[Election-Methods] D(n)MAC/RB

[fsimmons at pcc.edu]

Dear Jobst,

I think you are right:   Plain random ballot (as fall back) induces full cooperation at lower values of alpha than does a mixture of plain and approval random ballot, since the penalty is greater for failing to cooperate in the former case.

However, given a value of alpha for which both fall back methods (plain and mixed) induce full cooperation, it seems to me that the fallback mixture will give a higher probability of the compromise candidate A being elected, since the probabilities only differ in case of fallback, where random ballot favors A1 and A2, while random approval ballot favors the compromise A.

My Best,

Forest

----- Original Message -----
From: Jobst Heitzig 
Date: Tuesday, May 27, 2008 3:40 pm
Subject: Re: [english 94%] Re: D(n)MAC/RB
To: fsimmons at pcc.edu
Cc: election-methods at lists.electorama.com

> Dear Forest,
> 
> a quick calculation for your suggestion (please check!) gives:
> 
> Winning probability for A under full cooperation of the A1 and 
> A2 voters:
> (16+4*8)/81 + 8/27*1/2*2/3 = 56/81 = approx. 70% (OK)
> 
> Gain in expected utility for the A1 voters when reducing their 
> cooperation by an infinitesimal epsilon:
> epsilon*(
> 4*3*(2/3)²*1/3*(
> 1/2*1/3*(alpha1-alpha)
> + 1/2*1/3*(alpha2-alpha)
> + 2/3*(beta-alpha) )
> + 8/27*1/2*(
> (alpha1-alpha)
> + 1/3*1/(2/3)*(alpha1-alpha) )
> )
> = epsilon/27*(14alpha1+8alpha2+16beta-38alpha)
> 
> In this, alpha[1|2] and beta are the utilities of A[1|2] and B 
> for the 
> A1 voters. We may assume that alpha1=1 and beta=0.
> 
> The A1 voters have no incentive to reduce their cooperation as 
> long as 
> the above gain is <0, i.e. when alpha>(7+4alpha2)/19. The latter 
> is 
> always true when alpha>58%. Similarly, the A2 voters will fully 
> cooperate when they rate A at least at 58% the way from B to 
> their 
> favourite A2.
> 
> This is good, however with pure Random Approval as fallback it 
> is even 
> better, it seems: The gain then changes to
> epsilon*(
> 4*3*(2/3)²*1/3*(
> 2/3*(beta-alpha) )
> + 8/27*(
> (alpha1-alpha)
> + 1/3*1/(2/3)*(alpha1-alpha) )
> )
> = epsilon/27*(12alpha1+16beta-28alpha)
> which is negative even when alpha>3/7 only!
> 
> (Please double-check these calculations!)
> 
> Yours, Jobst
> 
> 
> fsimmons at pcc.edu schrieb:
> > Jobst,
> > 
> > After thinking about your recent example:
> > 
> > > 33: A1>A>A2 >> B
> > > 33: A2>A>A1 >> B
> > > 33: B >> A1,A2,A
> > >and the 66 A-voters try to cooperate to elect A by 
> unanimously approving
> > >of her, then they still get A only with a low probability of 16/81
> > >(approx. 20%) while A1 and A2 keep a probability of 64/243 
> (approx. 25%)
> > >each. A
> > 
> > I have two ideas for incremental improvement:
> > 
> > 1. For the fall back method, flip a coin to decide between 
> Random 
> > Ballot and Random Approval Ballot.
> > 
> > Note that if Random Approval Ballot were used exclusively, 
> then there 
> > could be insufficient incentive for the first two factions to 
> give 
> > unanimous support to A.
> > 
> > 2. Reduce the approval requirement from 4 of 4 to 3 out of 4 
> matches. 
> > The the fall back method would be used less frequently, since 
> the 3 of 4 
> > requirement is more feasible for a candidate approved on two 
> thirds of 
> > the ballots.
> > 
> > Of course, this doesn't solve the general problem, and I'm 
> afraid that 
> > any attempt to automate these kinds of adjustments might be 
> vulnerable 
> > to manipulation by insincere ballots.
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