[EM] “goal of a better election method”
Kristofer Munsterhjelm
km_elmet at t-online.de
Tue Feb 14 13:41:18 PST 2017
On 02/12/2017 12:48 PM, steve bosworth wrote:
> Re: “goal of a better election method”
>
> To all (but prompted by Sennet Williams and Rober Bristow-Johnson, and
> earlier by Michael Ossipoff, Richard Fobes, Kristofer Munsterhjelm,
> Steve Eppley,Jameson Quinn, and Kevin Venzke)
[...]
> I look forward to your feedback.
I suppose the concern that comes most clearly to mind is that MJ is not
meant to be a cardinal method, whereas using averages introduces a
cardinal (numeric) element.
To rephrase that, all that "plain" MJ knows about is that there's a
common standard among the people where
Excellent is better than Very Good
Very Good is better than Good
Good is better than Acceptable
Acceptable is better than Poor
and Poor is better than Reject.
It doesn't know what the common standard *is*, however.
To use a school grading metaphor, it knows that A is better than B, but
not that
A: 100%-97% of max attainable points
B: 97%-85%
(or what have you).
Suppose that we want to use MJ to determine the student with the best
performance. Then MJ is supposed to work equally well no matter whether
the grading scale is
A: 100%-97%
B: 97%-80%
(... etc)
of if the grading scale is
A: 100%-92%
B: 92%-77%
(... etc)
as long as all classes use the same grading scale (that's the "common
standard" qualification).
If you use averaging as part of an MJ method, then the ordinal
assumption is violated, because taking the mean doesn't just depend on
the relative order (i.e. A is better than B), but also on just how much
(A has a mean score of 98.5%, B has a mean score of 88.5% vs A has a
mean score of 96% and B has a mean score of 84.5%).
As a consequence, the method may produce better results in certain
scenarios (where the common standard is equally spaced), but it trades
that off by potentially producing worse results in other scenarios
(where the common standard is not equally spaced).
Or more simply put: making the scale cardinal, which you need to be able
to calculate means on the votes, adds more assumptions that may not be true.
For more on this particular objection, see B&L's "Election by Majority
Judgment: Experimental Evidence", p 45. and onwards ("Voting by Points
and Summing"). Quoting:
>> But, is it reasonable to use numerical scales in voting? The answer is a resounding
>> no, for several reasons.
(My source is
https://1984f707-a-62cb3a1a-s-sites.googlegroups.com/site/ridalaraki/xfiles/ElectionByMajorityJudgment(ExperimentalEvidence)Final.pdf)
>From an MJ point of view, only using an inferred numerical scale for
tiebreaking is surely better than using it everywhere (like in Range).
But the same theoretical arguments apply as soon as you're using a
numerical scale at all.
There's also a strategy argument, which could be generally argued like this:
- Either ties of the type where many candidates have the same median are
common or they're not.
- If they're not common, then a mean tiebreak is not going to change the
outcome often, so we can go with Bucklin or MJ's system.
- If they're common, then, since Range is more susceptible to strategy
than MJ, using a mean tiebreak will make HMJ considerably more
susceptible to strategy as well, and so should be avoided.
See this slide set by B&L for more on that:
http://igm.univ-mlv.fr/AlgoB/algoperm2012/01Laraki.pdf . In particular
they say:
"The unique [social grading function]s that are partially
strategy-proof-in-ranking are the order functions."
This means that the only grade-based voting methods that are partially
strategy-proof in ranking (which they define earlier) are order
functions, which are of the form "return the nth best grade" for some n.
(The median sets n=voters/2) Consequentially, to keep the SGF partially
strategy-proof-in-ranking, the tiebreak should also be an order function
with a different n, which it is in MJ and Bucklin, but not in HMJ.
B&L do not analyze the HMJ variant in itself, but they note:
"The unique aggregation functions that minimize the probability of
effective-manipulability are the middlemost. Point-summing-methods, f_1
and f_n maximize this probability."
which does suggest that using a point-summing method as a tiebreak will
weaken the method more than using another type of method.
In addition, averaging would make a method no longer
strategy-proof-in-grading. For the same reason that the median (being an
order function) makes MJ resistant to strategy, an order function tie
break makes MJ resistant to strategy in the tiebreak. B&L say:
"The unique strategy-proof-in-grading SGFs are the order functions.
If the mechanism is a point-summing method (the mean with respect to
some parametrization), for almost all profiles, all voters can manipulate."
(Last slide on "Strategy in Grading")
-
This is not to say that HMJ is a *bad* method. I would certainly choose
it if the alternative was, say, IRV or Borda. But using averaging does
undermine B&L's theoretical foundation of MJ (since it is no longer an
ordinal grade method), and if you accept their strategy-resistance
arguments, it also weakens MJ's resistance to strategy.
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