[EM] Re: simulating an Approval campaign/election

[Forest Simmons]

> Date: Wed, 2 Feb 2005 14:45:57 -0800 (PST)
> From: Rob LeGrand <honky1998 at yahoo.com>
> Subject: [EM] Re: simulating an Approval campaign/election
>
> Forest wrote:
>> Actually, DSV with Strategy A can sometimes converge to a stable
>> equilibrium even when there is no Condorcet Winner:
>>
>> 4900 C
>> 2400 B
>> 2700 A>B

Rob replied:
> You're right.  I forgot to disclose my assumption of strict
> preferences.  A better way to say what I meant:  When for every
> candidate X there is another candidate Y that is strictly preferred
> by a majority of voters to candidate X, there is no equilibrium
> under Approval when every voter uses strategy A.  Also, I forgot to
> mention that the results of my example election under the different
> DSV modes assume that all voters use strategy A.
>
>> [In my opinion this is unfortunate, since it would be better to
>> have B and C win with about equal probability in this example.]
>
> Why, do you think?  Candidate A can't win under Approval no matter
> what (unless some voters are completely irrational), so the A-first
> voters have strong motivation to approve B.  Plus, a majority
> strictly prefers B to C.


That's the way I thought about it until Kevin pointed out that making B 
the sure winner in this case would give high incentive for truncation 
among B supporters if their sincere preference happened to be B>A>C, thus 
demoting A from Condorcet Winner and promoting B to (wv) cycle winner.


Some folks solve this dilemma by going with the margins winner C for the 
truncated case.

I believe that a coin toss between B and C would be better in this case.

I wonder how good it would be in general to decide between the margins 
winner and the wv winner with a coin toss.

Here's another solution:  do a symmetric completion of the ballots before 
applying your DSV.  There would still be a cycle, and with strict 
orderings your DSV would cycle among the candidates, so that all had a 
chance at winning.

What would be their winning probabilities?

For a refinement of that idea use a refined symmetric completion:

4900 C becomes

2450 C>>A>B
2450 C>>B>A


Then interpret the first subfaction to mean that C, A, and B, 
respectively, have relative utilities of 3, 1, and 0.


How would this change the probabilities?


In some other election where a faction voted

1000 A=B>C

the refined symmetric completion would be

500 A>B>>C
500 B>C>>A

where the first of these subfactions would impute relative utilities of 3, 
2, and 0 to candidates A, B, and C, respectively.


For another approach, using "lotteries," see my recent posting under that 
title.

Forest