[EM] Possible PO replacement for MMPO

[Kristofer Munsterhjelm]

While trying to create a four-candidate monotone resistant set method 
using geometry, I found a pairwise counting approach that might preserve 
(or come closer to preserving) the FBC while avoiding plurality failures.

Call it "favors victor". (A>B)_fv is the number of voters who do not 
rank A strictly lower than B. This means that if a voter ranks A and B 
equal, that counts as a full point both to (A>B) and (B>A). As such, it 
generalizes the tied-at-the-top rule.

I'll leave it to the people with more experience with FBC methods to 
test what pairwise methods pass FBC with this counting approach. But it 
might be interesting to check minmax, Copeland, and Ranked Robin (since 
the tournament Borda count based on fv reduces to Mike Ossipoff's 
Summed-Ranks method which he said passes FBC.)
It's possible that it does allow methods to pass FBC but is too 
"approval-y" to be of use; that is, that it causes a race to an Approval 
bottom because equal-rank is more powerful than strict rank.

In any case, it does get rid of the MMPO "bad-example" problem,
x: A
1: A=C
1: B=C
x: B

where MMPO elects C. Here's a count using fv with x being a billion:

                             contributes to
		A>B	A>C	B>A	B>C	C>A	C>B
1e9: A>B=C	yes	yes		yes		yes
   1: A=C>B	yes	yes			yes	yes
   1: B=C>A			yes	yes	yes	yes
1e9: B>A=C		yes	yes	yes	yes

(e.g. A>B=C contributes to A>B because A is ranked higher than B, and to 
B>C and C>B because B is not lower ranked than C, and C is not lower 
ranked than B, respectively). That gives

			against
		A	B	C
	A	---	1e9+1	2e9+1
for	B	1e9+1	---	2e9+1
	C	1e9+2	1e9+2	---

So the maximum defeat scores (penalties) are
	A: 1e9+2
	B: 1e9+2
	C: 2e9+1
and the minmax outcome is A=B>C, as desired.

-km