[EM] IRV variants

[Jameson Quinn]

Forest: I think your system (Bubble IRV, in the sense of bubble sort?)
would have some good properties in terms of results. But honestly, I don't
really see the point. We have a number of systems which give good results.
To me, the point of designing new systems is to give good results while
gaining some combination of (in roughly descending importance)

   - Simplicity of explanation
   - Simplicity of counting (summability, etc.)
   - Simplicity of voting (ballot design, minimal strategy considerations,
   etc.)
   - Broad appeal (For instance, a method that would appeal to both IRV and
   Condorcet suppporters)
   - New, unexplored mathematical properties ("just for fun")

I think your Bubble IRV doesn't really give any of the first three, and not
too much of the last two.

It does, however, inspire me to present my own proposal, a simple
modification of David's IRV3/AV3. My only change from his system is that
equal ranking would be allowed. In the IRV part, equal-ranked-top would be
counted as a full first-place vote for both (all) of the top candidates.

What does this do to summability (one of the main advantages of David's
proposal over IRV)? It actually works fine. You'd keep three tallies: top
ranks, approvals, and a Condorcet matrix. Since with three candidates
there's only one true IRV elimination, equal-ranking doesn't cause the
logistical headaches it would with full IRV.

What does it do to simplicity of explanation? In my opinion, it's at least
as good as its predecessor IRV3/AV3. "Keep the three top approvals; discard
the one with fewest top ranks; and use preferences to see who'd win between
the remaining two." In a certain sense, that's actually simpler than even
giving a full explanation of IRV.

What does it do to simplicity of voting? It is much better for ballot
design. With equal-ranking allowed, you'd simply eliminate the problem of
spoiled ballots, which seems to me to be a real concern for IRV. And
allowing equality makes it possible to vote a ratings-style ballot, which
is cognitively easier.

How about strategy? Additional votes at level 2 or 3 would essentially be a
way to make the approval part of this method freer, without the arbitrary
limit of three approvals. Additional votes at top level would be a way for
a solid majority coalition to ensure that their principal enemy is
eliminated. The chicken dilemma still applies, so it would rarely (never?)
be "strategically optimal" from a first-order perspective to vote two
candidates at top ranking; but on the whole, I think it's good to allow
voters the option to explicitly say that they don't care about such
first-order strategic considerations between two candidates they consider
to be clones.

In terms of results, I think this system would tend to give the IRV winner,
with perhaps a small step in the direction of MJ. That's not my favorite
place to be; it still allows for self-perpetuating two-party domination.
However, I think that there's a real possibility that this system would
allow for smoother transitions if the set of top two parties changed at a
local level. So ... well, perhaps it's my "I invented it" bias, but I think
it's good enough.

Thoughts?

Jameson

ps. One more minor comment on Forest's Bubble IRV proposal, below:

2011/11/5 <fsimmons at pcc.edu>

> 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.


Wouldn't B be the underdog initially here? (Not that it matters to the
result or to the further analysis below.)


> 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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