[Election-Methods] D(n)MAC

[fsimmons at pcc.edu]

Dear Jobst,

It was too good to be true!

And I really messed up my example in my last message:

> > Suppose that we have 2Q>P>Q>0, P+Q=100, and factions
> > 
> > P: A>C>B
> > Q: B>C>A
> > with C rated at R% by all voters.
> > 
> > If R = Q/2^(n-1) + P, then (under D(n)MAC/RB) the common 
> strategy of each faction approving C on 
> > exactly Q ballots is a global equilibrium , so that the 
> winning probabilities for A, B, C become
> > 1-2Q%, 0, and 2Q%, respectively.
> > 
> > This can be thought of as implicit trading, since the second 
> faction moves C up to equal-first on all Q of 
> > its ballots, while the first faction moves C up to equal-first 
> on Q of its ballots, as well.
> > 
> > The expected "utilities" for the two factions are
> > 
> > EA = (100-2Q)+R*(2Q)%, and
> > EB = R*(2Q)%.

These expectations are not the ones given by my later analysis.  They should be

EA=100*pA+R*pC, and EB=100*pB+R*pC,

where pC=(2q)^3, pB=g*q, and pA=p^3-q^3+g*p, and
g=1-(p^3-q^3)-(2q)^3 = 1 - p^3 - 7q^3

> > 
> > For example, if P=60, Q=40, n=3, and R=70, the equation 
> R=Q/2^(n-1)+P is satisfied, so
> > 
> > the global equilibrium strategy is for both to approve C on 40 
> ballots, yielding the winning probabilities
> > 
> > 20%, 0, and 80%, respectively, 
> > 
> > so that the expectations are
> > 
> > EA= 76, and EB=56,

In this example, these expectations are 71.2 and 49.28, respectively.

Suppose we allow the second faction to have a different appraisal S of C, as you suggested.

Let's set to zero the partial derivative of EB with respect to y evaluated at (x,y)=(.4, .4), 
to see how low S could be without destroying the equilibrium at (.4, .4):

It boils down to  S = P/2^(n-1)+Q, which, for this example, is S = 55.

With this value of S, 

EB becomes 41.6, while EA remains at 71.2 .

It is easy to check graphically that this equilibrium is global, 
i.e. no defection from this equilibrium no matter how large or small by either faction
acting alone could improve its expectation.


Your most 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. AMP performs better here in giving A the complete 66% 
> probability, 
> but AMP is considerably more complex and non-monotonic...

just shows we still have work to do, i.e. no end to the fun.

The amazing thing to me is how a lottery method can be the surest way of getting
certainty in the outcome (assuming rational voters) in certain situations.

My Best,

Forest