<html><body style="word-wrap: break-word; -webkit-nbsp-mode: space; -webkit-line-break: after-white-space; "><div><div>On May 9, 2010, at 2:24 PM, Juho wrote:</div><blockquote type="cite"><div>On May 9, 2010, at 6:49 PM, Peter Zbornik wrote:<br><blockquote type="cite">an alternative way to count wins against equally ranked candidates,<br></blockquote><blockquote type="cite">would be to give both 0.5 wins per vote and not 0 as I did in my previous mail.<br></blockquote><blockquote type="cite"><br></blockquote><blockquote type="cite">Thus the current state: A>B gives 1:0. A=B gives ?:? (A=B is only allowed at the end of the ballot)<br></blockquote><blockquote type="cite">The proposal above: A>B gives 2:0. A=B gives 1:1 (linear transformation from A>B - 1:0 and A=B - 0.5:0.5).<br></blockquote><blockquote type="cite">The soccer proposal A>B gives 3:0, A=B gives 1:1 (linear transformation from A>B - 1:0 and A=B -1/3:1/3. This proposal would (as in soccer) encourage voters to differenciate between candidates, and candidates to try to gain more support than the other candidates. The soccer ranking is purely speculative though.<br></blockquote><br>All classical Condorcet methods can handle equal rankings and their impact has been analyzed quite well.<br><br>Usually the discussion focuses on how to measure the strength of the pairwise preferences. This is the next step after the matrix has been populated. The two most common approaches are margins and winning votes.<br><br>- Let's define AB = number of votes that rank A above B<br>- In margins the strength of pairwise comparison of a A against B is: AB - BA<br>- In winning votes the strength of pairwise comparison of a A against B is: AB if AB>BA and 0 otherwise<br><br>- Note that in margins ties could be measured either as 0:0 or as 0.5:0.5 since the strength of the pairwise comparison will stay the same in both approaches<br><br>The most common (/classical) Condorcet methods give always the same winner if there are three candidates and all votes are fully ranked. If there is no Condorcet winner then the candidate with smallest defeat will win. The strengths of the defeats (and therefore also the end result) may differ in margins and winning votes if there are equal rankings. These differences (and their naturalness with sincere vote and their impact on strategies) have been discussed a lot. I will not analyze the differences in detail here. In margins victory 50:40 has the same strength as victory 10:0. In winning votes victory 50:40 has the same strength as victory 50:0.<br><br>Juho</div></blockquote><br></div><div>"smallest defeat"? Oops - I will quote from wikipedia:</div><div><blockquote type="cite" class=""><span class="Apple-style-span" style="color: rgb(0, 0, 0); font-family: -webkit-sans-serif; font-size: 18px; line-height: 27px; "><div style="margin-top: 0.4em; margin-right: 0px; margin-bottom: 0.5em; margin-left: 0px; line-height: 1.5em; ">Ranked Pairs and Schulze are procedurally in some sense opposite approaches (although they very frequently give the same results):</div><ul style="line-height: 1.5em; list-style-type: square; margin-top: 0.3em; margin-right: 0px; margin-left: 1.5em; padding-top: 0px; padding-right: 0px; padding-bottom: 0px; padding-left: 0px; list-style-image: url(http://bits.wikimedia.org/skins-1.5/monobook/bullet.gif); margin-bottom: 0.5em; "><li style="margin-bottom: 0.1em; ">Ranked Pairs (and its variants) starts with the strongest defeats and uses as much information as it can without creating ambiguity.</li><li style="margin-bottom: 0.1em; ">Schulze repeatedly removes the weakest defeat until ambiguity is removed.</li></ul></span></blockquote></div></body></html>