[EM] [Fwd: IRV in action.....Improved IRV] (slightly corrected)

[Chris Benham]

-------- 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








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