[EM] Defection, nomination disincentive, MMPO

[Kevin Venzke]

Gervase,

Thanks for your reply.

 --- Gervase Lam <gervase.lam at group.force9.co.uk> a écrit : 
> > In "MinMax (Pairwise Opposition)" or "MMPO," a pairwise matrix
> > This method has flaws: It fails Condorcet,
> 
> I initially thought MMPO did fail Condorcet.  However, I thought I worked 
> out why it did not.  May be I am wrong.  Can you give an example of this?

It requires some equal rankings. (A CW who has majority-strength
victories over every candidate can't lose in MMPO.) I doubt I can make
an example off the top of my head, but suppose in the CW's closest
match, 49% of the voters oppose the CW. There is no particular reason
why another candidate, not the CW, couldn't have an "MPO" of less than
49%.

> > Majority, Plurality, and Clone Independence.
> 
> May be MMPO can be patched by Forest's de-cloning method before applying 
> MMPO.

I should note that MMPO (which is incidentally equivalent to what
Woodall calls "MinDAGS") fails Clone-Winner, not Clone-Loser. In other
words, a party can't gain by running a bunch of clones.

Perhaps some method could be used to get rid of clones, although I
would worry about doing something to cause Later-no-Harm to be failed
unnecessarily.

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.

Does MMPO fail Majority when fewer than four slots are permitted?
I presently doubt it...

I'm not sure if there's a way to satisfy Plurality:

7 A>B
3 B
8 C

This is an A-B tie. But A shouldn't be allowed to win according to 
Plurality, since A has fewer votes (of any kind) than C has first 
preferences.

We can't first eliminate candidates such as A, since that would give
B voters (etc.) incentive not to give a second preference to C. That
would clearly lead to a Later-no-Harm failure.

> 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.

> It can be seen that for an N candidate election, there can be up to (N-2) 
> tie-breakers.  However, assuming that enough ballots are cast, it could be 
> said any candidate who wins as a result of the tie-breaker is lucky.  
> Therefore, it could be said that any tie-breaker could do.

I don't know about this. Ties would be more common under MMPO. Even this
is a B-C tie:

25 A
24 A>B
24 B
27 C>B

So I would want to use a tiebreaker that satisfies Later-no-Harm. I
think FPP is a decent choice. FPP fails Clone-Winner, but MMPO already
fails that criterion.

So I would make B the decisive winner, based on how I'm interpreting
FPP (i.e., eliminating other candidates first).

Thanks for your thoughts.

Kevin Venzke


	

	
		
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