[EM] Single-candidate DMTBR idea
km_elmet at t-online.de
Thu Mar 10 15:33:39 PST 2022
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.
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