A long time ago, I said that my fpA-fpC three-candidate method passed UM

and Condorcet. I've now found out that that's not the case.

Consider this election:

2: A>B>C

2+x: B>A>C

1-x: B>C>A

3: C>A>B

With x=1, A is the CW; with x=0, we have a tie-cycle (A>B, B>C, C=A).

With an x slightly less than zero, the election is a proper ABCA cycle.

The candidate scores are:

A: fpA - fpC = 2 - 3 = -1

B: fpB - fpA = 3 - 2 = 1

C: fpC - fpB = 3 - 3 = 0

so the ordering is B>C>A no matter what x is within the allowed range.

I've been thinking about particular types of three-candidate Condorcet

methods. What I've been doing suggests that for a continuous method to

pass UM and Condorcet, it must return an A=B=C tie for every pairwise

tie election (i.e. where A=B, A=C, B=C). Furthermore, adding ballots to

turn A=B and A=C into A>B and B>C must make A win with certainty.

The reason is that if we're somewhere on the "A is the CW" side of the

(A>B, B>C, A=C) boundary, there exists an election where changing an

epsilon of a B>A>C ballot into B>C>A will get us *onto* that boundary.

Such a change leads to an UM violation if A doesn't win with certainty

afterwards.

(It also looks like the number of elections near the boundary with no

B>A>C ballots is so small compared to the number of elections with, that

any election that makes A win with certainty on the area of the boundary

you can reach by turning B>A>C into B>C>A must make A win with certainty

along the whole boundary. At least if it's continuous, although I've

only properly investigated a form of linear method.)

"Center-squeezy" methods like fpA-fpC fail by not returning an A=B=C tie

on every pairwise tied election. Methods like minmax pass the boundary

condition, but fail inside the ABCA cycle region itself.

What I'd have to show to have an actual impossibility proof is that it's

impossible to construct a method that both respects UM when going from

the "A always wins" region to the ABCA cycle region, and also respects

UM inside the region. I don't have that (yet, at least). But if I were

to guess, I think Condorcet and UM are incompatible. This is unfortunate

because it would otherwise be a good criterion to use to find

manipulation-resistant Condorcet methods.

----

Election-Methods mailing list - see

https://electorama.com/em for list info