[Election-Methods] RE : Corrected "strategy inCondorcet" section

[Paul Kislanko]

This is enough to convince me that approval is an appropriate method.

  _____  

From: election-methods-bounces at lists.electorama.com
[mailto:election-methods-bounces at lists.electorama.com] On Behalf Of Chris
Benham
Sent: Tuesday, August 14, 2007 5:11 PM
To: Juho
Cc: Forest W Simmons; Election Methods Mailing List
Subject: Re: [Election-Methods] RE : Corrected "strategy inCondorcet"
section




Juho wrote:


On Aug 2, 2007, at 6:44 , Kevin Venzke wrote:



  

1000 A>B, 1000 C>D, 1 D>B

      



  

Yes, I do think D is the proper winner.

Do you have a verbal (natural language) explanation why D is better  

than A and C. This scenario could be an election in a school. One  

class has voted A>B (A and B are pupils of that class), another class  

has voted C>D, the teacher has voted D>B. What should the teacher  

tell the C>D voting class when they ask "didn't you count our votes"?  

Maybe this is clear to you. Unfortunately not as clear to me. The  

teacher vote seemed to be heavier than the pupils votes :-).

  


I  agree with Kevin that D is the proper winner, but Winning Votes isn't my
favourite algorithm.
If we are sticking with Condorcet  "immune" methods and so are only
focussing on how to compare
(measure) defeat strengths, then I like Approval Margins (Ranking) if we are
using plain ranking ballots.

So interpreting ranking (above bottom or equal-bottom) as approval, we get
these approval scores:
D1001,   B1001,   A1000,  C1000

All the candidates have at least one pairwise defeat, and by AM  the weakest
is D's single defeat, C>D
by an AM of -1.
I also like  Approval-Sorted Margins(Ranking), which  is probably equivalent
to AM.

The initial approval order is  D=B>A=C.  The smallest approval gaps (zero)
are between D and B, and A
and C.  A pairwise ties with C but D pairwise beats B, so our first
modification of the order is D>B>A=C.
A pairwise beats B, so the second modification is  D>A>B=C.  B pairwise
beats C, so the third modified
order is D>A>B>C.  This order accords with the pairwise comparisons so is
the final order and D wins.

I also like eliminating (and dropping from the ballots) the candidate lowest
in this order and then repeating
the whole process until one remains. In this case that would give the same
winner, with the elimination order
just being the reverse of  the ASM(R) order.

The only candidate with any sort of claim versus D is C, and  C is pairwise
beaten by a more approved
candidate (B) so C is outside the "Definite Majority (Ranking)" set.

Chris Benham






-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://lists.electorama.com/pipermail/election-methods-electorama.com/attachments/20070814/56782bef/attachment-0003.htm>