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

[fsimmons at pcc.edu]

Dear Jobst,

Here's an example I worked out based on the scenario

33:  A1>A>A2>>B
33:  A2>A>A1>>B
33:  B ,

with A rated at 80% in both A factions, while the A1 and A2 factions rate A2 and
A1, 
(anti)respectively, at 50% .  

So both A factions slightly prefer A to a coin flip between A1 and A2.

Here's the method, based on approval ballots with favorites indicated.

Draw n=4 ballots at random.  If some candidate is approved on three or more of
the drawn ballots, then among these, elect the one with the most approval on all
99 ballots.

Otherwise, use another random ballot to pick the winner in one of two ways:
Method (1) elect the candidate marked favorite on this ballot.
Method (2) among the candidates approved on this ballot, 
elect the one with the most approval on the other 98 ballots.

Results:

Both methods have a global optimal strategy of full cooperation (i.e. full
approval of A) for both A factions.

The first method yields a probability of about 59.26% that the compromise A will
be elected.

The second method yields a probability of about  79% that A will be elected.

The first method yields an expectation of  about 62.22%  for  each of the A
factions.

The second method yields an expectation of  about 63.21%  for each of the A
factions.

Here's how I got this:

Let n=4, k=3, p=1/3, and q=2/3.

Then let 

c1=sum over j from 0 to (n-k) of binomial(n,j)*p^(n-j)*q^j , 

c2=sum over j from 0 to (n-k) of binomial(n,j)*x^(n-j)*(1-x)^j ,

c3=sum over j from 0 to (n-k) of binomial(n,j)*y^(n-j)*(1-y)^j,

c4=sum over j from 0 to (n-k) of binomial(n,j)*(x+y)^(n-j)*(x+y)^j, and 

g=1-(3*c1-c2-c3+c4) .

Then if x and y represent ballots (as fractions of the total 99) that approve A
in the respective A factions, we get

Prob(A1 is elected) is  c1-c2+g/3  for method 1, else c1-c2+g*(1/3-x) for method 2, 

Prob(A2 is elected) is  c1-c3 + g/3, for method 1, else c1-c3+g*(1/3-y) for
method 2,

Prob(B is elected) is  c1 + g/3, (both methods), and

Prob(A is elected) is c4 for method 1, else c4+g*(x+y) for method 2.

By symmetry, the  expectation for faction A2 equals the expectation for faction
A1 which is

E1(x,y) = Prob(A1) + 80%*Prob(A) + 50%*Prob(A2) . 

Plotting E1(x,1/3) for x from 0 to 1/3 shows (in both cases) a monotone
increasing function,
  
with respective slopes of 8/90 and zero at the right endpoints, in the
respective cases of methods 1 and 2.

I hope I have done this correctly.  It is very easy to get mixed up in these
kinds of calculations.

Forest