPAN style="FONT-WEIGHT: bold">Sent:</SPAN></B> Wednesday, 23 November 2011 9:01 AM<BR><B><SPAN style="FONT-WEIGHT: bold">Subject:</SPAN></B> Re: An ABE solution<BR></FONT><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></DIV></DIV></div></body></html>