[EM] Defection, nomination disincentive, MMPO

[Gervase Lam]

> Date: Fri, 17 Dec 2004 04:49:45 +0100 (CET)
> From: Kevin Venzke <stepjak at yahoo.fr>
> Subject: Re: [EM] Defection, nomination disincentive, MMPO

> Also, every example I've seen of MMPO's Majority failure involves
> the use of four slots. It's always this scenario:
>
> 20 A>B>C>D
> 20 B>C>A>D
> 20 C>A>B>D
> 13 D>A>B>C
> 13 D>B>C>A
> 13 D>C>B>A
>
> Majority (as well as Smith) requires that A, B, or C win, but D
> has the lowest MMPO score.

As Smith//MinMax exists, this could be extended to Smith//MMPO.  Though, I 
think this will make the method fail Later-No-Harm.

> > The most easily presentable tie-breaker I thought of using was to find
> > the next highest opposing votes for each of the tied candidates.  The
> > candidate with the lowest number of these opposing votes is the
> > winner.
> >
> > If there is still a tie here, then you go on to the next highest
> > opposing votes for the tied candidates.  And so on, if required.
>
> This occurred to me, but I'm worried about a Clone-Loser problem. It
> seems to me that a party could benefit from running clones, so that the
> opposition votes from the party's candidates have to be plowed through
> one-by-one during the tiebreaker.

Do you really mean Clone-Loser here?  If so, why not just use MMPO on the 
tied candidates?

If you meant Clone-Winner here then may be something like which of the 
tied candidates had the better pairwise result against the bottom most 
MMPO candidate can be used?  Unfortunately, this one is a bit artificial.

Or may be for each tied candidate, find their Pairwise Proposition?  The 
number of ballots that ranked Y>X is the Pairwise Opposition of candidate 
X used for MMPO.  Therefore, the number of ballots that ranked X>Y is the 
Pairwise Proposition of candidate X.  The candidate with the highest 
proposition from among the tied candidates is the winner.  However, I 
think this is bit naff as I think it would encourage Approval style voting.

Thanks,
Gervase.