[EM] Favorite Betrayal and Condorcet

Kevin Venzke stepjak at yahoo.fr
Sat Apr 16 15:57:32 PDT 2022

Hi Forest,

> It seems that my "proof" failed because I assumed that C's score could not
> change by raising B equal to C ... but that's only true if we're talking
> majority defeat: raising B to equal with C  cannot change a defeat of B by C (or
> a non-majority defeat of C by B) to a majority defeat of C by B.
>
> So let's try this fix:
>
> Elect the candidate that on the fewest ballots is outranked by any candidate
> that majority defeats it.

Stated like this, this does satisfy FBC, however it is not a Condorcet method.

If I dub this method MajBTP, MajBTP is very close to MDDA as you note. It has
similar properties, including a Plurality failure risk with 4+ candidates. SFC
and SDSC/MD both seem to be preserved.

It's pretty interesting that this works, and to consider how this resolves
differently from MDDA. One example:

40: A>B>C
35: B>C>A
25: C>A>B

A>B>C>A majority cycle. MDDA elects B as the approval winner. MajBTP effectively
picks the first pref winner A, as every second-ranked candidate has a maj loss
to the first preference.

> Or for lay person proposal completeness ...
>
> Lacking a Condorcet winner, elect the candidate that on the fewest ballots is
> outranked by any candidate that outranks it on a majority of ballots.

I might call this C//MajBTP. This can fail FBC in the same cases C//A does:

0.394: C=A>B
0.299: B=C>A  -->  B>A=C
0.179: B>A>C
0.126: A=B>C

A>C>B>A cycle, no majorities. A wins on approval.

When the .299 lower C, B becomes the CW.

Kevin