Strike my previous reply... Didn't notice that #6 pairwise beat #1, but pairwise lost to #2-#5. Here's a case where I'd actually like to see instead of the pairwise matrix the matrix that shows counts of votes for #1, #2, ... #5. In particular, which is the Bucklin winner? #6 loses or ties with every alternative except #1.
_____ From: election-methods-boun...@lists.electorama.com [mailto:election-methods-boun...@lists.electorama.com] On Behalf Of Andrew Myers Sent: Saturday, January 29, 2011 4:41 PM To: Election Methods Mailing List Subject: [EM] An interesting real election Here is an unusual case from a real poll run recently by a group using CIVS. Usually there is a Condorcet winner, but not this time. Who should win? Ranked pairs says #1, and ranks the six choices as shown. It only has to reverse one preference. Schulze says #2, because it beats #6 by 15-11, and #6 beats #1 by 14-13. So #2 has a 14-13 beatpath vs. #1. Hill's method ("Condorcet-IRV") picks #6 as the winner. -- Andrew 1. 2. 3. 4. 5. 6. 1. - 13 15 17 16 13 2. 9 - 13 14 17 15 3. 11 11 - 13 15 14 4. 9 10 10 - 14 13 5. 11 10 9 10 - 13 6. 14 11 11 13 10 -
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