[EM] Unmanipulable majority and Condorcet

[Forest Simmons]

On Fri, Jan 3, 2020 at 1:02 PM <
election-methods-request at lists.electorama.com> wrote:

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> Kristofer,


So it seems like UM and Condorcet are almost but not quite compatible ...
tantalizingly close but "no banana" as they say.

It reminds me of how tntalizingly close we can get to a method that
satisfies both the CC and the FBC:

Consider the method Majority Defeat Disqualification Top (symmetric
completion). This method satisfies the CC but not the FBC.

 Now change it so that the symmetric completion does not apply to equal
top. After this change it becomes MDDT(symmetric completion below top) and
no longer satisfies the CC but does satisfy the FBC.

Any Insights?

Forest

>
> Message: 2
> Date: Fri, 3 Jan 2020 08:53:12 +0100
> From: Kristofer Munsterhjelm <km_elmet at t-online.de>
> To: EM <election-methods at lists.electorama.com>
> Subject: [EM] Unmanipulable majority and Condorcet
> Message-ID: <0d74bf5d-fd16-852b-3f3d-18438b2bc65b at t-online.de>
> Content-Type: text/plain; charset=utf-8; format=flowed
>
> A long time ago, I said that my fpA-fpC three-candidate method passed UM
> and Condorcet. I've now found out that that's not the case.
>
> Consider this election:
>
> 2: A>B>C
> 2+x: B>A>C
> 1-x: B>C>A
> 3: C>A>B
>
> With x=1, A is the CW; with x=0, we have a tie-cycle (A>B, B>C, C=A).
> With an x slightly less than zero, the election is a proper ABCA cycle.
>
> The candidate scores are:
>         A: fpA - fpC = 2 - 3 = -1
>         B: fpB - fpA = 3 - 2 = 1
>         C: fpC - fpB = 3 - 3 = 0
>
> so the ordering is B>C>A no matter what x is within the allowed range.
>
> I've been thinking about particular types of three-candidate Condorcet
> methods. What I've been doing suggests that for a continuous method to
> pass UM and Condorcet, it must return an A=B=C tie for every pairwise
> tie election (i.e. where A=B, A=C, B=C). Furthermore, adding ballots to
> turn A=B and A=C into A>B and B>C must make A win with certainty.
>
> The reason is that if we're somewhere on the "A is the CW" side of the
> (A>B, B>C, A=C) boundary, there exists an election where changing an
> epsilon of a B>A>C ballot into B>C>A will get us *onto* that boundary.
> Such a change leads to an UM violation if A doesn't win with certainty
> afterwards.
>
> (It also looks like the number of elections near the boundary with no
> B>A>C ballots is so small compared to the number of elections with, that
> any election that makes A win with certainty on the area of the boundary
> you can reach by turning B>A>C into B>C>A must make A win with certainty
> along the whole boundary. At least if it's continuous, although I've
> only properly investigated a form of linear method.)
>
> "Center-squeezy" methods like fpA-fpC fail by not returning an A=B=C tie
> on every pairwise tied election. Methods like minmax pass the boundary
> condition, but fail inside the ABCA cycle region itself.
>
> What I'd have to show to have an actual impossibility proof is that it's
> impossible to construct a method that both respects UM when going from
> the "A always wins" region to the ABCA cycle region, and also respects
> UM inside the region. I don't have that (yet, at least). But if I were
> to guess, I think Condorcet and UM are incompatible. This is unfortunate
> because it would otherwise be a good criterion to use to find
> manipulation-resistant Condorcet methods.
>
>
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