[EM] Single-candidate DMTBR idea

[Kristofer Munsterhjelm]

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 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.


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). Okay, so C can't win 
because of second-order reasoning. 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