[EM] Single-candidate DMTBR idea

[Kristofer Munsterhjelm]

Earlier I had an idea that perhaps a good way to try to find a monotone 
DMTBR Condorcet method is to try to start with one that passes 
single-candidate DMTBR (DMTCBR?), i.e. if the CW has more than 1/3 first 
preferences and wins, then voters who prefer some other X can't make X 
win by burying the winner under some other candidate Y.

And just now I had a thought about a starting point for a method that 
could pass this for any number of candidates, not just three like 
fpA-fpC, and might do so in a monotone way.

If there are two candidates with more than 1/3 fpp, elect the one that 
pairwise beats the other. If there is only one candidate with more than 
1/3 fpp, elect that candidate. And if there are none, do something else 
that doesn't violate monotonicity given the rules above.

The idea here is that if W is the DMTC, then by definition of being the 
CW, W also beats everybody else. If the buriers' candidate X is the 
second candidate with 1/3 fpp, then since W wins, W must beat X 
pairwise. The X>W voters can't do anything about this because they're 
already maximally contributing to X>W. On the other hand, if X does not 
have 1/3 first preferences, then the buriers can't make X into the CW by 
using burial (for the same reason). Thus X can't win that way, either.

This piece of the puzzle is not in itself Condorcet... but I think an 
"elect the CW if there is one" hack would work to fix that particular 
problem. Since DMTCBR only provides protection if the winner happens to 
both be the CW *and* have >1/3 first preferences, electing the CW 
whenever there is one is no problem: if the CW doesn't have >1/3 fpp, 
then he's not within the domain of DMTCBR, and if he does, the base 
method would've elected him anyway.

For a suitable generalization of the starting point, perhaps this: 
Choose the two candidates with the most first preferences. If only one 
of them has more than a third of the first preferences, that candidate 
wins; otherwise, the candidate who beats the other pairwise wins. I 
don't know how you'd turn this into a social order in a natural way, though.

-km