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: [EMAIL PROTECTED] Cc: election-methods@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 > > > [EMAIL PROTECTED] 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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