It's a little tough to spot without the coloring that CIVS does, but #1
loses pairwise to #6. This makes #2 win according to Schulze. As Markus
points out, #2 is the candidate with the weakest pairwise defeat (13-9
vs the 14-13 defeat of #1 by #6).
-- Andrew
On 1/30/11 2:33 PM, Paul Kislanko wrote:
How is #1 not a Condorcet Winner, since #1 pairwise-beats every other
alternative?
------------------------------------------------------------------------
*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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