<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">Here is an example of my suggested new FBC-complying method performing better than ICT <BR style="RIGHT: auto">("Improved Condorcet, Top", a name coined by Mike Ossipoff for a method I defined).<BR></div>
<div style="RIGHT: auto">30: A=B<BR>30: B<BR>20: A<BR>10: C>A<BR>10: D>A<BR></div>
<div style="RIGHT: auto">According to the TTR (Kevin Venzke's "Tied at the Top Tule"),<BR> A>B 70-30 and B>A 60-40. A> C 50-10, A>D 50-10, B>C 60-10, B>D 60-10.<BR><BR>Only A and B are qualified by TTR, and ICT elects the qualified candidate with highest</div>
<div style="RIGHT: auto">Top ratings (we'll say these are Top-Middle-Bottom 3-slot ratings ballots, with default<BR>rating being Bottom). <BR></div>
<div style="RIGHT: auto">TR scores: B60, A50, C10, D10.<BR></div>
<div style="RIGHT: auto">So ICT elects B.<BR><BR>The first part of my new method is the same, so only A and B are qualified.<BR><BR>To determine the winner a different pairwise matrix is looked at to weigh defeats (while keeping<BR>the same TTR "direction").<BR><BR>So A>B 70-60 and "B>A" 60-70 (the 30 A=B ballots each give a whole vote to both A and B).<BR></div>
<div style="RIGHT: auto">A and B have no other pairwise "defeats", so (weighing them by Losing Votes) A's MinMax score is<BR>70 and B's is 60 so A wins.<BR><BR>A is rescued from the splitting of the A>B "faction''s vote by C and D being on the ballot.<BR><BR>As it does here, the new method is much more likely than ICT to elect the real Condorcet winner.<BR><BR>Chris Benham<VAR id=yui-ie-cursor></VAR><BR style="RIGHT: auto"><BR><BR>I wrote (Tues.20 Nov 2012):<BR></div>
<div style="BACKGROUND-COLOR: transparent; FONT-STYLE: normal; FONT-FAMILY: times new roman, new york, times, serif; COLOR: rgb(0,0,0); FONT-SIZE: 16px; RIGHT: auto">I have an idea for a not-very-sinple FBC-complying method that behaves like ICT with 3 candidates, but better<BR>handles more candidates and ballots with more than 3 ratings-slots or ballots that allow full ranking of the candidates.<BR> <BR>*Voters rank from the top however many candidates they wish. Equal-top ranking and truncation must be allowed.<BR> <BR>Use the "Tied-at-the-Top Rule" (invented by Kevin Venzke) to discover if any candidate/s pairwise beats (according<BR>to that rule's special definition) all the others, and if so to disqualify all those that don't.<BR> <BR><A href="http://wiki.electorama.com/wiki/Tied_at_the_top_rule">http://wiki.electorama.com/wiki/Tied_at_the_top_rule</A><BR> <BR>Then construct a pairwise matrix that is "normal" except that ballots
that equal-rank at the top any X and Y contribute<BR>a whole vote (in the X versus Y pairwise comparison) to each of X and Y. Ballots that equal-rank any X and Y in any<BR>below-top position contribute (in that pairwise comparison) no vote to either.<BR> <BR>The purpose of that matrix is just to determine Losing Votes scores. The directions of the defeats are determined by<BR>the Tied-at-the-Top rule (according to which X and Y can pairwise "defeat" each other.<BR> <BR>Elect the qualified candidate whose worse "defeat" (as identified by TTR and measured by Losing Votes with the above<BR>equal top-ranking rule) is the weakest.*<BR> <BR>I hope that inelegant waffle is at least clear.<BR> <BR>Chris Benham<BR></div></div></body></html>