2nd example of manipulation in Condorcet's method
Bruce Anderson
landerso at ida.org
Tue Jun 4 01:48:23 PDT 1996
On May 11, I posted an example of an election in which Condorcet's method, as
"officially" defined on this list (EM-Condorcet), was vulnerable to several
kinds of manipulation. This example seems to have been dismissed as applying
only to "open" elections, i.e., ones in which some voters can know, in advance,
how all the other voters have voted. There might be good reasons for arguing
that this example is inconsequential; but saying that it only applies to such
open elections is not one of them. To see this, take my May 11th example and
multiply (across the board) the numbers of voters involved by 1,000,000. Then,
instead of 101 voters, there would be 101,000,000 voters participating in the
election. Also, instead of the closest pairwise matchups being decided by 1
vote, every pairwise matchup would be decided by at least 1,000,000 votes.
Accordingly, if the (now) 2,000,000 voters with preference order B,C,A could
take (i.e., could pay to take) a poll that is known to be accurate to within
1,000,000 votes, then those 2,000,000 voters could manipulate the election as
indicated in that example.
True, 1,000,000 divided by 101,000,000 is about 0.99%. But, with enough money
to spend on the poll, is this level of accuracy necessarily impossible?
I don't understand how any voting system could be vulnerable to manipulation
under the restriction that each voter has absolutely no idea how any other voter
will vote. So it seems to me to be just a question of how accurate a poll any
one group of voters can take (and yet still have time to inform the voters in
that group of the analyzed results of that poll before those voters must cast
their ballots).
Further, while 0.99% accuracy is good enough to guarantee manipulative
capabilities in that example, other examples can be constructed in which poorer
accuracy can still guarantee vulnerability to manipulation. Here's one:
Suppose that 99,000 voters will vote in an upcoming election, that A, B, and C
are the only candidates under consideration, and that 93,000 of the 99,000 will
cast ballots as follows:
26,000: A,(B&C)
2,000: (A&B),C
20,000: B,A,C
14,000: (B&C),A
31,000: C,(A&B)
-------
93,000: subtotal.
Suppose that the remaining 6,000 voters belong to a well organized group, and
that each of these voters has a strong preference for B over C, a strong
preference for C over A, and a very strong preference for B over A; i.e.,
93,000: subtotal
6,000: B,C,A
-------
99,000: total # of voters.
Suppose that the voting method being used in the election is the version of
Condorcet's method as "officially" defined on this list (EM-Condorcet).
Accordingly, ranked ballots are used, with ties and truncations allowed.
Suppose that this highly organized group of 6,000 voters is planning to take a
very accurate poll of the general electorate in time to advise the members of
the group on how they should cast their ballots. How much accuracy is needed?
If the poll were perfect (and if I didn't make a computational error), the
possibilities are as follows:
If those 6,000 voters vote honestly, then their second choice, C, will win.
If those 6,000 voters do not vote at all in that election (e.g., they "stay
home"), then their first choice, B, will win.
If they decide to vote, but to truncate instead of voting honestly (i.e., to
vote:
2,000: B,(A&C)
instead of:
2,000: B,C,A,
even though they strongly prefer C over A), then B wins again.
Of course, if they truncate, they would be the only voters who don't express a
preference between A and C, and so they might decide to reverse C and A on their
ballots instead of voting honestly (i.e., to vote:
2,000: B,A,C,
instead of:
2,000: B,C,A,
even though they strongly prefer C over A). B wins in this case also.
In short, according to the EM-Condorcet voting method, if those 6,000 voters
vote honestly, then their second choice, C, wins; but if they don't vote, they
truncate, or they pair-reverse C and A, then their first choice, B, will win.
For comparison (assuming I did the calculations correctly):
Voting Stay Pair-
Method Honest Home Truncate Reverse
--------------- ------ ---- -------- -------
EM-Condorcet: C [B] [B] [B]
Condorcet(1\2): C C [B&C] [B]
Borda: B C B B
Hare: C C C C
Kemeny: C C [B&C] [B]
Nanson: C C C [B&C]
Plurality: C C C C
Pl-runoff: C C C C
Reg.-Champ.: C C C C
Candidates in brackets, i.e., [B] or [B&C], indicate places where strategic
advantage is gained by manipulation.
Further, every pairwise matchup, and every "votes-against" result used by the
EM-Condorcet voting method, is decided by at least 3,000 votes here. Thus, a
polling accuracy of 3,000 divided by 99,000, or about 3.03%, would seem to be
sufficient to allow those 6,000 voters to manipulate this election.
As before, this example certainly does not mean that the EM-Condorcet voting
method is necessarily worse than any other particular voting method. To me,
what it means is as follows. It is not good enough just to state that EM-
Condorcet satisfies some particular criterion and just to state that this
criterion is obviously important to satisfy. Instead, EM-Condorcet should be
treated like any other voting method. A claim that any particular voting method
satisfies any particular criterion needs, at a minimum, 1) a precise definition
of the voting method, 2) a precise definition of the criterion, and 3) a
valid proof that the voting method satisfies the criterion. Without such
proof, the "claim" should either be dropped or be explicitly labeled as a
conjecture which may or may not be true. Of course, examples that illustrate
the voting method, and a good discussion of the importance of the criterion
along with suitably illustrative examples, can be extremely helpful and are
invariably quite desirable. But examples, by themselves, are insufficient to
justify the validity of such a claim.
Bruce
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