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 ---- Election-Methods mailing list - see http://electorama.com/em for list info
