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Warren Smith | 1 Sep 2005 01:59
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range voting on plurality machines

is actually quite pleasant.  Try the demo at
   http://math.temple.edu/~wds/crv/quickdemo/PresRadio.html

(and let me know if there are any browser problems. I have heard a rumor
there is a bug in early versions of netscape which may cause this
demo to be unattractive...)
wds
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Warren Smith | 1 Sep 2005 02:05
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FBC for Condorcet

do you have a counterexample showing favorite betrayal for 
Condorcet (winning vote, equality-rankings permitted, partial orders not
permitted)
?

(And if you do not, isn't that a good reason NOT to prefer DMC?)            
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Warren Smith | 1 Sep 2005 02:51
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favorite betrayal in Condorcet(wv, =permitted, no partial votes)


The original general-purpose 19-voter FBC example from the Center for Range Voting web page
http://math.temple.edu/~wds/crv/IncentToExagg.html:
8:B>C>A
6:C>A>B
5:A>B>C
B wins under Condorcet Voting [Ranked Pairs variant, winning votes,
equality-ranking permitted] according to Eric Gorr's calculator at
http://www.ericgorr.net/condorcet/.  (Incidentally, I think this is also
the same as what has been called "Steve Eppley's MAM method.")

Modification I: In original, 3 of the 6 C>A>B voters switch to A>C>B:
8:B>C>A
3:C>A>B
3:A>C>B
5:A>B>C
New result: AB tie.  This is an improvement from the switched voters' point of view.

Modification II: In original, 3 of the 6 C>A>B voters switch to A=C>B:
8:B>C>A
3:C>A>B
3:A=C>B
5:A>B>C
B wins.  (This is no change.)

CONCLUSION: these 3 voters, by "betraying" their favorite third-party candidate C
so that they could strategically exaggerate the major-party candidates A and B
to "top and bottom", with C "middle", caused the election result to improve.
But if they had only partially betrayed C by voting A=C>B, then that would not
have provided enough power to change the election result.  A full-power betrayal
(Continue reading)

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Eric Gorr | 1 Sep 2005 03:02
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Re: favorite betrayal in Condorcet(wv, =permitted, no partial votes)

> B wins under Condorcet Voting [Ranked Pairs variant, winning votes,
> equality-ranking permitted] according to Eric Gorr's calculator at
> http://www.ericgorr.net/condorcet/.  (Incidentally, I think this is also
> the same as what has been called "Steve Eppley's MAM method.")

It is not.

A MAM calculator can be found at:

http://www.duniho.com/remote-mamcalc.php

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Warren Smith | 1 Sep 2005 04:03
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my previous post about FBC and condorcet

seems to depend on having an exact tie.  Therefore, this
counterexample is not as impressive as I thought.
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Abd ulRahman Lomax | 1 Sep 2005 04:04

Problems with Range and suggested solutions

It came back to me yesterday what I -- and some others -- have seen 
as the biggest problem with Range voting, and also a solution that I 
think I also expressed somewhere. But I tend to write way too much 
and sometimes I think good ideas have been buried in fluff.

Range rewards those who exaggerate, it may weaken the vote of those 
who do not use the full range of ratings.

Consider two candidates within an election, A and B, and two voters.

Voter 1 rates A and B as 99 and 20. One might analyze this vote as 
indicating that the voter clearly approves of A and considers B as 
possibly better than Genghis Khan.

Voter 2 rates A and B as 0 and 99.

In Range, voter 2's ratings outweigh those of voter 1.

This is essentially a kind of violation of one-person, one-vote.

It is fixable by using granularity-2 range, i.e. Approval, or 
partially by normalizing the votes. I think basic normalization 
should be done regardless in Range, that is, the maximum vote cast 
should be normalized to 1 and the minimum vote cast to 0. However, in 
an election with more than two candidates, this would not solve the 
problem, because there might be a third candidate who was truly 
awful, and Voter 1 might rate that candidate as zero, leaving the 
same problem in the pairwise race between A and B.

So to go the distance, I'd suggest that Range ballots be analyzed
(Continue reading)

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Abd ulRahman Lomax | 1 Sep 2005 03:31

Duverger's "law"

At 04:54 PM 8/31/2005, Warren Smith wrote:
>For
>evidence about Duverger law see
>    http://math.temple.edu/~wds/crv/Duverger.html
>and if anybody wants to supply me with some more datapoints (e.g. 
>from 2002-2005 elections)
>then I can add them to the picture there.

