[EM] Another Approval-Condorcet compromise method

[Forest Simmons]

More thoughts on Universal Approval:

If there are N candidates, and two of them are A and B, then there are
2^(N-2) subsets that contain both A and B. Therefore ...

1. The maximum number of points that a candidate could get from one
pairwise comparison on one ballot would be 2^(N-2). 

2. If m and n are the totals in the (A,B) and (B,A) positions of the total
pairwise comparison matrix for the election, and p and q are the same
respective entries divided by the product of 2^(N-2) and the number of
voted ballots, then p is the probability that A would be approved over B
in a randomly chosen subset of candidates (containing both A and B) by a
randomly chosen voter.  The number 1-p-q is the probability that neither A
nor B would be preferred above the other by a random voter in a random
subset (including both A and B).

3. The number of computations to produce the comparison matrix for each
ballot is on the order of (N^2)*2^(N-2), which is not bad for a dozen or
so candidates.

4. Polls for predicting the outcome could be conducted as a MonteCarlo
simulation of the election:

Each randomly polled voter is presented with a different random subset of
the candidates and asked which ones she would support if everybody else
dropped out of the election.

Forest