[EM] Circular paradox (Re: maybe a new variant of Condorcet)
Peter Barath
peb at freemail.hu
Sun May 13 15:09:53 PDT 2007
>>I call a subset of candidates a quasi-clone set, if:
>>
>>1. they don't make up the whole set of candidates
>>2. for every candidate out of the set they are in
>>the same winning relation with (all beat / all tie /
>>all lose)
>This is similar to Forest Simmons' "beat clone sets" he uses in his
>Dec. 2004 "sprucing up" process idea.
>your proposal is very similar to Mike Ossipoff's subcycle rule.
I apologise for not reading the whole archive, but it's
volume is bigger now than that of the Holy Bible.
I realized that the mentioned proposals are really
close to mine, but not exactly the same. But I have
to admit that my proposal still fails the Pareto
criterion.
Well, I think it's just natural that if you are new
in the field, you dream of working out something new.
So, I kept thinking in the last weeks, hoped for a new
method, then for at least a new criterion, and finally
got a paradox - only hope that it's new and interesting
for some extent.
The beginning is: if in a Condorcet voting three candidates
are in circular ambiguity, let's say A > B > C > A, it only
can be caused by circular sets of votes:
ABC
BCA
CAB
We can call a set of ballots a circular set, if there is
a permutation wich if applied on each ballot, we get
the same set of ballots. Here an appropriate permutation
is: move the first element to second place; move the
second element to third place; move the third element to
first place. So it makes ABC into CAB, makes BCA into ABC,
and makes CAB into BCA, which are the same ballots, only
in different order.
It seems obvious that these ballots give wholly symmetrical
positions to the candidates, so we can eliminate them.
Unfortunately, this usually works only on three candidates.
These three ballots
ABCDE
CDEAB
EABCD
also makes a circular ambiguity, but there is no circular
set to eliminate - it would need 5 ballots.
So as a method it fails, but still, I hoped for a
criterion: eliminating a circular set must not change
the result. Disappointment again.
Let's see the following 9 ballots:
"the nice group"
BCDA
CDAB
DABC
"the ugly group"
DCAB
BADC
CDBA
"the completter"
ABCD
"the two others"
DCBA
CDBA
The "nice group" and the "completter" together makes
a circular set:
BCDA
CDAB
DABC
ABCD
so we can eliminate them, and remains:
DCAB
BADC
CDBA
DCBA
CDBA
in which D is the Condorcet and also the Instant Runoff
winner. But the "ugly group" and the "completter" also
makes a circular set:
DCAB
BADC
CDBA
ABCD
which is not so easy to see, but use the following
permutation: move the first element to third place,
the second to fourth, the third to second, and
the fourth to first. It transforms each ballot
into the next one in this set.
Removing them, remains:
BCDA
CDAB
DABC
DCBA
CDBA
in which C is the Condorcet and also the Instant Runoff
winner. So this criterion seems logical at the first
sight but useless in the second one (Borda complies with
it, but otherwise works poorly).
Peter Barath
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