Jonathan Lundell wrote:
On Apr 30, 2009, at 8:38 AM, Kristofer Munsterhjelm wrote:
Kathy Dopp wrote:
STV has *all* the same flaws as IRV but is even worse.
It is unimaginable how anyone could support any method for counting
votes that is so fundamentally unfair in its treatment of ballots and
produces such undesirable results.
The reason is very simple: the Droop Proportionality Criterion. The
DPC ensures that a group of voters greater than p times (the number of
voters)/(the number of seats + 1) can get p representatives on the
council.
As the number of seats increases, the actual result within each group
becomes less important, whereas that the DPC is held becomes more
important. Therefore, STV works well.
Other multiwinner methods fulfill the Droop Proportionality Criterion,
as well, but they're not very well known. Schulze's Schulze STV
(reduces to Schulze, which is Condorcet, when there's only one winner)
as well as QPQ also meet this criterion.
According to my tests, QPQ is better than STV, which in turn is better
than Condorcet modifications to STV. I haven't tested Schulze STV,
since it requires a lot of space for very large assemblies.
The precise scores are (lower is better):
Remind us, please, what your scores are.
Each candidate and voter gets assigned a number of binary opinions or
issues ("yes" or "no" for each). Each voter ranks the candidates so that
those that agree with him on more issues get ranked above those that
agree with him on fewer.
An "opinion profile" with regards to a subset of the electorate is
simply a vector of k fractions (for k issues): the first is what
proportion of the set say "yes" for the first issue, the second is the
same for the second, and so on.
Then a method's raw (un-normalized) score is just the RMSE of the
opinion profile of the council elected by that method (provided the
ballots consistent with the ranking mentioned before) and the opinion
profile of the people. The more like the people the council is, the
lesser the RMSE, and the better the score.
In order to remove randomization effects (perhaps some opinion
configurations are harder to fulfill than others), I make a bunch of
random councils. The best one (most proportional) gets assigned score
zero, while the worst one (least proportional) gets assigned score one.
The normalized score is just the unnormalized score normalized between
those two extreme points.
I do this lots of times (1000 times for the Schulze STV one) with
different council, voter, and opinion numbers, and then the mean is
simply the mean of the normalized scores, and the median is similarly
the median. Most truly proportional rules have mean scores below 0.25.
0.3 to 0.4 have semiproportional rules (D'Hondt without Lists), 0.5 and
above is very majoritarian.
I think the small state bias arises from that the various opinions are
completely uncorrelated. I'm not sure, though. For that matter, I don't
know if the small state bias is real, but I'm guessing so from that
IRV-as-multiwinner (n winners, just pick the n last eliminated) and SNTV
gets quite good scores.
A smaller simulation (only assemblies of few seats, so that Schulze
resolves within reasonable time) gives these results:
Mean Median Method name
----------------------------------------------
0.12374 0.01416 QPQ(Sainte-Laguë, multiround)
0.12754 0.02213 QPQ(Sainte-Laguë, sequential)
0.14783 0.0316 Schulze STV
0.15264 0.04725 STV
0.15984 0.05199 STV-ME (Plurality)
Note that these scores are lower. I think this is because I only checked
small councils (since otherwise, Schulze STV would take forever, being
exponential in the number of seats).
In the context of groups with solid support, this should mean "If more
people stop supporting group B, or switch their support from group A
to group B, then A should not get fewer seats in the assembly".
There's a typo here, right? Should be "switch their support from group B
from group A".
Yes. More accurately:
"If A gains more supporters (as a percentage of the population) than
does B, then A should not lose seats while B gains them or stays the
same". In the standard monotonicity case, B loses x while A gains that x.
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