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

[Joshua Boehme]

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:
> Suppose, for any given election, we relabel candidates so that they are in 
> sorted order by first preferences, i.e. A has the most first preferences, 
> then B, then C, and so on.
> 
> A wins in the honest election.
> 
> Then the election is manipulable if, when every X>A voter ranks X first, the 
> winner changes from A to X.
> 
> For a three-candidate election, specifically, we can manipulate if
>      fpA > fpB and (B>A) > fpA,
> because then all B>A voters rank B first and fpB becomes equal to B>A which 
> is greater than fpA.
> 
> Now consider the impartial culture with c candidates and n voters. Every 
> first preference count is a binomial with expectation n/c, while every 
> pairwise preference has expectation n/2.
> 
> Since n/2 > n/c when we have more than two candidates, in the limit of n 
> approaching infinity, we should have (B>A) > fpA with probability one. Thus 
> Plurality is almost always manipulable under impartial culture with enough 
> voters.
> 
> (What's needed to make this formal rather than just intuitive is an argument 
> that uses the variance of the binomials to show that p((B>A) > fpA) = 1 in 
> the limit of n approaching infinity.)
> 
> I suspect that certain manipulability under IC is true of at least every 
> three-candidate weighted positional system closer to Plurality than Borda, 
> but proving that would be pretty tough. I did some calculations that, if I 
> didn't mess them up, shows that it's *not* true for Antiplurality.
> 
> -km
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