[EM] Some chance for consensus (was: Buying Votes)

[fsimmons at pcc.edu]

Jobst,

This reminds me of your two urn method based on approval ballots:

Initialize with all ballots in the first urn.

While any ballots are left in the first urn ...

    find the approval winner X of these remaining ballots

    circle candidate X on all of the ballots in the first urn that approve candidate X, and then transfer them to the second urn.

End While

Elect the circled candidate on a randomly drawn ballot from the second urn.

It looks like your newest method is a variation where "Approval" is interpreted as "positive rating," and X is circled only on those ballots that rate X sufficiently high* relative to the number of (remaining) ballots that do not approve X.   

If 99% of the (remaining) ballots do not approve X, then X is circled only on those ballots that rate X above 99%.  If less than 1% of the (remaining) ballots do not approve X, then even a ballot that rates X at a mere 1% would get a circle around X.

The exact relation between the required rating relative to the lack of approval (on the remaining ballots) can be played with to get variations of this method.

In this method there is no need to rate any candidate that the voter cannot conceive of as a compromise.  Therefore it seems quite natural to consider positive rating as some level of approval.

*Some provision must be made for ties and for the case where no ballot rates the current X high enough to get transfered into the second urn.

Does that capture the idea?

Forest


----- Original Message -----
From: Jobst Heitzig 
Date: Thursday, November 6, 2008 3:37 pm
Subject: Re: [EM] Some chance for consensus (was: Buying Votes)
To: fsimmons at pcc.edu
Cc: gregory.nisbet at gmail.com, election-methods at lists.electorama.com, Raph Frank , Kristofer Munsterhjelm 

