[EM] MMPO tiebreakers that don't violate FBC.

[MIKE OSSIPOFF]

First of all, no one has found an example in which my most recent MMPO tiebreaker
fails FBC.
 
"Solve MMPO's ties by MMPO".
 
I'll call that tiebreaker 1.
 
Tiebreaker 2:
 
Whenever there's a tie, the winner is the candidate whose next largest pairwise opposition is the
least. ("next largest" after the pairwise opposition with which s/he is tied with other
candidates)
 
Tiebreaker 3:
 
If there is a tie, the winner is the candidate with the most top-ratings.
 
MMPO with tiebreaker 1 is MMPO1
 
MMPO with tiebreaker 2 is MMPO2
 
MMPO with tiebreaker 3 is MMPO3.
 
 
MMPO2 and MMPO3 meet FBC.
 
No one has found an example in which MMPO1 fails FBC.
 
MDDTR meets FBC, CD, and Later-No-Harm (LNHa); and fails Mono-Add-Plump.
 
MMPO2 and MMPO3 meet FBC, CD & LNHa, and Mono-Add-Plump; and "fails" in Kevin's 
MMPO bad-example:
 
9999: A
1:    A=C
1:    B=C
9999: B
 
In this example, MMPO elects C. 
 
MCA, MTA, MDDTR, MDDA, ABucklin and MDD,ABucklin return a tie between A & B.
 
How bad is this result of MMPO's?
 
Notice that nearly all of the A voters are indifferent between B and C. 
And the one A voter who isn't indifferent prefers C to B.
 
Likewise nearly all of the B voters are indifferent between A and C.
And the one B voter who isn't indiffefrent prefers C to A.
 
So how bad can c be? How bad can MMPO's result be?
 
So, the choice between MMPO and MDDTR is a choice between failing Mono-Add-Plump, vs 
"failing" in kevin's MMPO bad-example. Only polling will tell which "failure" is
more likely to appear bad to potential petition-signers and enaction-voters.
 
When we regard the pairwise comparison between C and A, and the pairwise comparison
between C and B, in isolation, it's easy to forget that we aren't actually holding those
two pairwise elections. We're used to looking at pairwise contests, because we're used to
pairwise-count methods. But no one is saying that the one voter who votes C over A is
more important than the greater number who vote A over c. I suggest that the 
strong pairwise defeats against C look more important than they are, because we're
used to looking at pairwise count methods, with the result that we actually start
believing that there was a a 2-candidate election between A and C, and one between B and C.
 
As I said, MDDTR and MMPO might be "controversial", because of failure of Mono-Add-Plump,
or "failure" in Kevin's MMPO bad-example. So, I don't consider them to be the easiest or
best public proposals. It would be necessary to talk to some people before being sure 
that they'd be winnable proposals.
 
A 3-slot version of SFC should be written, to tell of the SFC-like guarantee offered by
MDDTR. (Strictly speaking, SFC can only be passed by full ranking methods.)
 
By the way, when people object to "random-fill incentive" for MDDTR, maybe they're
forgetting that MDDTR is a 3-slot method.
 
And, if MMPO were proposed as a 3-slot method, that would avoid the "random-fill incentive"
criticism of it too.
 
 
Mike Ossipoff