[EM] IRV variants

[Ted Stern]

On 05 Nov 2011 17:46:49 -0700, Forest Simmons wrote:
>
> Dear EM Folks, I’ve been very busy, while watching postings on the
> EM list out of the corner of my eye.  I was very happy to notice
> Mike Ossipoff’s interesting contributions.  In particular his
> promotion of a variant of IRV where equal rank counts whole strikes
> me as promising in the context of the “chicken problem.”
>
> On a related note I have been thinking about how to make a monotone
> variant of IRV.  Perhaps the two ideas could be combined without
> sacrificing all of the other nice features that IRV(=whole) seems to
> have.
>
> It is well known that certain kinds of elimination methods cannot
> satisfy the monotonicity criterion.  The basic variant of IRV is of
> that type, namely what I call the “restart with each step type” of
> elimination method.  This means that when one candidate is
> eliminated, the next stage starts all over again without learning
> from or memory of the eliminated candidate.  Range elimination
> methods that renormalize all of the ballots at each stage are of
> this type, too, since the renormalization is an attempt to eliminate
> the effect of the eliminated candidates on the remaining stages of
> the process.
>
> But some methods of elimination that do not suffer from the “restart
> problem” turn out to be monotone.  For example, approval elimination
> where the original approvals are kept throughout the whole
> elimination process; trivially the highest approval candidate is the
> last one left.
>
> 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?
>
> Thanks,
>
> Forest
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It strikes me as needlessly complex.

Equal Rank whole-vote IRV, with some kind of Borda or score based
elimination, would be much simpler.

However, any quota-based ranked system (and any form of single-winner
IRV has a quota of 50%) will be subject to one of the Woodall mono-*
failure modes.  So even if you manage to satisfy one of the
monotonicity criteria, you will fail participation in some respect.

Ted
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