<html><body style="word-wrap: break-word; -webkit-nbsp-mode: space; -webkit-line-break: after-white-space; "><div>True, I missed the "majority" part.</div><div><br></div><div>Juho</div><div><br></div><br><div><div>On May 24, 2010, at 10:38 AM, C.Benham wrote:</div><br class="Apple-interchange-newline"><blockquote type="cite"> <div bgcolor="#ffffff" text="#000000"> <br> <div>Juho wrote (23 May 2010):<br> <br> <snip><br> </div> <pre>><i> 1. Rank the candidates. Truncation is allowed. Equal ranking is not
</i>><i> planned for (but we could come up with something).
</i>><i> 2. Label the candidates A, B, C, ... Z in descending order of first
</i>><i> preference count.
</i>><i> 3. Let the current leader be A.
</i>><i> 4. While the current leader has a majority pairwise loss to the very
</i>><i> next candidate, set the current leader to the latter candidate. (In
</i>><i> other words step 4 must be repeated until there is no loss or no other
</i>><i> candidates.)
</i>><i> 5. Elect the current leader.
</i></pre> <div> <blockquote type="cite"> <pre>How about this example and LNH.
6: A>C
5: B>A
2: C>B
2: C
Candidate names indicate the order in first preferences. B beats A. C
beats B. C wins.
6: A>C
5: B>A
2: C>B
2: C>A
Two "C" voters have changed their vote to "C>A". B does not beat A. A
wins. The "C" voters were harmed when they included their later
preferences.
Juho
</pre> </blockquote> </div> <br> Juho,<br> <br> The key word you missed in the definition is "majority". In both your elections there are 15 ballots, so a <br> "majority pairwise loss" requires a winning score of at least 8. <pre>In both cases the FPP winner A wins, in the first because B's pairwise score against
A is 7, a pairwise win but not a "majority" pairwise win (and so of course not a
"majority pairwise loss" for A).
Chris Benham
</pre> <br> </div> </blockquote></div><br></body></html>