Thanks, I'll try to reproduce this. However, this seems rather complex. How
nonmonotonic is Nanson/Baldwin Method?
From: James J Faran <[EMAIL PROTECTED]>
To: John Wong <[EMAIL PROTECTED]>
CC: [EMAIL PROTECTED]
Subject: Re: [Election-Methods] How is the Nanson and/or Baldwin
non-monotonic?
Date: Thu, 20 Sep 2007 09:56:09 -0400
OK, here's a description on how to generate a Nanson example. A similar
method should generate a Baldwin example.
Start with a profile that gives the Borda ranking A>B>C:
ABC: 2
BAC: 1
This gives Borda scores A:5, B:4, C:0, and A beats B pairwise, so A is
the Nanson winner. We want to have enough voters to move (increasing
ranking of A) without changing the ranking of C so that, after the move,
C then has a better Borda score than B. This requires 5 voters of type
either CBA or BAC. Adding 5 voters of each type doesn't change the
ranking, so take the profile
ABC: 7
ACB: 5
CAB: 5
CBA: 5
BCA: 5
BAC: 6
Borda: A 35, B 34, C 30. A still wins Nanson, but if we switch,
increasing A's rankings (5 CBA voters become 5 CAB voters),
ABC: 7
ACB: 5
CAB: 10
CBA: 0
BCA: 5
BAC: 6
Borda: A 40, B 29, C 30. B is now eliminated, but A beats C pairwise, so
A is still the Nanson winner. We need to make C beat A pairwise without
messing up the Borda rankings, so we add Condorcet triplets of the
correct type. A is beating C by 3, so we need to add 4.
Original Profile:
ABC: 11
ACB: 5
CAB: 9
CBA: 5
BCA: 9
BAC: 6
Borda: A 47, B 46, C 42. C is eliminated and A beats B pairwise 25-20.
New Profile (5 CBA voters become 5 CAB voters):
ABC: 11
ACB: 5
CAB: 14
CBA: 0
BCA: 9
BAC: 6
Borda: A 52, B 41, C 42. B is eliminated and C beats A 23-22.
We could also take 5 BAC voters and make them ABC voters and get another
example.
You should be able to find a Baldwin example by a similar technique.
On Thu, 2007-09-20 at 00:55 -0700, John Wong wrote:
> How is the Nanson and/or Baldwin non-monotonic? I've been trying to
develop
> an example where they are non-monotonic, but I'm having trouble.
>
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