1 has a path to 6 at least as strong as 6's path to 1, namely 1>3>6, at 15-11
and 14-11. It
seems a little odd, to me at least, that 6's path to 1 should benefit 2 but not
6 itself.
Starting from the top seems the only way of ensuring that the path that orders
the two
candidates relative to each other is the one which actually contributes to the
final outcome.
--- On Sat, 1/29/11, Andrew Myers <an...@cs.cornell.edu> wrote:
From: Andrew Myers <an...@cs.cornell.edu>
Subject: [EM] An interesting real election
To: "Election Methods Mailing List" <election-meth...@electorama.com>
Date: Saturday, January 29, 2011, 4:40 PM
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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