Duverger's law isn't a law. From wikipedia:
While there are indeed many FPTP systems with two parties, there are 
significant 
<http://en.wikipedia.org/wiki/Counterexample>counterexamples: 
<http://en.wikipedia.org/wiki/Scotland>Scotland has had until 
recently first-past-the-post and similar systems but has seen the 
development of several significant competing political parties. Many 
commentators regard the 
<http://en.wikipedia.org/wiki/United_Kingdom>United Kingdom's 
<http://en.wikipedia.org/wiki/Liberal_Democrats_%28UK%29>Liberal 
Democrat party, since the 2005 
<http://en.wikipedia.org/wiki/General_Election>General Election, as 
forming a 'third party' and creating a three-party system. 
<http://en.wikipedia.org/wiki/Canada>Canada and 
<http://en.wikipedia.org/wiki/India>India have multiple regional 
parties. Duverger himself did not regard his principle as absolute: 
instead he suggested that first-past-the-post would act to delay the 
emergence of a new political force, and would accelerate the 
elimination of a weakening force - 
<http://en.wikipedia.org/wiki/Proportional_representation>proportional 
representation would have the opposite effect.
(Continue reading)

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Abd ulRahman Lomax | 1 Sep 2005 03:41

wiki

At 06:12 PM 8/31/2005, Jobst Heitzig wrote:
> > Re your DFC wiki page, I hate it.
>
>Thanks again. You're a very emotional man it seems. I hope you have your
>pills in reach since I heard that hating is not very healthy.

Warren, it seems to me, doesn't actually "hate," the kind of hate 
that would indeed be unhealthy in the way suggested. He is, however, 
often impolitic with his criticism.

> > I also think I want there to be a concise clear and unambiguous 
> description of DFC,
> > and that wiki isn't it.
>
>Feel free to assemble one. By the way, I told you the "concise clear and
>unambiguous description" you demand in one of my last emails after you
>had mentioned DFC. Actually, the randomness is deliberately included in
>the design since (i) no deterministic method will provide for group
>strategy equilibria when there is no Condorcet Winner in the usual sense
>- only randomized methods can do so, and (ii) it will provide for some
>kind of "proportional representation" in the long run over a number of
>DFC decisions.

I'll disagree that "only randomized methods can do so," since there 
are other alternatives that are neither deterministic or randomized, 
beginning with the simple one of holding some kind of runoff.

Perhaps someone could take Mr. Heitzig's mail, with its concise 
definition, and use it to clean up the wiki page. I'd do it if I had time.
(Continue reading)

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Adam Tarr | 1 Sep 2005 05:02
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Re: favorite betrayal in Condorcet(wv, =permitted, no partial votes)

Two responses:

1)  As I and others have said before, it is absurd that you attempt to draw conclusions about the long-term effects of an entire election method from one example involving a three-way tie.

2)  The reason this example does what it does is because, in the case of an EXACT tie, the tiebreaker considers the margin of defeat.  So yes, you can get margins-condorcet-esqe incentives to bury your favorite in the case of an EXACT tie.  In a real, public election, this is utterly irrelevant since the probability of an exact tie is vanishingly close to zero.

-Adam


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Chris Benham | 1 Sep 2005 06:19

medians and Heitzig's approval-voting strategy

Warren,

>--aha.  So by "median candidate" you do not mean what I thought you meant
>(namely, in an N-canddt election, the top-quality floor(N/2) are above median)
>but rather median in the prior distribution of probabilities of winning.
>
>But wait, that would be even more insane, since the policy of
>voting only for the candidates with above-median prior election 
>probability, would be a policy that would completely disregard the
>quality of the candidates.
>
My understanding of Weinstein's approval strategy is this:
"Approve your favourite (or equal favourites). If the remaining (so far 
unapproved) candidates are on more
than one of your preference-levels, then approve the candidate/s on your 
next-from-the-top  preference-level if
you consider that the probability that one of the candidates you prefer 
less than this/these candidate/s  will win
is greater than the probability that one of the candidates you prefer 
more will win.  And so on."

This strategy seems sane to me, and probably right for voters who only 
have a ranking.

Chris Benham
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Gmane