[EM] New Condorcet/RP variant
ciphergoth at gmail.com
Mon Nov 8 02:04:17 PST 2004
On Fri, 05 Nov 2004 11:30:05 -0800, Steve Eppley
<seppley at alumni.caltech.edu> wrote:
> Paul C asked:
> > Does Eppley still read this list?
> Yes, sometimes. By the way, I prefer that my friends
> call me Steve.
Steve it is.
> I chose the tiebreaker for complete satisfaction of clone
> independence, monotonicity and strong Pareto.
My current method is definitely not clone independent. But that
raises a question
Is clone independence strictly more important than determinism?
I've defined a variant of my method which is strictly more
deterministic than MAM and retains clone independence (and all the
properties I've already proven for my original method, like
monotonicity and strict Pareto). The variant is also strictly less
deterministic than my original proposal. By "strictly more
deterministic" I mean that the deterministic ballot sets for my
variant are a strict superset of MAM's deterministic ballot sets,
where a deterministic ballot set is a ballot set where the voting
method chooses a particular ranking with probability 1. I'll describe
the variant in a later post if people are interested; what I'm
interested in at the moment is whether people would consider it to be
an improvement given the properties listed here.
Where the variant is deterministic, it always makes the same decision
as my original method. But where it is nondeterministic, my original
method would have sacrificed clone independence to deterministically
choose one of the winners
Consider these ballots
10 A > B1> B2
10 B1> B2 > C
10 C > A
where we consider B1, B2 to be clones. With the variant, A and C have
a 1/3 chance of winning, just as if the clones are not present. The
original treats the clones as "extra information" used to resolve
ties, and declares A > B1 > B2 > C the winner.
Is that desirable behaviour? Is it thought likely that supporters of
A would anticipate the tied circular ranking, and strategically
nominate a clone of B to break the circular rank, and fool voters into
ranking B1 and B2 together even though B2 is really sponsored by an A
supporter? Or is it better, where there would otherwise be a
nondeterministically resolved tie, to say that if A does better than
two alternatives and both alternatives do better than C then that's
enough information to resolve the tie in favour of A > B1 > B2 > C?
\/ o\ Paul Crowley
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