[EM] Re: "Weighted Pairwise" proposal

[Chris Benham]

James,
On Thurs.Jun.10 you asked for comments on your version of Weighted 
Pairwise, which you had just amended to more
closely resemble normal Condorcet (Winning Votes).
You wrote:

>Here I am suggesting a new version of the weighted pairwise method which
>I proposed. The change I'm proposing is that defeats where the winning
>side constitutes a majority (of the valid vote) should always be
>considered to be stronger than a defeat where the winning side does not
>constitute an actual majority, regardless of the winning marginal
>utilities involved.
>
>To sum up:
>	If there are two pairwise defeats that are both by a majority, the one
>with the higher marginal utility is stronger.
>	If there are two pairwise defeats such that neither is by a majority, the
>one with the higher marginal utility is stronger.
>	If one pairwise defeat is by a majority and another isn't, then the
>defeat by a majority is necessarily stronger.
>  
>
><beginning of new definition>
>
>Ballots:
>1. Ranked ballot. Equal rankings are allowed.
>
>2. Ratings ballot. e.g. 0-100, whole numbers only. Equal ratings allowed.
>Note: You can give two candidates equal ratings while still giving them
>unequal rankings. However, if you give one candidate a higher rating than
>another, then you must also give the higher-rated candidate a higher
>ranking.
>
>Tally:
>1. Pairwise tally, using the ranked ballots only. Elect the Condorcet
>winner if one exists.
>If no Condorcet winner exists:
>
>2. Determine the direction of the defeats by using the ranked ballots for
>a pairwise comparison tally.
>
>3. Determine the strength of the defeats by finding the weighted magnitude
>as follows. Well say that the particular defeat were considering is
>candidate A beating candidate B. For each voter who ranks A over B, and
>*only* for voters who rank A over B, subtract their rating of B from their
>rating of A, to get the marginal utility. The sum of these winning
>marginal utilities is the total weighted magnitude of the defeat. (Note
>that voters who rank B over A, or rank them equally, do not contribute to
>the weighted magnitude; hence it is never negative.)
>
>4. Now that the directions of the pairwise defeats have been determined
>(in step 2) and the strength of the defeats have been determined (in step
>3), you can choose from a variety of Condorcet completion methods to
>determine the winner. Beatpath and ranked pairs are my preferred choices.
>
>5. There is one situation in which a defeat with lesser weighted magnitude
>is considered to be stronger than a defeat of greater weighted magnitude:
>If the winning side of one defeat constitutes a majority (of the valid
>vote), and the winning side of another defeat does not constitute a
>majority, then the majority defeat is necessarily considered to be
>stronger. Otherwise, the weighted magnitude is always the determining
>factor in relative defeat strength.
>
><end of new definition>
>

CB: You gave this example to explain the need for the amendment:

>45: A 100 > B 40 > C 0
>10: B 100 > A 90 > C 0
> 5: B 100 > C 90 > A 0
>40: C 100 > B 40 > A 0
>
>	If everyone votes sincerely, then B is a Condorcet winner, and wins in
>weighted pairwise. However, using the original weighted pairwise rules,
>the A voters can steal the election for A by truncating.
>
>altered votes:
>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
>
>	A now wins with my original proposal.
>
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.

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.

>Sincere Expectation Standard
>Given that a voter has no knowledge about how others will vote, a
>sincere vote must be at least as likely as any insincere vote to
>give results that are in some way better in the eyes of the voter.
>
>Or expressed as a more rigid criterion:
>-----
>Sincere Expectation Criterion (SEC)
>Consider a voter with a preference order between the possible
>outcomes of the election.  Let us call his sincere ballot, X.  Now,
>assuming that every possible legal ballot is equally likely for every
>other voter, there must be some justification for the vote X over any
>other way to fill out the ballot, which I will call Y.
>This justification is given by the following comparisons:
>
>The probability of X electing one of the voter's first choices vs.
>the probability of Y electing one of these choices
>
>The probability of X electing one of the voter's first or second
>choices vs. the probability of Y electing one of these.
>
>The probability of X electing one of the voter's first, second or
>third choices vs. the probability of Y electing one of these.
>... And so on through all the voter's choices
>
>X must either do better in one of these comparisons than Y, or equal
>in all.  Otherwise the sincere vote can not be justified.
>----
>In other words, there must be some justification for voting sincerely
>even if the voter does not know how any one else is voting.
>
>
>  
>
Winning-Votes fails this criterion because if your sincere preference
is A > B=C, it is more likely to be to your advantage to vote A > B > C or 
A > C > B.  It can back-fire, but the insincere vote is more likely
to get you what you want, so unless you have detailed knowledge about
how everyone else is voting, the insincere vote is better.

http://lists.electorama.com/pipermail/election-methods-electorama.com/1998-September/002082.html

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.


Chris Benham


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