## Ballots with cycles (reposts of excerpts + new thoughts)

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• Andrew Myers: Suppose that in a Condorcet system, we allow people to submit a ballot that has an arbitrary preference relation, so any two alternative A and B
Message 1 of 2 , Mar 5, 2008
Andrew Myers:
Suppose that in a Condorcet system, we allow people to submit a
ballot that has an arbitrary preference relation, so any two
alternative A and B can have either A<B, A=B, or A>B. There can
therefore be cycles in the graph of preferences, like A<B<C<A.

One reason why we might want to set up the system this way is that we
can protect voter privacy better by separating different preferences
during the tallying process.

The question is whether this creates new strategic voting
opportunities. I have not been able to construct a scenario where it
makes strategic voting more powerful.

Juho Laatu:
This makes it a bit easier to intentionally generate a loop among say
three candidates (A,B,C) of the competing party. My vote could be
X>A, X>B, X>C, A>B, B>C, C>A, where X is my own party candidate. If
many X supporters vote systematically this way there is a chance that
the candidates of the competing party will all lose to each others,
and that might make X the winner in some Condorcet methods like
[Simpson-Kramer] minmax if the race is otherwise very tight between
the two parties.

WDS illustrates to make it more concrete:
compare
#voters their vote
3: A>B, B>C, C>A
with each candidate defeated by a 3:0 majority,
versus

1: A>B>C
1: B>C>A
1: C>A>B
with each candidate defeated by a 2:1 majority (not as severe).

The former is more powerful in the Simpson-Kramer minmax system
(in which the only thing that matters is the strength of your
most-severe defeat) as a way to hurt the ABC party.
Indeed if there were 49% X-voters and 51% ABC voters then
X would normally lose to each of ABC, indeed is the Condorcet loser.
However with the X-voters strategically playing the cycle game, X
suffers only tiny-margin defeats while A,B,C each suffer severe
pairwise defeats -- therefore X now wins the election under
Simpson-Kramer.

WDS:
permitting cycles in ballots has the big advantage that it makes
Condorcet possible even on a "dumb" totalizing machine (it is beyond
the capability of such a machine to detect cycles and thus forbid
illegal votes, but if we make them be LEGAL votes, then no problem).
If there are few enough candidates that the labor of providing an
all-pairs vote is not ridiculously large, then Condorcet would

CONCLUSION:
Hence it is worth investigating this further - can you prove a theorem
- for various natural kinds of Condorcet methods that are based on
pairwise matrices (Schulze? Tideman?) - that cyclic votes are never
strategically useful?
• I m not totally sure, but think ballots with cycles might also be strategically useful with Schulze-beatpath style of Condorcet voting. Here s a sketch of
Message 2 of 2 , Mar 5, 2008
I'm not totally sure,
but think ballots with cycles might also be strategically useful
with Schulze-beatpath style of Condorcet voting.

Here's a sketch of artificial scenario to which may be an example.

Again, suppose it is again X on one side, versus A,B,C
(and perhaps D,E,F... too) in the opposed party, and
normally there would be a moderate-strength cycle X>A>B>X.

Hence, normally, it would not be clear a priori who is going to win:
A,B, or X.

And aside from that:
Due to random fluctuations and a small fraction of
independent voters, normally X would
slightly defeat some of the ABCDEF party candidates pairwise,
while being defeated by a slight margin by the others.

However, suppose a big chunk of the ABCDEF-voters systematically vote
ABCDEF in a <-directed cycle.

In that case, Schulze says "aha, I see there are strong beatpaths
leading from each of {A,B,C...} to each of the others, and hence there
is a beatpath from each of A,B,C... to X whose strength in each case
is the same as the strongest pairwise defeat of X."

Schulze also says "X has a beatpath to each of A,B,C,... whose
strength is the same as the strongest pairwise victory by X, EXCEPT
since the B-A margin is weak, this only works for some not
others.

Hence Schulze concludes (with better probability) that X does not win.

hence the cylcing strategy paid off for the ABC.. party.

Warning: I'm not sure this really works.
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