[EM] Holy grail: a "condorcet compliant" cardinal method (MCA/Bucklin variant)
jameson.quinn at gmail.com
Sun Sep 5 08:39:03 PDT 2010
Here's my latest Bucklin variant, which, pending the results of the
naming poll <http://betterpolls.com/do/1189>, I'm calling RMCA (because of
the catchy music). (Of course, if it's OK to appropriate the name MCA, the
editorial headline writes itself...)
Start with two-rank Bucklin ballots: Preferred, Approved, or Unapproved. The
highest majority preferred, if any, wins it. If not, find the highest number
of approved-or-preferred ("approval winner", AW), and the highest range
score ("range winner", RW), counting 2/1/0 for P/A/U. If those are the same
candidate, that candidate wins; otherwise, those two go into a runoff. (If
either of these measures gives an exact tie, then the two tied candidates go
The first (to me, surprising) result is that any Condorcet winner which is
determinable from the ballots must get into the runoff. Proof: Say that the
AW is not the CW. Then the number of ballots n with CW>AW is greater than m
with AW>CW. On a ballot where X beats Y, X has a range advantage of either 1
(X>Y) or 2 (X>>Y). Sf n2 where CW>>AW is greater than (m2 where
AW>>CW)+(n-m), then the CW is the RW. And if n2 < m2, then there are more
ballots which approve the CW and not the AW than the reverse, which
contradicts the assumption that the AW is the AW. QED.
As a practical matter, this system achieves the mathematically impossible:
it's an O(N) summable Condorcet method! Of course, the loophole is if the
honest CW exists, but the ballots don't show it. However, that possibility
can be overcome by a simple, rational voting strategy. Say Vernon Voter
knows that Cathy Candidate is a possible CW. He should not rank any other
[serious] candidates at the same rank as her; that is, he should give only
her the middle "approved" rank unless she is his honest favorite or worst
[among the serious candidates]. If all voters do that, the CW will be
visible from the given ballots. (In fact, I suspect but have not proven that
if all voters use this strategy, the CW is guaranteed to win without a
In my mind, this method is truly the holy grail. Consider:
1) "Condorcet compliant"
2) IIA and clone-proof (assuming IAs and clones do not shift the ranking of
the non-clone winner on any ballot, which is an acceptable interpretation of
those criteria for the cardinal case.)
(That is, as a cardinal system which isn't "universal" in Arrow's ordinal
sense, it "breaks" Arrows theorem; yet, unlike other cardinal systems, it
still can be said to obey [a weakened version of] the Condorcet criterion,
which [in its strong version] is ordinally-based.)
3) O(N) summable (can be run on any voting machine, very susceptible to
verification and security techniques such as partial or sampled recounts)
4) Pretty simple to explain; certainly, more intuitive than Schulze or even
5) Good Bayesian Regret; perhaps even better than Range under some
circumstances. (Hasn't been measured yet, but it should come out similar to
Chris Benham's prior version of MCA, which is one of the few systems which
beats Range in some conditions)
6) No excuse for favorite betrayal, or indeed any non-semi-honest votes.
7) A simple-to-explain voting strategy which only requires knowing one
frontrunner, not two; and which does not even require "serious
semi-betrayal" of any favorites (that is, ranking a different serious
candidate even with your favorite.)
8) Others may disagree, but I think it has just the right amount of runoffs.
Essentially (provably???), there would only be runoffs if there's no known
honest CW; in which case, further deliberation by the voters is completely
9 (???) I think it even deals with a 3-way Condorcet cycle, if which all
voters vote the 3 candidates separately (one-per-rank-level), by electing
the same candidate as Minimax, Schulze, or RP. (!!) (I'll investigate this
further, and share my results either way.)
I'm pretty excited about this one. Those are some amazing characteristics.
So please, take your best shot. What's wrong with this system? Can you give
some contrived election where it gets the arguably "wrong" answer?
(ps to Chris Benham: I'm sending a copy of this directly to you, even though
I suspect you'd read it off of one of the lists I'm using anyway, to ask you
if it's OK to use the name "MCA" for this system. It's almost identical to
your version of MCA, but using Range-2 instead of
multiple-approval-majorities to decide who goes to the runoff gives it the
extra provable "ballot-Condorcet" compliance. If you say no, no offense
taken, there's plenty of names.)
Of course, Approval also will trivially elect any CW visible on the
ballots, but since, unlike Approval, RMCA has a simple voter strategy to
ensure that a given candidate will be a ballot-CW iff they are an honest CW,
RMCA's ballot-Condorcet is much stronger than Approval's.
OK, I am aware of the contrived elections where there are >5 candidates
and 4 voters, and all voters have perfect knowledge of each others'
preferences, and 3 of the voters have cyclic preferences over 3 "clone-sets"
while the other voter's preferences are nearly-orthogonal, caring
almost-but-not-quite only about distinctions within the clonesets. The
equilibrium is a mixed strategy for the cyclic voters, and a non-semi-honest
vote from the orthogonal voter. Such a scenario could happen with this
system. However, such scenarios require perfect information, and are
exponentially improbable with more than 4 voters; so I'm willing to say that
there are no reasons for a non-semi-honest-vote.
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