[EM] D'Hondt without lists

[Forest Simmons]

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 Phragmén's method elects ABCDEFGJK.
> Phragmén'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.
>
> Phragmén 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 Phragmén 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. Phragmén'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






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