[EM] ] An ABE solution

[Chris Benham]

 "I also think that FBC and (voted ballot) Condorcet are not incompatible for a 3-slot ballot."
 
Jameson,
 
Thanks for sharing this (based-on-who-knows-what)  thought. Your problem with Kevin Venzke's
June 2005  3-candidate demonstration that FBC and Condorcet are incompatible (pasted below)
is....what?
 
Chris Benham
 http://lists.electorama.com/pipermail/election-methods-electorama.com/2005-June/016410.html
 
Hello,

This is an attempt to demonstrate that Condorcet and FBC are incompatible.
I modified Woodall's proof that Condorcet and LNHarm are incompatible.
(Douglas R. Woodall, "Monotonicity of single-seat preferential election rules",
Discrete Applied Mathematics 77 (1997), pages 86 and 87.)

I've suggested before that in order to satisfy FBC, it must be the case
that increasing the votes for A over B in the pairwise matrix can never 
increase the probability that the winner comes from {a,b}; that is, it must
not move the win from some other candidate C to A. This is necessary because
if sometimes it were possible to move the win from C to A by increasing
v[a,b], the voter with the preference order B>A>C would have incentive to
reverse B and A in his ranking (and equal ranking would be inadequate).

I won't presently try to argue that this requirement can't be avoided somehow.
I'm sure it can't be avoided when the method's result is determined solely
from the pairwise matrix.

Suppose a method satisfies this property, and also Condorcet. Consider this 
scenario:

a=b 3
a=c 3
b=c 3
a>c 2
b>a 2
c>b 2

There is an A>C>B>A cycle, and the scenario is "symmetrical," as based on
the submitted rankings, the candidates can't be differentiated. This means
that an anonymous and neutral method has to elect each candidate with 33.33%
probability.

Now suppose the a=b voters change their vote to a>b (thereby increasing v[a,b]).
This would turn A into the Condorcet winner, who would have to win with 100% 
probability due to Condorcet.

But the probability that the winner comes from {a,b} has increased from 66.67%
to 100%, so the first property is violated.

Thus the first property and Condorcet are incompatible, and I contend that FBC
requires the first property.

Thoughts?

Kevin Venzke





________________________________
From: Jameson Quinn <jameson.quinn at gmail.com>
To: Chris Benham <cbenhamau at yahoo.com.au> 
Cc: "fsimmons at pcc.edu" <fsimmons at pcc.edu>; MIKE OSSIPOFF <nkklrp at hotmail.com>; em <election-methods at electorama.com> 
Sent: Tuesday, 22 November 2011 8:30 PM
Subject: Re: [EM] An ABE solution


I agree it's silly to create complicated rules for a two-slot ballot. But, though Forest didn't quite say so, I also think that FBC and (voted ballot) Condorcet are not incompatible for a 3-slot ballot. 

Jameson


2011/11/22 Chris Benham <cbenhamau at yahoo.com.au>

Forest,
> 
>"When the range ballots have only two slots, the method is simply Approval, which does satisfy the 
>FBC."
> 
>When you introduced the method you suggested that 3-slot ballots be used "for simplicity".
>I thought you might be open to say 4-6 slots, but a complicated algorithm on 2-slot ballots
>that is equivalent to Approval ??
> 
>"Now consider the case of range ballots with three slots: and suppose that optimal strategy requires the 
>voters to avoid the middle slot.  Then the method reduces to Approval, which does satisfy the FBC."
> 
>The FBC doesn't stipulate that all the voters use "optimal strategy", so that isn't relavent.
>
>
>http://wiki.electorama.com/wiki/FBC
> 
>http://nodesiege.tripod.com/elections/#critfbc
>
>Chris  Benham
> 
> 
> 
> 
>
>
>
>
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