[EM] Single-candidate DMTBR idea

Forest Simmons forest.simmons21 at gmail.com
Sat Mar 12 22:47:36 PST 2022

El sáb., 12 de mar. de 2022 10:02 p. m., Forest Simmons <
forest.simmons21 at gmail.com> escribió:

> El sáb., 12 de mar. de 2022 4:45 p. m., Kristofer Munsterhjelm <
> km_elmet at t-online.de> escribió:
>> On 3/12/22 6:58 PM, Forest Simmons wrote:
>> > Kristofer,
>> >
>> > 45 ABC
>> > 35 BCA
>> > 25 CAB
>> >
>> > Each of the A and B factions has more than a third of the votes.
>> > Candidate A defeats B pairwise.
>> >
>> > Almost every respectable method except TACC (as well as most
>> > non-respectable methods) agree that candidate A should have the
>> greatest
>> > winning probability.
>> >
>> > But some nagging doubt persists ... whence the Condorcet Cycle?
>> >
>> > A general scalene triangle has a longest side, and the endpoints of
>> that
>> > side are further from each other than they are from the vertex V
>> > opposite that side, which means that the V faction favorite cannot be
>> > the rational, sincere last choice of any of the three factions.
>> >
>> > And yet the voted ballots in our above three faction example give each
>> > candidate a turn at last place.
>> >
>> > Somebody's lowest preference is either mis-triangulated or
>> mis-represented.
>> I don't get what you mean here. Certainly it's possible for honest
>> Condorcet cycles to exist. Warren gave an example ...
> Warren and I gave an example an of a set of four factions in a single
> plane that induces a cycle geometrically. This cannot happen with only
> three factions ... if there is a cycle, then the preferences are
> inconsistent with the lengths of the sides of the triangle.
> A three dimensional issue space can give rise to a cycle but those
> preferences are not based on metrics/distances between factions.
> of candidates
>> evaluated on three issues, say corruption, domestic policy, and foreign
>> policy. Each faction cares primarily about one dimension, and the
>> candidates have positions on each issue that leads to a cyclical
>> majority (e.g. candidate A is incorruptible, has awful domestic policy,
>> and okayish foreign policy).
>> Honest cycles also exist in 2D spatial models, e.g. Poundstone's example
>> https://www.rangevoting.org/PoundstoneCondCyc.png.
> But not when limited to 3 ballot factions.
>> If the cycle is false, though, then any faction could have done the
>> burial. You say the A faction is the only group that has anything to win
>> by conducting the burial, so they must have done it. However, there's a
>> bit of battle-of-wits logic here. Suppose that the method did elect C by
>> this logic. Then it's possible that the C voters, knowing this,
>> engineered the cycle (honest is C>B>A) in order to push their winner
>> from B (their second choice) to C (their first).
> The burial of C by A isn't the only possibility ... just the most likely
> to succeed ... so the one most in need of checking.
> If the check confirms B>C as sincere, then B wins, so the burial of B did
> not pay off for C.
> Okay, so C can't win
>> because of second-order reasoning.
> C can and will win if it was buried by A, because it is a finalist in the
> B vs C honesty check.
> Don't make this too hypothetical. The proposed implementation is tweaked
> DMC:
> The Implicit Approval order is A>B>C.

Whoops ... actually B>A>C is the IA order.

The nominal DMC winner W is A,
Not A, but B=W.

> because A
[actually B]

> is not pairwise defeated by any candidate with greater implicit approval.
> The other finalist is C, the candidate that would be the DMC winner if W
> did not defeat it pairwise.
> The tweaked method elects the sincere winner between W and C, which is C
> if C was the sincere CW, else A, the DMC winner with ballots taken at face
> value.
> The tweaked method changes the DMC winner only if it detects and elects a
> better DMC winner (a sincere CW in this three faction example) while
> exposing an insincere order reversal.
> Of course, this tweaked version of DMC is not the only possibility for
> exploiting the potential for a sincere, binding, binary choice between two
> finalists.
> I already suggested a tweaked version of TACC that makes use of the same
> device.
> Do any other applications come to mind?
> -Forest
> And A can't win because of
>> first-order reasoning. So B must win, right? But then it's possible that
>> the B faction knew this (honest: B>A>C) and buried A to make B win.
>> So my point would be that since there's a Condorcet cycle, any Condorcet
>> method (no matter who wins) will be open to burial. One could argue that
>> the sensible methods do the right thing and elect the candidate whose
>> defector coalition has to be the largest for this to be a successful
>> burial: a method that elects A is fooled by a faction of 45 voters
>> executing burial, but if the method were to elect C, it could be fooled
>> by a faction of 25 voters, which is worse.
>> In a way, that's what DMTBR says: there's no way for a Condorcet method
>> to be absolutely immune to burial, so the best thing we can do is to
>> make some set of candidates immune to being buried by candidates outside
>> of that set, and then try to make that set as small as possible. And I
>> suspect that 1/3 is the best possible...
>> At least without doing something clever with UD or repeated balloting.
>> I'm not sure how a second ballot question would help, because there's no
>> reason for an A>B>C burier to not also "bury" by indicating B>C where
>> honest is C>B... so I may be missing something. Duple rules (like Random
>> Pair) are IIRC only strategy-proof if the pair is decided independently
>> of the voters' input.
>> In the vein of DSV, imagine that I take some Condorcet method plus top
>> two and make the DSV procedure fill in the second ballot information so
>> that it's consistent with (or strategically advantageous given) the
>> first ballot ranking. Then either the combined method is not Condorcet
>> (and it's not surprising that it would resist burial better), or it's
>> subject to the same limitations as above, I would think...
>> I would guess the answer is that the combined method isn't Condorcet,
>> because there would be a tension between burying the honest CW so that
>> the second round consists of your favored candidate and someone who's
>> going to lose - and burying too far which means that someone intolerable
>> wins the second round. Perhaps most UD solutions are like Approval:
>> there may be a Nash (or core) equilibrium around the honest CW, but the
>> setting benefits whoever has got the most complete information, and the
>> potential backfire can get very unpleasant indeed.
>> -km
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