[EM] D'Hondt without lists

Olli Salmi olli.salmi at uusikaupunki.fi
Wed Aug 21 21:44:46 PDT 2002

At 03:50 +0300 18.8.2002, Forest Simmons wrote:
>Here's an example of a Condorcet-like PR method:
>In the first round put the Ranked Pairs winner C1 into the winners'
>In the second round compute a modified pairwise matrix by giving weight
>1/2 to each pairwise preference in which C1 is preferred above both of the
>candidates in the pairwise comparison.
>Then augment the winners' circle by the Ranked Pairs winner computed from
>the modified pairwise matrix.
>In each subsequent round for each ballot re-compute the pairwise matrix
>for the remaining candidates by giving a weight of one over one more than
>the number of candidates in the winners' circle that are ranked above both
>of the candidates being compared.

This is not bad. Pairwise comparisons ensure that lower perferences are
taken into account. But it's not a pleasure to count.

I have come up with a something that is so simple it didn't occur to me
earlier. In a single-winner case it's clear that the candidate with the
Droop quota is elected. The Droop quota is used in the Hagenbach-Bischoff
method which is a variation of d'Hondt, so it should be OK here. If we
transfer votes with the value 1/n for the nth candidate if a candidate has
the Droop quota, and eliminate like STV does, we get a system that works
much the way I want. It's Tom Round's Highest Average STV but without
resurrection of eliminated candidates. It doesn't have the drawbacks that
the Finnish method and the Swedish method have.

All the ballot papers of an elected candidate are transferred. The transfer
value and the number of retained ballots are inherent in the ballot paper.
If a Meek style reduction of the quota is necessary it can be done without
redistribution of the votes so a manual count is possible. I'm not sure it
is needed though, because non-transferable ballot papers seem to take care
of themselves and the example I tried produced a quota that was far too
low. Ballots are "transferred" to elected candidates, like in Meek.

If we consider cases where all the voters vote according to the
recommendations of their parties the results are the same as in party list
d'Hondt. If a candidate is eliminated, the votes are transferred down the
party list and the whole party is eventually eliminated. The votes for the
candidates are d'Hondt priority numbers, and if the last candidate is
declared elected because all others have been eliminated, his or her vote
is still a regular priority number, so the system fully accords with
d'Hondt. STV is equivalent to Largest Remainders, because the candidates
retain the quota and if all other candidates have been eliminated, the ones
with the largest remainders are elected. Meek may be different.

This method doesn't take into account lower preferences any more than STV
does but I don't think it does it any worse, either.

It should be possible to use the method with Sainte-Lagu‘ divisors. If you
want to compare it with STV, Sainte-Lagu‘ should probably be used because
it gives results that are similar to Largest Remainders.

The method could even be used as a practical system for small non-partisan
elections in our political culture, which has been using d'Hondt since
1906. For larger or partisan elections list d'Hondt is more manageable.

I think I've understood something I haven't understood before. Now I'm
waiting for glitches to turn up.

Olli Salmi

For more information about this list (subscribe, unsubscribe, FAQ, etc), 
please see http://www.eskimo.com/~robla/em

More information about the Election-Methods mailing list