I think you've hit on an even more general result than you realized. It
looks like it applies much more broadly than impartial culture. If my
intuition is right, we can sacrifice a lot of the assumptions of impartial
culture and still prove the result.

If we have something like the following, with O() meaning order notation...

1. P(fpA = fpB)->0 as n->infinity
2. O(sqrt(Var(B>A - fpA))) < O(E[B>A - fpA])
3. O(sqrt(Var(A>B - fpB))) < O(E[A>B - fpB])
4. E[B>A - fpA] > 0
5. E[A>B - fpB] > 0

then I think you get there regardless of the underlying distribution, and
even if voters aren't all independent. #1 provides your first inequality and
Chebyshev's inequality plus the rest gives you the second (possibly with A
and B flipped consistently throughout for a particular realization).

Note: when writing out the exact variances for a particular case, B>A and
fpA usually covary negatively, and likewise for A>B and fpB.

You can construct voting models where voters anti-coordinate in weird ways
that violate these conditions, but my guess is that realistic models are
generally going to meet them.

The above isn't a rigorous argument, but it gestures at the intuition. You
can probably weaken the assumptions further, too -- you don't need #3 and #5
if fpA > fpB with probability going to 1, for example.

On 4/17/25 4:43 PM, Kristofer Munsterhjelm via Election-Methods wrote:

On 2025-04-20 18:36, Joshua Boehme via Election-Methods wrote:

Very nice. I think you're right. (Strictly speaking, we can probably
remove the first condition, because if fpA=fpB, then as long as B>A >
fpA in expectation, we can still change a tie between fpA and fpB into a
decisive victory for B, which is an improvement for B.)

However, other obvious voting model - impartial anonymous culture or IAC

• fails the criteria above. As a result, most of the methods I've been
  looking at converge to a fractional manipulability value for three
  candidates for IAC.

The reason IAC has fractional values is that it's essentially a
Dirichlet model, so the manipulability is the volume of the polytope
where the conditions hold, divided by the total volume of the polytope
(under constraints that the number of voters sum to some given n).

Suppose spatial is the most realistic model. Then IC (apart from its
soft spot for Antiplurality) seems to generally be more punishing than
spatial, while IAC is less punishing. Here are IAC n->infty values for
some methods, from https://www.jstor.org/stable/41106091:

Plurality:      7/24 ~= 0.2917
Antiplurality: 14/27 ~= 0.5185
IRV/runoff:    1/9  ~= 0.1111
Coombs:        31/72 ~= 0.4306

I've verified this with Monte-Carlo for Plurality. Since it's no longer
almost surely manipulable, one needs to check B and C separately.

I'm not sure if spatial converges to edge values (0 or 1). JGA's results
(https://mpra.ub.uni-muenchen.de/32200/1/MPRA_paper_32200.pdf table 4)
at first seems to indicate the manipulability converging (e.g. to 0.46
for approval), but the step from 639 voters to 2559 voters doesn't seem
to fit the pattern. So I'm unsure.

For spatial, the easiest way to start is probably by fixing the
candidate locations in opinion space and then looking at the
distribution of Euclidean distances from random multivariate normals to
each of these fixed points. But converting that to something like E[ABC]
is still going to be tough (e.g. https://stats.stackexchange.com/q/167133).

Alternatively, perhaps something around: suppose that the candidate
positions are fixed, then we have polytopal sets in opinion space for
each ordering. S_{A>B>C} would contain the points in opinion space
closest to A, second closest to B, and furthest away from C. Then the
fraction of the voters who vote ABC would be the integral of the normal
distribution within S_{A>B>C} - call this the Gaussian volume of
S_{A>B>C}, or Vol(ABC) with E[ABC] = n * Vol(ABC). Then there's nothing
special about each such set (in the expectation of every arrangement of
candidates), so that one would expect volumes of more regions in total
(e.g. B>A = BAC + BCA + CBA) to be greater than those of fewer (e.g. fpA
= ABC + ACB), which would suggest it's likely for Plurality to almost
always be manipulable. But this is very handwavy and I don't feel like I
could justify it formally. We don't get the variances this way, so it
could well be wrong.

On an aside, I think I've also managed to determine that minmax is
almost always manipulable under impartial culture with three candidates.

-km

On 4/21/25 3:43 PM, Kristofer Munsterhjelm wrote:

This could still lead somewhere interesting...

Let A be the honest frontrunner of a given plurality election (i.e., fpA >
fpB for all B != A).

The election is manipulable iff there exists a B != A such that B>A >= fpA.

That is, it is NOT manipulable iff for all B != A, fpA > B>A.

This is an interesting situation, as it's akin to being a Condorcet winner
but even more so. A>B > fpA, so this condition implies A is a strong
Condorcet winner, but the converse doesn't hold (e.g., a centrist candidate
could be everyone's #2 pick but no one's #1).

In other words, in a plurality election, there's either a "super" Condorcet
winner decisively above all the other candidates, or it's manipulable by
strategic voters.

(Maybe this was already a known result somewhere, though.)

On 2025-04-22 03:40, Joshua Boehme via Election-Methods wrote:

You're right, disregarding ties (e.g. if fpA = fpB and A>B = B>A, then B
still can't manipulate). And we can disregard ties in the limit, at
least for impartial culture.

I have been playing a bit with three-candidate IRV, and it seems like
it's always manipulable when there's a Condorcet cycle. And your result
shows that Plurality is also always manipulable when there's a Condorcet
cycle. Do there exist methods that aren't? Do there exist /majoritarian/
methods that aren't?

Maybe Antiplurality answers the first question in the positive. In a
sense, it feels like the second is false, at least with no equal-rank or
truncation. Suppose A won. Find some B that beats A pairwise (there
always exists one: either A is not in the Smith set or B beats A in the
top cycle). Then make all the B>A voters vote B first. Since there's no
equal-rank or truncation, B>A must be a majority, so now a majority
ranks B first. Then, by the majority criterion, B must win, and we're done.

A nice corollary if I'm right is: suppose that f_M(v,c) is the expected
manipulability of method M over the impartial culture with v voters and
c candidates. Then lim c->inf lim v->inf f_M(v, c) = 1 if method M is
majoritarian, because this is lower-bounded by the probability of a
Condorcet cycle, which converges to one for the impartial culture.

(That nearly every election is a cycle given enough candidates was shown
for IAC by Quas,
https://www.worldscientific.com/doi/abs/10.1142/S0219493704000912, and
another result shows that impartial culture is at least as cycle-happy:
https://www.jstor.org/stable/41106568. I suspect this holds for normal
spatial models with at least two dimensions, too, but that the
convergence is slow - I have no proof, just going on a hunch based on
JGA's numbers.)

I've also played a bit more with spatial models, and I think their
method manipulability values converge to nontrivial values (neither one
or zero) as the number of voters approaches infinity but the number of
candidates is reasonably low. Actually finding these manipulation values
exactly is hard work even in the simplest cases, and beyond me for most
models (normals in any dimension, or more than one dimension for better
integrable cdfs).

-km