[EM] minx(fpA-fpC) DMTBR counterexample?

[Kevin Venzke]

Hi Kristofer,

This method is a little tricky to double-check so I have a little less
confidence than usual in offering scenarios...

Le samedi 16 avril 2022, 06:05:45 UTC−5, Kristofer Munsterhjelm <km_elmet at t-online.de> a écrit :
> Define the fpA-fpC scores for three candidates like this:
>     - If W is the CW, then W gets score equal to |V|, the number of voters;
> and the other two get score 0
>     - If the Smith set is size 2, both get |V|/2 and the final candidate gets 0
>     - Otherwise the scores are as in fpA-fpC. If the cycle is A=B>C>A, then
> B is counted as if A beat him pairwise.

That last sentence is an interesting choice. Since Schwartz is {B} I might
treat B the same as a CW.

Also what about when there are two ties? A=B, B=C, C>A. It's a size 3 Smith set.

> Now define minx(fpA-fpC) as the following:
>     - fpA-fpC restricted to a three-set is the fpA-fpC election with every
> candidate but those three eliminated.
>     - A's score is the vector of three-candidate fpA-fpC scores of every
> three-set that includes A.
>     - The candidate with the greatest leximin score wins.
> 
> Clearly, this method is monotone because raising A can't decrease A's
> score on any three-set containing A, due to three-candidate fpA-fpC
> being monotone. And it's summable because we only need the Condorcet
> matrix, as well as three Plurality count values for every three-set.

Hmm, but try this one:

0.345: C>D>B>A
0.320: A>C>D>B
0.243: B>A>C>D
0.090: D>B>C>A  -->  D>C>B>A

It seems to me this moves the win from C to A.

Defeats are A>C>B>A, A>D, C>D, D>B. They're unchanged by the vote.
Changing B>C to C>B seems to help A by reducing B's effective first pref count
for the three-set {A,B,C}.

> Can anyone find a DMTBR counterexample for minx(fpA-fpC)? I haven't been
> able to for either four or five candidates. Since DMTBR plus Condorcet
> and either monotonicity or summability is a high bar in itself, I want
> to be *very* sure before proclaiming that I've accomplished something
> spectacular.

How about this:

0.389: D>A>C>B
0.334: A>D>B>C  -->  A>B>C>D
0.162: B>A>D>C
0.114: C>B>D>A

Initially D is the CW and has over a third of first prefs. I think this vote
change elects A.

Lowering D creates a B>D win. The other wins are A>B, A>C, C>B, D>A, D>C.
D's worst score becomes -.107 (from {B,C,D}) while A's worst is -.055 (from
{A,B,D}).

Let me know if something seems off.

Kevin