Forest Simmons wrote (5 Nov 2011):
Now here’s what I propose for an IRV variant:
1. Use the ranked ballots to find the pairwise win/loss/tie matrix M.
This matrix stays the same
throughout the process.
2.Initialize a variable U (for Underdog) with the name of the
candidate ranked first on the fewest number
of ballots, and eliminate U from the ballots.
3.While more than one candidate remains, eliminate candidate X that is
ranked first on the fewest
number of ballots after the previously eliminated candidates’ names
have been wiped from the ballots (as
in IRV elimination) and then replace U with X, unless U defeats X, in
which case leave the value of U
unchanged.
4. Elect the pairwise winner between the last value of U and the
remaining candidate.
Note that a simplified version of this where you just eliminate the
pairwise loser of the two candidates
ranked first on the fewest number of ballots in NOT monotone. We have
to remember the previous
survivor and carry him/her along as "underdog challenger" to make this
method monotone.
Note also that this method satisfies the Condorcet Criterion. So we
gain monotonicity and CC, but what
desireable criteria do we lose? It still works great on the scenario
49 C
27 A>B
24 B
Candidate A starts out as underdog, survives B, and is beaten by C, so
C wins. But if B supporters
really prefer A to C they can make A win. On the other hand if the A
supporters believe that the B
supporters are indifferent between A abd C, they can vote A=B, so that
B wins.
When I have more time, I'll sketch a proof of the monotonicity.
Comments?
Yes. I don't get it. (I am confused by your explanation of the algorithm).
How do you think this is better than your latest version of Enhanced DMC?
I think a good method is the IRV-Condorcet hybrid that differs from IRV
only by before any and each elimination
checks for an uneliminated candidate X that pairwise beats all the other
uneliminated candidates and elects the
first such X to appear.
That of course gains Condorcet, and it keeps IRV's Mutual Dominant Third
Burial Resistance property.
So if a candidate X pairwise beats all the other candidates and is
ranked above all the other candidates on more than
a third of the ballots then (as with IRV) X must win and a rival
candidate Y's supporters can't get Y elected (assuming
they can somehow change their ballots) by Burying X.
Does your method share that property?
49 C
27 A>B
24 B
Candidate A starts out as underdog, survives B, and is beaten by C, so
C wins.
From what I think I do understand of your algorithm description,
doesn't candidate B start out as "underdog"?
Chris Benham
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