Mike R wrote:
> Steve Eppley wrote:
-snip-
>> The obvious question is, why prefer Kemeny's method?
>> What criteria does it satisfy that other methods fail
>> that are more important than the criteria other methods
>> satisfy that Kemeny fails?
>
> I like Kemeny-Young is because it has many of what I
> consider "must have" voting characteristics (Condorcet and
> Extended Condorcet, especially), plus it removes voting
> cycles, is resistant to voting manipulation, and it orders
> all of the candidates (useful when more than one candidate
> can win or when an elected candidate cannot serve).
I'm not sure what Mike means by some of those terms.
(Extended Condorcet? Cycle "removal"? Voting
manipulation "resistance"?) I wonder if he can tell
me if MAM, described at www.alumni.caltech.edu/~seppley,
meets those criteria. MAM does satisfy the above
criteria that I do understand: It chooses within
the top cycle, hence is Condorcet-consistent, and
it socially orders all the candidates. (Like Kemeny,
MAM's ordering satisfies local independence of
irrelevant alternatives, and, unlike Kemeny,
MAM satisfies a stronger criterion, immunity
from majority complaints.) Also, MAM can be
tallied in a time that's a small polynomial
function of the number of voters and the
number of candidates.
-snip-
> If there is a method you prefer, you might show me
> an example of how Kemeny gets it wrong compared
> to your favorite method.
-snip-
Kemeny finds the "best" social ordering by finding
the ordering that maximizes the sum of sizes of
majorities that agree with it. That makes it
vulnerable to clone manipulation. MAM is equivalent
to analogous method that finds the "best" social
ordering by using the maxlexmax comparitor rather
than the sum comparitor, so it's independent of clones.
Here's an example to consider, involving Kemeny's
vulnerability to clone alternatives. (I hope I have
the example straight. I whipped it up fairly fast.)
9 voters, 3 alternatives:
4: A>B>C
3: B>C>A
2: C>A>B
There are 3 majorities:
6 voters rank A over B.
7 voters rank B over C.
5 voters rank C over A.
Nearly every method elects A here, yes? Kemeny and MAM
pick A>B>C as the social ordering. Kemeny picks A>B>C
because it agrees with 13 (the majority of 6 who voted
"A>B" + the majority of 7 who voted "B>C"). Note that
the social ordering B>C>A agrees with only 12 (the 7
"B>C" majority + the 5 "C>A" majority).
(There's another definition of Kemeny that finds the social
ordering that maximizes the sum of pairwise preferences
that agree with it, rather than the sum of sizes of
majorities that agree with it, but its principle is
nearly equivalent, and a similar example illustrates
its problem with clones.)
Now add in one or more clones of C. Each clone added
in adds another majority of size 5 to the Kemeny-best
social ordering that ranks C over A than to the
Kemeny-best social ordering that ranks A over C.
Thus, adding a clone of C, say C', causes either
B>C>C'>A or B>C'>C>A to become the Kemeny-best
social ordering, changing the outcome from A to B.
On informational grounds this is illogical, since
no information about A or B is gained when voters
also rank alternatives that are nearly identical
to C. Worse, it creates incentives to manipulate
the outcome by strategic nomination, and it could
even lead to routine farces where each faction
nominates as many (clone) alternatives as they can
find that will pairwise-beat the alternative(s)
that will pairwise-beat their favorite(s).
(Using the other definition of Kemeny above, 2 or 3
clones of C would be added in part two of the example.
Adding 2 leads to a tie between A and B, whereas
adding 3 leads to electing B outright, assuming
I haven't made a math mistake. With some more
thought, I believe I could construct a "tighter"
example where adding 1 clone leads to a tie
between A and B and adding 2 clones leads
to B winning outright.)
If one considers immunity from majority complaints
to be a reasonable criterion, that's another reason
to prefer MAM over Kemeny (and other methods).
--Steve
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