[EM] IRV variants

[Jameson Quinn]

I just realized I didn't give my IRV3/AV3 variant system a name. I think
I'll call it 3-2-1 voting, because it is a pretty natural way (in my mind
at least) to eliminate down in that fashion.

2011/11/6 Jameson Quinn <jameson.quinn at gmail.com>

> 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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>> info
>>
>
>
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