Equal defeats in RP(m)

[Blake Cretney]

On Thu, 03 May 2001 00:27:09 -0000
"MIKE OSSIPOFF" <nkklrp at hotmail.com> wrote:

> I'd asked:
> 
> >Someone has designated Ranked Pairs(margins), and so it's necessary
> >to ask for a complete definition of it. In particular, how exactly
> >does it deal with equal defeats, in all the kinds of situations
> >in which they can occur and in all the procedural questions they
can
> >raise in an RP(m) count?
> >
> >Obvioiusly it would be inappropriate for that definition to be
> >supplied after the ballots are posted.
> 
> Blake replied:
> 
> I've been assuming that no specific tie-breaker is proposed for the
> purposes of this election.  Unless I'm mistaken, that is the case
for
> most of the methods suggested.  For example, IRV, plurality,
Approval,
> etc.
> 
> I reply:
> 
> If someone designated IRV, we'd have to ask how to deal with
> equally lowest candidates. Ties in Approval? You apparently missed
it,
> but we discussed what to do when a designated method returns a tie.
> 
> It's the internal ties during a method's procedure for which
solutions
> should be specified before the ballots are posted.

Oh, I thought you were just talking about RP(m) as a proposal.  I
wasn't considering the whole voter's choice issue.

To be fair, the method should be specified by someone who hasn't seen
the results (as you mention), and so, unless there are any objections,
I'm going to specify the method to be used to designate a RP(m) winner
for voters choice, when there are ties.  It's the same one Zavist and
Tideman gave, and which I mentioned earlier.

In either the locking and skipping definition, or the definition from
my signature, the problem is solved if you have some way to order the
victories of equal margin.  So, you pull out one ballot at random. 
You randomly resolve any ties in ranking on this ballot.  Now you have
a complete Tie Breaking Ranking of the Candidates, which is called a
TBRC.

Let's say you have two different victories, like
A vs. B and C vs. D
this involves the candidates A, B, C, and D.  One of these candidates
is ranked highest on the TBRC.  Whichever victory involves that
candidate is considered higher after the tiebreaker.  The same
candidate can appear in both victories, as long as it isn't highest. 
For example, given the TBRC A>B>C>D>E, C>B would beat D>E, since
B is ranked highest on the TBRC.

But it might be, that the highest candidate in both victories is
the same candidate.  Not surprisingly, you look at the
candidates the victories don't share, and see which is higher of
those.  That victory is considered higher after the TBRC.  For
example, with the tiebreaker A>B>C>D>E, C>D would beat E>C, since D is
higher than E in the TBRC.

Next, you consider pairwise ties.  Score those as a victory for the
higher ranked over the lower ranked candidate.  So, with tiebreaker
A>B>C>D>E, B=E would cause B to be ranked over E.  Once there are no
pairwise ties, and no victories are considered equal, the normal
definitions give a result.

This procedure meets my Pairwise Dominance Criterion, which states
that if X pairwise beats Y, and X does at least as well in Y in every
other pairwise contest, whether measured by votes for or against, that
Y should not win.

p(X,Y) is the number of voters who rank X over Y.  
For all X, if exists Y s.t. (for all Z, p(Y,Z)>=p(X,Z) and
p(Z,Y)<=p(Z,X)) and p(Y,X)>p(X,Y)
then  X must not win

---------------------------------------------------------------------
Blake Cretney   http://www.fortunecity.com/meltingpot/harrow/124/path

Ranked Pairs gives the ranking of the options that always reflects 
the majority preference between any two options, except in order to
reflect majority preferences with greater margins. 
(B. T. Zavist & T. Tideman, "Complete independence  of clones in the 
ranked pairs rule", Social choice and welfare, vol 6, 167-173, 1989)