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

[Kristofer Munsterhjelm]

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

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