[EM] Proportional Representation beyond STV?

Alex Small asmall at physics.ucsb.edu
Tue Jul 8 23:05:02 PDT 2003


Chris Benham said:
> My second idea is this:  Elect the CW (completed however), and then
> depending on how many seats there are to be filled, fractionally mark
> down some of the ballots according to their contribution to electing the
>  winner, and taking into account the "wasted" vote. Repeat  until the
> desired number of candidates are elected. The details of  exactly which
> ballots to mark down by exactly how much  I haven't yet thought about,
> but I shouldn't think it is a huge problem.


Some sort of fractional system seems necessary when generalizing
single-winner methods (be it IRV, Condorcet, Approval, etc.) to
multi-winner races.  I'm not sure how to do this in Condorcet, although I
seem to recall some posts on this subject a while ago.  My best guess:

Each time a candidate is elected, determine a "runner-up", i.e. the
candidate who would have been elected had our latest winner not also been
in the running.  Anybody who preferred the newly elected candidate to the
runner-up has his ballot marked down by a factor n/(n+1) where n is the
number who are now elected.

Of course, I haven't thought this through very carefully yet.

I put PR systems into 3 main categories:

1)  Fractional systems:  Generalizations of single-winner methods that
involve marking down ballots after a candidate has been elected.  STV,
Proportional Approval Voting, and Condorcet PR all fit into this category.

2)  Plurality systems:  SNTV, cumulative voting, limited vote, etc.  Just
give the voters N votes, specify the maximum number of votes they can give
a single candidate, and elect the top S candidates, where S is the number
of seats.

3)  List systems.

Personally, I prefer simplicity over sophistication for public elections,
so my preference is list systems for explicitly partisan multi-winner
races, and plurality systems for ostensibly non-partisan multi-winner
races.  But I see virtues in fractional systems when the number of winners
is 2 or 3.



Alex





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