[EM] A quick and dirty strength of victory measure for LIIA methods

[Kristofer Munsterhjelm]

This came up in the Electorama call, and I thought I'd elaborate:

Here's a score/support level concept that isn't too theoretically sound, 
but might be interesting. Suppose that we want to know not just who won 
according to a LIIA method, or the order of finish, but also how 
comfortable their victory is, or what their support level (of some kind) is.

Then if the method passes LIIA, we'd like the support values to have 
some kind of preserved property so that, if we remove the loser (or the 
winner), the relative scores between the remaining candidates don't change.

A simple idea I tinkered with, that should pass this property, is as 
follows:

Let the social order from some LIIA method be A>B>C>D, and let the 
unknown support values (strength of victory, candidate scores) be x_A, 
x_B, x_C, and x_D.

Then set

	x_A / x_B = (A>B)/(B>A)
	x_B / x_C = (B>C)/(C>B)
	x_C / x_D = (C>D)/(D>C)
	||x||     = 1

An example: Using the 2009 Burlington election gives the following set 
of equations:

	x_Montroll / x_Kiss = M>K/K>M = 4067/3477
	x_Kiss / X_Wright = K>W/W>K = 4314/4064
	x_Wright / x_Smith = W>S/S>W = 3975/3793
	x_Smith / x_Simpson = S>Si/Si>S = 5573/721
	x_Simpson / x_WriteIn = Si>Wr/Wr>Si = 3338/165
	x_Montroll + x_Kiss + x_Wright + x_Smith + x_Simpson + x_WriteIn = 1

This ends up being approximately:

	x_Montroll =  28.30%
	x_Kiss     =  24.20%
	x_Wright   =  22.80%
	x_Smith    =  21.75%
	x_Simpson  =   2.81%
	x_WriteIn  =   0.14%

The intuition here is that A's score relative to B's should be the same 
as their relative pairwise victories. (So if A beats B 10% more strongly 
than B beats A, A's score should be 10% higher than B's.)

In addition, to preserve LIIA[1], the relation between A and B's scores 
should only depend on pairwise contests involving A and B. Eliminating A 
would not alter the order of the remaining candidates, so the social 
order would simply be B>C>D; and the equation set would be reduced to

	x_B / x_C = (B>C)/(C>B)
	x_C / x_D = (C>D)/(D>C)
	(...)

But since there are n candidates and only n-1 such relations, this only 
sets the support levels up to some constant of proportionality. Here 
I've chosen to make total support sum up to 100%, but that does 
compromise its mathematical elegance. In addition, it's not cloneproof. 
Imagine adding a bunch of clones of ay, Montroll, with the voters 
ranking them in a random order. Then we get

	x_M1 / x_M2 = 1
	x_M2 / x_M3 = 1
	...
	x_M(k-1) / x_M(k) = 1

which will lower everybody's support numbers as Montroll's support is 
distributed over all his clones and the other candidates still need to 
have a lower score.

But it might still be interesting.

I think the most important part is that, if we're to transplant LIIA, 
then the ratings/support values can only depend on the pairwise 
victories along the path of the social order. I.e. we can't involve 
Montroll>Wright when determining Montroll's score. The only other bit 
that stays the same is the total number of voters (and only then, 
sometimes). Perhaps it can be used to solve the clone dependence 
problem? Or maybe one has to relax the LIIA-ish property.

-km

[1] Formally speaking, this property is: if you're basing the scores 
from the social order (order of finish) of a method that passes LIIA, 
and you get the scores for two overlapping candidate sets (both of which 
have been selected by repeatedly eliminating losers or winners from a 
larger election), then you can determine the scores for the election 
containing every candidate in the union. (Ideally, I'd want the scores 
to not change at all, so that if candidate A is in both elections, A has 
the same score in both; but I'm not sure that's possible.)