Condorect sub-cycle rule
David Marsay
djmarsay at dra.hmg.gb
Fri Oct 3 05:08:10 PDT 1997
Markus has kindly provided another example.
100 voters, 6 options (I don't want to know about examples with more
options!)
BCDFEA 25
CDFEAB 24
ABFECD 20
EABCDF 15
EBCADF 8
ECADBF 4
ECABDF 4
I tally pair-wise majorities
AB 67-33=44
AC 35-65=-30 BC 68-32=36
AD 51-49=2 BD 72-28=44 CD 100
AE 20-80=-60 BE 45-55=-10 CE 49-51=-2
AF 51-49=2 BF 76-24=52 CF 80-20=60
DE 49-51=-2 DF 80-20=60 EF 31-69=-38
I hope that that is right.
I assume that there are lots of cycles, so I start 'at the top'.
T=100:CD
T=60:CDF,EA
T=44:CDF,EABDF
T=38:gives FE, which creates a
cycle. Thus FE must be ignored.
T=36:gives BC, leading to
EABCDF, which I claim is the 'obvious' solution. It would be
reasonable to declare a tie between CE, on the basis that the
evidence is unclear, but if pushed E must be chosen, particularly as
it has a pair-wise majority of 2 over C.
Note that one has a non-uniform threshold, with some evidence being
accepted even though it has less support that some other evidence in
the same 'base cycle' that is discounted.
I f I had started at the bottom, for T<36 I would have the above
solution as remaining, eliminating only contrary evidence. At T=36 I
would have a cycle (EABCDFE), and so might be tempted to reject BC as
having the least evidence. However, deleting it does not resolve the
'impossible opinion'. For this T=38. What next? Rejecting all
evidence below the threshold for each cycle is simple, but some of it
may be salvaged. In essence, one can carry on with the top-down
approach to break ties.
Thus, I suggest:
1) guess a threshold (e.g., 0)
2) check and increase where necessary to remove cycles
3) reduce where necessary to resolve ambiguities.
It is simpler to describe a pure top-down approach, but the above
mixed approach may be quicker. With computerisation, it hardly
matters, I guess.
I don't see how one can have one form of
words that is both appealing to voters and complete and unambiguous.
Starting with a low threshold seems intuitive and follows Condorcet,
but introduces complications.
Here's an attempt, following Condorcet:
"Ballots that express a preference for X over Y are tallied, for each
X and Y. A threshold, T, is found such that the preferences that
tally T or more form a consistent set. This will typically be about
half the number of voters. There may be some ties, in which case
these are broken using any remaining consistent evidence, with
precedence being given to that evidence with the highest tallies.
Typically, though, this stage is not necessary"
If abstentions are allowed then the method needs to be modified.
"In rare cases it can happen that there is a non-chosen option that
has a slight pair-wise majority over the chosen option. This is due
to there being no clear overall view held by the electorate. If the
existence of such an option is considered to grounds for rejecting
the selected option, then similar or stronger grounds could be held
against any of the options. Thus, although the method cannot always
select a clear winner, its results will stand up to scrutiny."
Unless we can find some way making Condorcet acceptable, the details
are of purely academic interest.
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Sorry folks, but apparently I have to do this. :-(
The views expressed above are entirely those of the writer
and do not represent the views, policy or understanding of
any other person or official body.
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