[EM] Re: "Weighted Pairwise" proposal
James Green-Armytage
jarmyta at antioch-college.edu
Tue Aug 31 23:13:47 PDT 2004
>CB: I just point out that the problem you are addressing doesn't seem to
>exist in the "Automated-Approval Margins"
>(AAM) method.
>AAM eliminates non-members of the Schwartz set, and then each ballot
>approves (of the remaining candidates) those
>candidates they rate above average (and half-approves those they rate
>exactly average). The rankings are used to determine
>the results ("directions") of the pairwise comparisons, but the margins
>between the (derived) approval scores are used to
>measure their "strengths". On this basis pick the winner using Ranked
>Pairs.
>In second part of your example above (in which the A voters are
>insincerely truncating) the candidates thus derived
>("automated") approval scores are A55, B55, C45. Using these scores to
>rank the pairwise comparisons, we get
>A>C 55-45 = +10, B>A 55-55 = 0, C>B 45-55 = -10; giving the final order
>B>A>C. The sincere CW wins.
I don't understand how B got an automated approval score of 55. When I
did it, I got a score of 15 instead of 55. Here are the ballots again:
>
45: A 100 > B 0 = C 0
10: B 100 > A 90 > C 0
5: B 100 > C 90 > A 0
40: C 100 > B 40 > A 0
>
When you say "rate above average", what exactly do you mean? By
"average", do you mean the average of all the possible ratings, i.e. 50?
Do you mean the arithmetic mean of all the candidates marked on that
ballot? Or do you mean the median of all the candidates marked on the
ballot. I'm guessing it's the arithmetic mean of the candidates. If so,
let's take one of the people who voted C100>B40>A0. The arithmetic mean of
the candidates on this ballot is 46.67. So, only candidate C is rated
above the mean.
Therefore, I don't know where you got a score of 55 for B. The 45 ABC
voters don't contribute to his score, the 10 BAC voters and 5 BCA voters
do, and the 40 CBA voters, I think, do not.
Anyway, if I'm right about the example, then I think this invalidates
what you said about AAM being immune to this sort of problem / situation.
>Weighted Pairwise seems to needlessly have the same problem as plain
>ranked-ballot WV Condorcet, that is zero-information
>random-fill incentive and therfore not complying with Blake Cretney's
>excellent "Sincere Expectation" Standard/Criterion.
Chris, thanks for this food for thought!
The random fill incentive in winning-votes is an interesting issue. I
think that when you say it applies to weighted pairwise, you specifically
mean a version of weighted pairwise which contains the provision that a
majority beat is necessarily stronger than a minority beat.
That provision, and the maximization-in-scale provision, are optional,
i.e. not essential to the method. Maybe they both have small pluses and
minuses. The majority/minority thing might give a small random fill
incentive... and the maximization thing might make the method slightly
non-monotonic.
(By the way, I suspect that AAM is probably non-monotonic for the same
reason (having an extra candidate in the Schwartz set may be beneficial
for my candidate, even if the extra candidate is only in the Schwartz set
because he beats my candidate, because it affects the ratings scores in
general), but I don't have a non-monotonicity for AAM or for maximization
in scale yet, so I could be wrong on both counts.)
Not that monotonicity is necessarily so important, but you know, it's
nice to have.
Anyway, I hope that you are satisfied that the only random fill incentive
in weighted pairwise would come from the majority/minority criterion. Let
me reason out why I think it wouldn't come into play otherwise:
First of all, random-filling can only help you if there is a top cycle.
When there is a top cycle in weighted pairwise, the ratings are the
operative determinant of defeat strength.
Let say that my sincere rankings are A>B=C. It seems pretty clear that in
most cases I would want to rate A at 100, while rating B and C at 0. This
way, I maximize the strength of the only two preferences I care about,
namely A>B and A>C. It seems like it would usually be foolish to waste
some any of my ratings range on a B>C preference or a C>B preference,
since I genuinely don't care who wins there.
>
>CB: This is a weaker version of No Zero-Information Strategy, which says
>that with no knowledge (or guess) about how any
>other voters are voting, regardless of how the voter rates the
>candidates the voter's best "strategy" is give a full sincere ranking.
>This standard is met by AAM.
>
Can you prove this? I'm interested to see what a proof of
0-info-0-strategy would look like.
my best,
James
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