[EM] DH3 and honest Condorcet winners

Kristofer Munsterhjelm km_elmet at t-online.de
Sat Jul 17 10:31:34 PDT 2021

In the classical DH3 scenario, the dark horse is universally loathed by
the voters. This suggests a smart-aleck way of making almost every
method DH3-proof: just eliminate every candidate that don't have at
least one first preference. Since the dark horse is eliminated outright,
there's no point in attempting a burial - it simply won't work.

DH3 resistance thus shares with clone independence the feature that it's
easy to make a brittle method that technically complies with the
criterion. But add some very slight amount of noise (a few voters who
rank other candidates between the clones, or a few voters who rank the
dark horse first) and the hack fails.

Now suppose that we have an election after every faction has gone on its
burial spree, and there were initially some voters who ranked the dark
horse first. It would look something like this:

34: A>X>B>C
33: B>X>C>A
32: C>X>A>B
 5: X

But this is just a three candidate version of the Left, Center, Right
scenario. If X is a genuine consensus candidate, then X should be
elected and a failure to do so is an instance of center squeeze. But if
this is the aftermath of a DH3 scenario, then X must absolutely not be

If the method's strategy is to eliminate candidates who are ranked first
by too few voters, then it has to have some kind of threshold -- how
many X voters distinguish a DH3 scenario from one with a centrist
candidate. In addition, it'll fail the Condorcet criterion because you
can have a CW that has zero first preference votes.

So this means that a Condorcet method can't defend against the worst DH3
outcome by simply rejecting the dark horse - because it can't tell the
dark horse apart from a true centrist winner once the buriers have all
done their damage.

Thus, the way a Condorcet method resists DH3 must be by making it either
pointless or harmful to the faction to try to engineer a cycle to begin
with. If A wins, and a faction who prefers B buries A under X, then if
the winner changes, it mustn't be to B.

Just how to do that is not an easy feat - but at least the argument
above shows what *doesn't* work. (Not if you want to pass Condorcet, at

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