Ollie, thanks for your insights and examples. The main example below has led me to consider another, perhaps better, way of scoring the head-to-head PR methods that I have been working on lately.
See below. Forest On Sun, 17 Nov 2002, Olli Salmi wrote in part: <snip> > Cassel gives the > following example with 9 seats: > 4200 ABCDEFGHI > 1710 ABCJKLMNO > > Thiele's method elects ABCDEFGHI, while Phragm�n's method elects ABCDEFGJK. > Phragm�n's examples are similar but shorter. He finds it bad that the > smaller group is penalized for voting for common candidates. The smaller > party would be tempted not to vote for a common candidate, and in that case > even the larger party might prefer not to vote for one, especially if the > difference between the sizes of the parties was smaller. Both parties might > be willing to accept a common candidate as chair of a committee but under > Thiele's method it might happen that neither party would want to vote for > him or her. > > Phragm�n finds justification for his method from the idea of branching > lists, which have common candidates only at the top of the lists. > > DEFGHI > / > ABC > \ > JKLMNO > > With such a branching list Phragm�n thinks that the whole combination > should be given as many seats as the total vote entitles. Three of those > seats should go to the common candidates, while the rest should be > apportioned between the branches according to d'Hondt's rule, in the order > D 4200, E 2100, J 1710, F 1400, G 1050, K 855. Phragm�n's method gives the > same result. > > If I've understood the method correctly and if there's no mistake in the > following calculations, ordinary non-sequential PAV with d'Hondt's rule > gives the same result as Thiele's method. The two outcomes above get the > following satisfaction scores: > > ABCDEFGHI > 4200*(1+1/2+1/3+...+1/9)+1710*(1+1/2+1/3) > > ABCDEFGJK > 4200*(1+1/2+1/3+...+1/7)+1710*(1+1/2+...+1/5) > > 4200*(1+1/2+1/3+...+1/7)+1710*(1+1/2+1/3) is the same in both, so to > determine which is greater we only need to count > > ABCDEFGHI > 4200*(1/8+1/9)=991 2/3 > > ABCDEFGJK > 1710*(1/4+1/5)=769 1/2 > > So ABCDEFGHI would be preferred. I haven't calculated the other > possibilities, but if the order of a party's candidates is irrelevant there > aren't many of them. > I get the same result, assuming that there is no preferred order among the candidates on the lists. Yet there is an implicit order, since H is chosen before I, and J before K. There must be a decision about the order of the candidates on the list if they are not listed randomly. So here's where a Condorcet Flavored PR method could be put to use. Let's compare the two candidate subsets head-to-head: ABCDEFGHI and ABCDEFGJK first differ in the eighth and fourth rank according to the 4200 and 1710 factions, respectively. Suppose we say that the larger faction has used up 7/9 of its vote, and the smaller faction has used up 3/9 of its vote, the amount of agreement down to the first disagreement, as a fraction of the number of seats. Then the first subset (ABCDEFGHI) is preferred by 4200 voters with 2/9 of their voting strength left, and the other subset (ABCDEFGJK) is preferred by the other 1710 voters, with 6/9 of their voting strength left. Since 4200*2/9 is smaller than 1710*6/9, the smaller faction gets its way on this one. The method implicit in this example would not be practical in a typical nine seat election if there were thirty candidates, but it could be used to make head-to-head comparisons among the results of various competing methods. In the present case, PAV did give each faction at least a proportional share (i.e. 3/9 is greater than 1710/5910), but it didn't take full advantage of the serendipitous ABC overlap. Forest ---- For more information about this list (subscribe, unsubscribe, FAQ, etc), please see http://www.eskimo.com/~robla/em
