On May 9, 2010, at 11:22 PM, Dave Ketchum wrote:
On May 9, 2010, at 2:24 PM, Juho wrote:
On May 9, 2010, at 6:49 PM, Peter Zbornik wrote:
an alternative way to count wins against equally ranked candidates,
would be to give both 0.5 wins per vote and not 0 as I did in my
previous mail.
Thus the current state: A>B gives 1:0. A=B gives ?:? (A=B is only
allowed at the end of the ballot)
The proposal above: A>B gives 2:0. A=B gives 1:1 (linear
transformation from A>B - 1:0 and A=B - 0.5:0.5).
The soccer proposal A>B gives 3:0, A=B gives 1:1 (linear
transformation from A>B - 1:0 and A=B -1/3:1/3. This proposal
would (as in soccer) encourage voters to differenciate between
candidates, and candidates to try to gain more support than the
other candidates. The soccer ranking is purely speculative though.
All classical Condorcet methods can handle equal rankings and their
impact has been analyzed quite well.
Usually the discussion focuses on how to measure the strength of
the pairwise preferences. This is the next step after the matrix
has been populated. The two most common approaches are margins and
winning votes.
- Let's define AB = number of votes that rank A above B
- In margins the strength of pairwise comparison of a A against B
is: AB - BA
- In winning votes the strength of pairwise comparison of a A
against B is: AB if AB>BA and 0 otherwise
- Note that in margins ties could be measured either as 0:0 or as
0.5:0.5 since the strength of the pairwise comparison will stay the
same in both approaches
The most common (/classical) Condorcet methods give always the same
winner if there are three candidates and all votes are fully
ranked. If there is no Condorcet winner then the candidate with
smallest defeat will win. The strengths of the defeats (and
therefore also the end result) may differ in margins and winning
votes if there are equal rankings. These differences (and their
naturalness with sincere vote and their impact on strategies) have
been discussed a lot. I will not analyze the differences in detail
here. In margins victory 50:40 has the same strength as victory
10:0. In winning votes victory 50:40 has the same strength as
victory 50:0.
Juho
"smallest defeat"? Oops - I will quote from wikipedia:
Ranked Pairs and Schulze are procedurally in some sense opposite
approaches (although they very frequently give the same results):
Ranked Pairs (and its variants) starts with the strongest defeats
and uses as much information as it can without creating ambiguity.
Schulze repeatedly removes the weakest defeat until ambiguity is
removed.
I think that should lead to the same conclusion. In a three candidate
cycle there are three cyclic defeats. Ranked Pairs should keep the two
strongest ones. Schulze should eliminate the weakest one and keep the
remaining two.
The philosophy of Minmax is a bit different since it does not aim at
approving or ignoring defeats (until there is a transitive order) but
just finds the candidate that has smallest opposition against her.
Juho
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