Methods of elimination in quota preferential STV

[LAYTON Craig]

The problem with these types of systems is that they require electors to
rank ALL candidates.  While this is perfecly reasonable for single-winner
elections, in PR the number of candidates make it impracticable & innacurate
as the number of informals is too high.  In our state PR elections for 6
members, we always have well over 50 candidates.  Implementation of any PR
systems requiring the ranking of all candidates would prove difficult, and
unpopular with voters.

-----Original Message-----
From: DEMOREP1 at aol.com [mailto:DEMOREP1 at aol.com]
Sent: Thursday, 5 October 2000 10:32
To: election-methods-list at eskimo.com
Subject: RE: Methods of elimination in quota preferential STV


A *democratic* legislative body exists only because the Electors (voters) 
cannot generally appear in person.

The legislators are the agents (i.e. representatives) of the Electors 
(voters).

The notion that a legislator must have only 1 vote in a legislative body is 
one of many political fictions.  A legislator should have a voting power in 
the legislative body equal to the number of voters that he/she represents.

The general case of choosing M winners among N choices in proportional 
representation elections is somewhat complex.

Example - elect 3 of 10

Vote YES on each acceptable choice and use number votes.

All of the combinations of the following would be done (i.e. an expanded
head 
to head matrix) using number votes (1, 2, etc.) ---

Any 3 test winners     versus 1 test loser       (6 other test losers)

A first choice vote for each of the other test losers gets transferred to 1 
of the 3 test winners or the test loser.

Obviously a test winner may or may not win in ALL of his/her matches.

Example

       YES  Place votes
     1    2     3     4    etc.

A  A1  A2  A3  A4  
B  B1  B2  B3  B4
C  C1  C2  C3  C4
D  D1  D2  D3  D4
E  E1  E2   E3  E4
etc.

Assume that D is a Condorcet winner (using both YES and NO votes).

One tiebreaker is to accumulate YES place votes to get the earliest Droop 
Quota (or the highest votes with truncated votes).

D2 (or D3, etc.) votes (YES or NO) would go to D (taking away first (second,

etc.) YES choice votes from the others).

Assume B gets a Droop Quota by summing the 1st to 3rd place B YES votes 
(reducing the first choice YES votes for the other candidates).

Assume E gets the highest YES votes (due to lots of truncated votes) (with D

and B getting more votes).

Another simpler (but somewhat inaccurate) tiebreaker would be simple IRV-- 
the candidate with the lowest number of first choice (plus transferred)
votes 
loses. Repeat until 3 candidates remain.

Another simpler tiebreaker would be to drop the choice with the lowest YES 
votes (repeatedly) and transfer the votes to the voter's next YES choice (if

any).

Votes from losers would transfer to an elected choice using the number votes

(whether or not a voter voted YES or NO for that elected choice).

Each winner should have a voting power equal to the final number of votes 
that he/she receives.

The complexity of the above is a reason why single winner math is only 
somewhat less complex.