Since it's taking longer than I expected to write a long reply to Mike's
post, I'll give some criterion failures for Condorcet//FPP while I work
on the reply.
Independence from clones (as I understand it) is defined as follows
(Tideman, 1987, "Independence of Clones as a Criterion for Voting Rules"):
Let X denote the set of alternatives.
Call Y (subset of) X a set of exact clones if, for all y,z (in) Y,
every voter ranks y equal to z.
Call Y (subset of) X a set of clones if, for all y,z (in) Y and all
x (in) X\Y, every voter who ranks x
over y also ranks x over z and every voter who ranks y over x also
ranks z over x.
For all Y (subset of) X which is a set of clones, all x (in) X\Y
such that Y union {x} is not a set of
exact clones, and all Y' which is a strict subset of Y, the
probability that x is elected
must not change if alternatives in Y' are deleted from all votes.
Or, less formally (but still somewhat, if I got it right):
Call Y a clone set if all candidates in Y have the property that:
- every voter who ranks a candidate in Y over some other candidate q,
ranks all candidates in Y over q,
- and every voter who ranks some candidate q over a candidate in Y,
ranks q over all candidates in Y.
Then eliminating a strict subset of Y should neither increase nor
decrease the probability that r is elected the winner, where r is any
candidate outside Y not equal-ranked with Y by every voter who ranks
both Y and r.
While the formal proof does not explicitly say so, I think that "ranks A
above B" means "ranks A strictly above B", since Tideman makes use of
"ranks y equal to z" in his definition of a set of exact clones, and he
does not mention equal-rank otherwise in the conditions for the
independence from clones criterion.
-
On to some criterion failures:
If there is no truncation or no equal-rank, then (I've been told) ICT is
equal to Condorcet//FPP. Hence, finding an example where Condorcet//FPP
fails independence from clones with no equal-rank or truncation suffices:
70: A > B > C
68: B > C > A
66: C > A > B
There's no CW, so A wins.
Now clone A into A1 and A2:
35: A1 > A2 > B > C
35: A2 > A1 > B > C
68: B > C > A1 > A2
66: C > A2 > A1 > B
Now B wins. Cloning A made A lose (or B win).
And some other failures while I'm at it:
Reversal symmetry:
1: A>B>C
1: C>B>A
1: B>A>C
1: C>A>B
There's no CW and reversing every ballot leads to the same ballot set.
Hence, the outcome must be A=B=C (since the method can't distinguish
between this set and its reverse), but C has a Plurality count of 2 and
thus wins.
Condorcet loser, majority loser, mutual majority, Smith, Schwartz,
Landau, others?:
2: A>B>C>D
2: B>C>A>D
3: D>C>A>B
There's no CW. The Smith set is {ABC}. 4 voters, constituting a
majority, rank the candidates {ABC} above D. Yet D has the greatest FPP
count and thus it wins.
I also think this example shows C//FPP failing Minimal Defense, but I'm
not sure - I don't know MD that well. It might also show that C//FPP
fails GSFC, but I'm even less sure of that.
Independence from Smith-dominated alternatives:
Start with the Condorcet loser example above, but without D. Then C
wins. Now add D, which is not in the Smith set. Then D wins, so the
method fails ISDA.
Mono-add-top, Participation:
51: A>B>C>D
50: B>A>C>D
50: C>B>A>D
51: D>C>A>B
1: D>B>C>A
There's no CW, so D wins by an FPP count of 52. Now two more D>B>C>A
voters appear:
51: A>B>C>D
50: B>A>C>D
50: C>B>A>D
51: D>C>A>B
3: D>B>C>A
B is now the CW and wins.
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