[EM] Favorite Betrayal and Condorcet

[Forest Simmons]

Kevin,

Thanks for your clarifications, insights and insightful examples.

-Forest

El sáb., 16 de abr. de 2022 3:54 p. m., Kevin Venzke <stepjak at yahoo.fr>
escribió:

> 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
>
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