Yes, notice that if someone now votes 2 > 1 > 6, the Schulze method
picks 1 over 2, which is the opposite of what the new voter wanted.
But I'm not really trying to criticize the Schulze method here --
perhaps the most salient thing is how rarely situations like this come
up in Condorcet voting.
-- Andrew
On 7/22/64 2:59 PM, Aaron Armitage wrote:
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 /<[email protected]>/* wrote:
From: Andrew Myers <[email protected]>
Subject: [EM] An interesting real election
To: "Election Methods Mailing List" <[email protected]>
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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