[EM] Intuitive argument that FPTP manipulability approaches certainty in impartial culture

[Kristofer Munsterhjelm]

On 2025-04-22 03:40, Joshua Boehme via Election-Methods wrote:
> 
> 
> On 4/21/25 3:43 PM, Kristofer Munsterhjelm 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.
> 
> 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).

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.

> 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.

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