[EM] Re: Testing 1 2 3
Dgamble997 at aol.com
Dgamble997 at aol.com
Tue Jan 13 14:56:05 PST 2004
Kevin Venzke wrote in part:
>> 2/ Lack of real world comparison data. I adjusted the votegenerator part
of
>> the spreadsheet as a result of looking at the data sets it was producing
and
>> comparing them with the South Australia state election of 1997 ( 3
parties-
>> Liberal/National, Labor and Australian Democrat made a reasonable showing
) >and
>> the Queensland state election of 1998 ( Liberal/National, Labor and One
Nation).
>> The distribution of second preferences in the votegenerator data sets was
>> much more variable than those in real elections and votegenerator was
adjusted >as
>> a result of this.
>This sounds like you're forcing your generator to assume a one-dimensional
>spectrum. If it were me, I wouldn't do that, because it limits the
applicability
>of any conclusions you draw.
No, I think you misunderstand. If for the set of 1st Preference votes:
A = 40 B = 25 C = 35
You give the following second preference distributions:
A= 0 A>B= 100 A>C = 0 B = 0 B>A= 50 B>C=50 C=0 C>A=0 C>B= 100
you are modelling a completely one dimensional spectrum.
This is a distribution of second preferences based on the South Australian
state election of 1997:
A>B= 90 A>C= 10 B>A= 39 B>C= 61 C>A= 16 C>B= 84
A is Liberal/National B is Australian Democrat C is Labor
A second preference distribution where voters were voting on more than one
dimension would possibly look something like this:
A= 20 A>B= 40 A>C=40 B= 20 B>A= 50 B>C= 30 C=50 C>A= 10 C>B= 40
With the original votegenerator there was a great deal more random variation
away from the mean values than I found in real Australian elections. For
example in the SA 1997 state election the Australian Democrat 2nd preference split
varied from 26% Liberal: 74% Labor to 48% Liberal: 52% Labor. With the
original votegenerator I was getting variations of levels of for example party A
second preference vote from 10%B:90%C to 90%B:10%C- too wide a variation.
>Off the top of my head, I might use some crude method to find two
front-runners
>from the ballots, and then form the Approval ballots as though every voter
>believes that only those two candidates have a chance at winning. Given
voter
>utilities for each candidate, optimal Approval ballots can be found very
>easily.
The strategic voter situation is not really that difficult all voters use the
strategy of "vote for every candidate you prefer to the front runner, if you
prefer the second place candidate to the front runner vote for that candidate
as well" based on a single,accurate and sincere Approval poll. 99% of the
time, by following this strategy the Condorcet winner is elected if there is one.
The non-strategic situation is more difficult, it is currently based on the
idea that the voters like everybody they rank and therefore approve everybody
they rank.
>You don't have to produce the sets, just the matrices. Open up "mspaint"
and
>plot four points, ABCD, and draw different colored arrows (representing
defeats
>of different strengths) among them, and see what winners are picked by RP, >
Schulze,
>Raynaud, etc. It's interesting, I swear.
>Also, cycles are not so rare if voters give a short ranking. Take the
favorite:
>49 A
>24 B
>27 C>B
I actually took 49 A, 24 B and 27 C>B put it into the model and ran it 4
times and then compared the result under RP winning votes and RP winning margins.
You are right and I was wrong, it does make a difference. Of the 200 seats
there were Condorcet cycles in 40 of them. Under winning votes B won all 40 of
the cycle seats, under winning margins A won 13 of the seats, B won 20 of them
and C won 7.
The input preference sets were 49 A, 24 B and 27 C>B.
The output overall vote was 47.73 A, 25.15 B and 27.13 C>B.
The results under Ranked Pairs winning margins were A 105, B 70 and C 25.
The results under Ranked Pairs winning votes were A 92, B 90 and C 18.
The Approval result with all voters using strategy was identical ( for some
good mathematical reason) to the winning votes result.
David Gamble
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