[EM] [Fwd: IRV in action.....Improved IRV] (slightly corrected)
Chris Benham
chrisbenham at bigpond.com
Mon Apr 7 07:08:02 PDT 2003
-------- Original Message --------
Subject: IRV in action.....Improved IRV
Date: Sun, 06 Apr 2003 01:12:50 +0930
From: Chris Benham <chrisbenham at bigpond.com>
To: jgilmour at globalnet.co.uk, "election-methods-electorama.com"
<election-methods-electorama.com at electorama.com>
In response to a quite plausible, poltical spectrum-based 5 candidate
example of IRV performing poorly ,James Gilmour stated :
"I recognise the problem very well, but what is the practical solution?
I am very sympathetic to Condorcet, but there must be serious questions
about the public acceptability of some of the results it is likely to
produce."
This has prompted me to unveil my idea for Improved IRV (an
IRV-Condorcet hybrid), which I think he might like.
At each step where IRV eliminates the candidate with the lowest tally
of votes, this method instead eliminates, from the set of candidates
whose vote tallies are below average and also not above 25%, the
Condorcet loser . If there is a circular tie for this spot all the tied
candidates are eliminated.
Repeats of this step may result in the field being condensed to three
candidates, each with more than 25% (of the votes left in play). In that
case if any of the three pairwise beat both the others then that candidate
wins.If not then eliminate the candidate with the lowest tally.
The big weakness of IRV lies in how it decides which candidate to
eliminate. To me the normal IRV rule is too arbitary. The figure I
use of "above 25%" represents a majority of an "STV quota", and so
to me doesn't seem too arbitary.
I think that this proposed method should appeal to those who like IRV
and are repelled by the pure-Condorcet possibility that a candidate
with almost half of the first-preference votes could lose to a
candidate who gets no first-preference votes.
The example I referred to at the top:
10: FR > R > C > L > FL
10: R > FR > C > L > FL
15: R > C > FR > L > FL
16: C > R > L > FR > FL
15: C > L > R > FL > FR
13: L > C > FL > R > FR
11: L > FL > C > R > FR
10: FL > L > C > R > FR
With 5 candidates, the average number of votes is 20. So FL and FR
with 10 votes each make up the initial "set of candidates whose vote
tallies are below average and also not above 25%" , so they runoff : FR
d. FL 51-49, so FL is eliminated (and FL's 10 votes are transferred to L.)
New tallies are L: 34 C: 31 R: 25 FR: 10
With 4 candidates, average number of votes is 25%, so FR is only
candidate for elimination so is eliminated.(and FR's 10 votes are
transferred to R.)
New tallies are R: 35 L: 34 C: 31
All tallies are above 25%, so no more candidates for elimination.
C pairwise beats L 66-34, and C pairwise beats R 65-35 ; so C
wins.
Admittedly in this example the count was easier than it might have been,
because the set of candidates for elimination never had more than two
members.
Chris Benham
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://lists.electorama.com/pipermail/election-methods-electorama.com/attachments/20030407/40ebc1d1/attachment-0002.htm>
More information about the Election-Methods
mailing list