[EM] Is it that easy? (Condorcet is almost incompatible with 1/n burial resistance for n>3)

[Forest Simmons]

On Fri, Jun 30, 2023, 12:30 PM Kristofer Munsterhjelm <km_elmet at t-online.de>
wrote:

> Let a candidate be a dominant mutual nth candidate if he has more than
> 1/n of the share of first preferences and also pairwise beats everybody
> else.
>
> Let a method pass "strong" dominant mutual nth candidate burial
> resistance if W is the dominant mutual nth candidate and voters who
> prefer some A to W can't increase the probability of A winning by
> burying W. (Ordinary DMnCBR requires that A becomes the sole winner.)
>
> Then the maximum n compatible with this stronger form and Condorcet is
> 3, i.e. mutual third.
>
> A very simple example that my linear programming solver found:
>
> 1: A>B>C
> 1: B>A>C
> 1: C>A>B
>
> A is the CW and thus wins. Furthermore, A is a dominant nth candidate
> for all n > 3 because he has exactly a third of the first preferences.
> But then
>
> 1: A>B>C
> 1: B>C>A (burying A under C)
> 1: C>A>B
>

Because of the perfect symmetry, it is not possible to set up a sincere
runoff that excludes the candidate least likely to be the sincere CW.

But under standard game theoretic assumptions ... of perfect mutual
knowledge of sincere preferences, etc... any of the following ballots would
elicit "votes" to elect A:

A vs (B vs C)
B vs (A vs C)
C vs (A vs B)

In the first of these, since C is the sincere CL, rational voters know that
if the "runoff" reaches the B vs C decision, the winner will be B. But a
majority of voters sincerely prefer A to B .... therefore their rational
choice is for that majority to express A>(B vs C), etc.

For the second ballot ...  if the second stage is reached, A will win.
Otherwise B will win.  Since A is sincerely preferred over B, a majority of
the rational voters will vote B<(A vs C), etc.

For the third ballot, a majority of ballots will start C<(A vs B) because
no matter the result of A vs B, it will be preferred over C, etc.


> and every anonymous and neutral voting method must return a perfect tie,
> which increased B's chance of winning.
>
> Now I say "almost" because we don't know that the strategists prefer a
> 33% chance of either candidate to 100% of A, we only know that they
> prefer B to A. So in that sense, no, it isn't that easy. But it seems
> very close.
>
> Maybe it's possible to get an actual proof by perturbing with some
> epsilons and using resolvability and monotonicity to say that the method
> must elect the candidate with the most first preferences, and then
> letting B be it... e.g. something like
>
> 10000: A>B>C
> 10001: B>A>C
> 10000: C>A>B
>
> and A wins by being a CW and nth dominant for n > 3.0001, i.e. 3 +
> epsilon that can be made arbitrarily small; then
>
> 10000: A>B>C
> 10001: B>C>A
> 10000: C>A>B
>
> by Tideman's resolvability almost suffices (since the perfect tie is
> broken by one additional voter).
>
> What do you think, is there any simple way to do the last step?
>
> -km
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