[EM] Cardinal primitives

[Kristofer Munsterhjelm]

The minimum number of candidates that we meaningfully distinguish by
using an indifference lottery is three, as in that the voter can provide
information/ratings (A: 1, B: p, C: 0) if he ranks A>B>C and values a
random draw with p probability of A, 1-p probability of C, equal to
getting B with certainty, risk aversion notwithstanding.

If we use only two candidates at a time, then we just get ranked
ballots, because a voter will either prefer A to B, B to A, or consider
them equally good: it's not possible to get information about "how much
better" A is than B without having an absolute scale, I think.

So if we want to make a method that's only based on lottery data, i.e.
that doesn't make assumptions about commensurability, then perhaps we
could construct this from looking at different three-tuples, where every
candidate but three are eliminated and then the ratings are normalized.

This gives a primitive, call it (A, B, C) to the three voters' scores.
And no matter how it's extended, it will get the good centrist/bad
centrist examples right, since those only involve three candidates -- if
the normalization gets it right, that is.

Maybe something interesting could be built on top. I tried but I
couldn't reliably extend STAR to be cloneproof, so I'll just give some
ideas:

"Copeland": For all three-tuples (A, B, C) let WLOG A be the normalized
Range/cumulative winner, B be second, and C third. Then give one point
to A. The winner of the method is the candidate with the most points.

"Copeland STAR": Same as Copeland, except give a point to the pairwise
winner of A and B instead.

"Beatpath": Let there be a beatpath A->B->C->D from A to D if A is the
winner in (A, B, C) and D is the loser in (B, C, D), for some other
candidates B and C; or the beatpath is A->X->D if there exists some (A,
X, D) so that D is the loser. Then do something with these beatpaths.
Does there exist a beatpath winner like Schulze? What is the beatpath
"value"? The minimum of the margins between consecutive candidates in
the beatpath, perhaps?

"Ranked Pairs": Let A >~ B be a relation, and A >~ B has magnitude m for
A and B if, when we choose a candidate X so that the margin between A
and B's scores in (A, X, B) is maximized, that margin between A and B's
score is m. For all pairs A, B, add this pair with associated weigth A
>~B to a list. Sort the list from greatest to least magnitude and do
Ranked Pairs as usual. (Would also work with River or Warren's Maxtree.)

Anything else? It might be a good idea to check what normalization would
mitigate the Burr dilemma the most for three candidate elections. I've
been going on the hunch that l_2 normalization would do so, but I have
nothing formal to back that up with.

-km