> Hi again,
> 
> here's another, somewhat more stable method which also achieves 
> the 
> following:
> 
> > ...
> > provides for strategic equilibria in which C is elected with 
> 100%, 55%,
> > and 100% probability, respectively, in the following situations:
> > 
> > Situation 1:
> > 55% A(100)>C(70)>B(0)
> > 45% B(100)>C(70)>A(0)
> > 
> > Situation 2:
> > 30% A(100)>C(70)>B,D(0)
> > 25% B(100)>C(70)>A,D(0)
> > 45% D(100)>A,B,C(0)
> > 
> > Situation 3:
> > 32% A(100)>C(40)>B,D(0)
> > 33% B(100)>C(40)>A,D(0)
> > 35% D(100)>C(40)>A,B(0)
> > 
> > (All these being sincere utilities)
> > ...
> 
> 
> The idea is that for each possible compromise option C, a voter 
> indicates, by a rating on her ballot, how many voters she 
> requires to 
> transfer their winning probability to C before she will do so, too.
> 
> 
> This is the method:
> 
> 1. Each of the N voters rates on her ballot each option between 
> 0 and 100.
> 
> 2. For each option X, put
> s(X) = no. of ballots rating X at zero.
> 
> 3. Put all ballots into a first urn labelled U1, and have 
> another urn 
> labelled U2, initially empty.
> 
> 4. For each option X, in order of ascending s(X), do the following:
> 
> 4.1 Find the smallest number R for which f(R) >= 100, where
> f(R) = R + 100 * (no. of ballots in U1 rating X above R) / N.
> 
> 4.2 For each ballot in U1 rating X above R: Mark X on that 
> ballot and 
> move the ballot from U1 to U2.
> 
> 5. On each of the ballots that remained in U1, mark the option 
> with the 
> highest rating on that ballot and also put the ballot into U2.
> 
> 6. Draw one ballot at random. The option marked on that ballot wins.
> 
> 
> (By "above", we mean strictly above and not equal, of course.)
> 
> If there is only one compromise option C besides some polar 
> favourite 
> options A,B,... , the method essentially simplifies to this:
> - Find the smallest R such that at least 100-R percent of the 
> ballots 
> rate C above R.
> - Then draw a ballot at random. If it rates C above R, C wins, 
> otherwise 
> the option with the largest rating of that ballot wins.
> 
> 
> Let's look at the three example situations:
> 
> Situation 1 (sincere utilities):
> 55: A(100)>C(70)>B(0)
> 45: B(100)>C(70)>A(0)
> 
> Take any pair of numbers x,y such that
> 
> 0 <= x <= 55,
> 0 <= y <= 45,
> x + y > 55,
> x > 3y/7, and
> y > 3x/7.
> 
> It is easy to check from the above true ratings that then all 
> voters 
> would gain if x of the A-voters and y of the B-voters 
> transferred their 
> winning probability to C. The voters can make sure this happens 
> in the 
> suggested method by voting this way:
> 
> 55-x: A(100)>C(0)=B(0)
> x: A(100)>C(101-x-y)>B(0)
> y: B(100)>C(101-x-y)>A(0)
> 45-y: B(100)>C(0)=A(0)
> 
> The method would begin with C (receiving the smallest numbers of 
> 0-rates), find R=100-x-y, and mark C on all the x+y ballots, 
> resulting 
> in these winning probabilities: A:55-x, B:45-y, C:x+y.
> 
> No voter has an incentive to reduce her C-rating since that 
> would 
> immediately move R to 100 and C's winning probability to 0.
> 
> So, for each such pair (x,y), the above way of voting is a 
> strategic 
> equilibrium, the socially best of whose is the one where x=55 
> and y=45, 
> C wins with certainty, and the ballots look like this:
> 
> 55: A(100)>C(1)>B(0)
> 45: B(100)>C(1)>A(0)
> 
> 
> Situation 2 (sincere utilities):
> 30: A(100)>C(70)>B,D(0)
> 25: B(100)>C(70)>A,D(0)
> 45: D(100)>A,B,C(0)
> 
> Here the only difference is that we require x+y>30 instead of 
> 55. For 
> each pair (x,y) with...
> 
> 0 <= x <= 30,
> 0 <= y <= 25,
> x + y > 30,
> x > 3y/7, and
> y > 3x/7.
> 
> ...the following is an equilibrium way of voting which gives C a 
> winning 
> probability of x+y:
> 
> 30-x: A(100)>B,C,D(0)
> x: A(100)>C(101-x-y)>B,D(0)
> y: B(100)>C(101-x-y)>A,D(0)
> 25-y: B(100)>A,C,D(0)
> 45: D(100)>A,B,C(0)
> 
> Here, the method starts with either X=C or X=D since one of them 
> has the 
> smallest s(X). Both ways, C will eventually be marked on the x+y 
> ballots 
> (at R=100-x-y) and D on the 45 ballots (at R=55).
> 
> In particular, the socially optimal equilibrium is
> 
> 30: A(100)>C(46)>B,D(0)
> 25: B(100)>C(46)>A,D(0)
> 45: D(100)>A,B,C(0),
> 
> resulting in the winning probabilities 55% for C and 45% for D.
> 
> 
> Situation 3 (sincere utilities):
> 32: A(100)>C(40)>B,D(0)
> 33: B(100)>C(40)>A,D(0)
> 35: D(100)>C(40)>A,B(0)
> 
> Here we consider triples of numbers x,y,z such that
> 
> 0 <= x <= 32,
> 0 <= y <= 33,
> 0 <= z <= 35,
> x + y + z > 35,
> x + y > 3z/2,
> y + z > 3x/2, and
> z + x > 3y/2.
> 
> In that case, the following is a voting equilibrium:
> 
> x: A(100)>C(101-x-y-z)>B,D(0)
> y: B(100)>C(101-x-y-z)>A,D(0)
> z: D(100)>C(101-x-y-z)>A,B(0)
> 32-x: A(100)>B,C,D(0)
> 33-y: B(100)>A,C,D(0)
> 35-z: D(100)>A,B,C(0)
> 
> As in situation 1, the socially best equilibrium is when all 
> voters 
> cooperate by rating C at 1, making C the sure winner.
> 
> 
> It seem the advantage of this method is that it is more stable 
> than the 
> first one, having a lot more desirable equilibria which. 
> However, the 
> socially optimal equilibria require a somewhat strange way of 
> voting in 
> which you understate your true rating of the compromise in order 
> to make 
> sure not to give others an incentive to reduce their rating and 
> thereby 
> lessen the compromise's winning probability...
> 
> What do you make of this?
> 
> Yours, Jobst
> 
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