<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html;charset=ISO-8859-1">
<title></title>
</head>
<body>
<meta http-equiv="Content-Type" content="text/html;charset=ISO-8859-1">
<title></title>
<meta http-equiv="Content-Type" content="text/html;charset=ISO-8859-1">
<title></title>
<title></title>
<tt>Previously,on Tuesday July 8, 2003 as part of
the "Proportional Representation</tt> beyond STV?" thread,<br>
I wrote:<br>
It seems to me that STV is impeccable in principle, EXCEPT for the horrible,
arbitary feature it has in common with IRV of eliminating the candidate
who happens to have the lowest total at the time it is convenvenient to eliminate
a candidate. So why not eliminate the Condorcet loser instead?<br>
I then went on to make the mistake of suggesting that ALL the votes should
count equally in determining the CL among the candidates without a quota,
instead of only those votes and fractions of votes not tied up in already
made quotas.<tt> So this is the corrected version: Ranked ballots, equal
preferences ok, truncation ok,truncation treated just like equal prefernces.
Elect those candidates with a Droop quota (1/number of seats to be filled
+ 1) of first preferences (equal first preferences count as fractions). Distribute
the surplus preferences ("the overflow") of those candidates all-at-once</tt>
as fractions, elect any who now have a quota and in the same way distribute
the new surpluses and so on. If at the end of this process there are still
seats unfilled, NOT counting those votes and fractions of votes that make
up the quotas of those already elected candidates, determine among the remaining
candidates the Condorcet Loser and eliminate that candidate. Distribute that
candidate's preferences, elect any who have a Droop quota, distribute any
surplus and so on until all the seats/positions have been filled.<br>
This from James Green-Armytage which showed up my error:<br>
<pre>Example #1
100 votes
4 seats
Quota: 20 (Droop)
20: A, B, C, D, E
10: B, A, C, D, E
10: B, C, A, D, E
10: C, B, D, A, E
10: C, D, B, E, A
9: D, C, E, B, A
9: D, E, C, B, A
19: E, D, C, B, A
3: F
"The outcome using regular STV is ABCE. That is also the outcome using
CPO-STV, or sequential STV. I hope that you can agree that this seems like
the most fair outcome. The outcome using the method you suggest, however,
is ABCD.
The imaginary political reality behind this example is a simple spectrum
from A to B, where voters vote for candidates nearest to their first
choices on the spectrum before they vote for candidates who are further
away from their first choices. F is kind of an outlier in his own world.
The F voters don't give a damn about any of the other candidates.
Anyway, the difference between the two outcomes is whether D or E gets
elected. Plain STV, CPO-STV, and sequential STV favor E over D because
s/he quite simply has more votes. D, however, is closer to the center, and
so is favored by the method you suggest."
My corrected version also elects ABCE.
Here is an example taken from the webpage about "sequential STV"
<a class="moz-txt-link-abbreviated"
href="http://www.electoral-reform.org.uk/publications/votingmatters/15P4.htm">www.electoral-reform.org.uk/publications/votingmatters/15P4.htm</a>
</pre>
<span class="bodytext">
<pre> 104 AEBCD
103 BECDA
102 CEDBA
101 DEBCA
3 EABCD
3 EBCDA
3 ECDBA
3 EDCBA
Elect 2.
Two seats to be filled, 422 votes so Droop quota is 422/3 = 140.666666666.
No candidate has a quota,so we eliminate the Condorcet Loser.
A<E 104-318, A<B 107-315, A<C 107-315, A<D 107-315 so A is CL and is eliminated.
This only results in E's tally rising from 12 to 116, still short of a quota.
D<E 101-321, D<B 209-213, D<C 104-318 so D is the new CL and is eliminated.
Now the ballots reduce to:
211 EBC
6 ECB
103 BEC
102 CEB
Now E has a tally of 217, a quota plus a surplus of 76.33333333. E is elected.
The surplus is distributed thus: B gains 76.33333 x 211/217 = 74.222732, and
C gains 76.33333 x 6/217 = 2.110599.
This gives
E: 140.66666 (a Droop quota)
B: 177.22273
C: 104.11059
Now B has a quota, so the final result is EB.
For the time being I leave open the question of which Condorcet (loser)tie-breaker
to use. I can't see any reason why this method shouln't be monotonic.
Chris Benham
</pre>
</span><br>
</body>
</html>