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

[Kristofer Munsterhjelm]

On 7/1/23 03:05, Forest Simmons wrote:
> 
> 
> On Fri, Jun 30, 2023, 12:30 PM Kristofer Munsterhjelm 
> <km_elmet at t-online.de <mailto:km_elmet at t-online.de>> wrote:

>>     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.

That's true. It's a Condorcet cycle, so it has the property that if a 
majority could decide who to go for, it could get that candidate elected.

However, in this context, my point was more that suppose that the method 
were to pass, say, the stronger form of mutual dominant quarter burial 
resistance as well as Condorcet. Then the burial example must not 
deprive A of his status as sole winner. But because the after-burial 
election is symmetrical, the method has no way of knowing who was the 
honest CW. Hence this is impossible.

Then I just tried to go from the (possibly too) strong criterion to what 
we actually care about - that X>A buriers shouldn't be able to make X 
win. But I can only do that with the additional assumption that in the 
case of a perfect tie, any X-first ballot will make X the winner.

-km