[EM] Final Conclusion on strong FBC!!!!!!

[Alex Small]

Forest Simmons said:
> Alex, how do you generalize to more than three candidates?  Is it still
> Top 2 or does it become Median Rank and Above?

Most methods should satisfy strong FBC for 4+ candidates if you stipulate
that the top 2 ranks are treated equally.  N-candidate Condorcet would
satisfy strong FBC if the top 2 on a list were treated as equal when doing
pairwise comparisons.  N-candidate Borda would pass if the top 2 ranks
each got N-2 points, the 3rd rank got N-3 points, etc.  Top 2, top 3,..top
N-1 etc. would also pass for N candidates.  (Top m just gives one point
each to the m candidates highest on each ballot, and zero points to all
others.)

However, all of these methods pass strong FBC only by pretending that
there is no 1st place/2nd place distinction.  I proved that with 3
candidates, any method making such a distinction fails strong FBC.  With
4+ candidates there are more available manipulations involving candidates
lower on the list.  My proof doesn't rule out the possibility that, with
so many "lower choice manipulations" available, one might design a method
that distinguishes between #1 and #2 without giving any incentive to
betray #1.

This is not to say that it is in fact possible, just that I haven't proven
it.  Since my method involves checking 8 different cases, it could be that
I'm missing some key geometric or algebraic principle underlying it all. 
Exploiting some underlying principle might make the generalization
straightforward, and also make the proof simpler.

Otherwise, it would be great if somebody could prove the following
conjecture:

If for N candidates it is impossible to design an election method that
removes all incentives for favorite betrayal but distinguishes between
first and second choices, then for N+1 candidates it is impossible to
design a method meeting those criteria.

Of course, maybe that conjecture is false.





Alex


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