[EM] MJ, Approval, Score, and Chicken Dilemma

[Jameson Quinn]

Michael Ossipoff recently claimed to have shown that MJ had a bigger
problem with the chicken dilemma than approval or score. Though he'd made a
basic mistake (apparently he thought that MJ used a coinflip as a
tiebreaker), it's still an interesting question, and worthy of sensible
discussion.

So, let's take a realistic chicken dilemma scenario (using XYZ because I'll
use ABCDF for grades in MJ):

30: X>Y>>Z
25: Y>X>>Z
45: Z>>X=Y

Why do I use 30, 25, 45 instead of 27,24,49? Because the tight margins in
the latter case means that even 1 or 2 percent of voters who prefer X>Z>Y
or Y>Z>X would upset the whole result. Since in reality there will always
be such people (for instance, the existence of Nader>Bush>Gore voters is
well-documented), I think it's best not to set a scenario too close to the
edge. On the other hand, I don't want to be accused of making the scenario
too easy, so I'm not using my usual numbers of 35, 25, 40.

Obviously, under Score, or MJ, with honest voters, X will win. Under
approval, if all voters approve everything to the left of >>, then X and Y
will tie; but if some small, constant fraction approve only their favorite
instead, then X will win.

It's worth looking a little harder at what happens with MJ. Say the grades
are A/B/C/D/F and the X voters all give Y a B, while the Y voters all give
X a B. Now the median for both X and Y will be B. But the tiebreaker will
suss out the difference and X will win.

That will happen using either the MJ tiebreaker, which involves removing
median votes from both tied candidates simultaneously until they have a
different median; and the CMJ tiebreaker, which involves adjusting each
median by the ratio of [the excess of votes above versus below the median]
to [twice the votes at the median]. The former is like¹ using the smallest
trimmed mean possible which shows a difference; the latter is like² using
the smallest trimmed mean possible which encompasses all the votes at
median.

So, what happens if a few voters are strategic? In approval, since the
"honest" result is a tie, it only takes a tiny number of strategic Y voters
to swing the result from Y to X. Similarly, in range, if the "honest" vote
for the X and Y factions is to vote the second choice relatively high —
say, 90/100 — then this is almost the same as the situation in approval; it
takes just a tiny 0.6% fraction of strategic Y voters to overcome X's
advantage.

In MJ, the median points for X and Y (50%) are closer to the 60% border
with C than they are to the 30% or 25% (respectively) borders between
factions. Thus, even 1 strategic voter who lowers the other candidate from
an honest B to a semi-strategic C will succeed. It would appear, then, that
MJ is at least as bad as approval, and certainly worse than range.

However, I think this appearance is false, an artifact of the
oversimplistic scenario. In real life, the X faction (for instance) will
never be perfectly homogeneous. Instead of all of them grading Y at B, they
will split between giving Y a B or a C, with perhaps a few of them giving Y
an (honest) A or D or F. Thus the grades for Y will be something more like:
25%: A (from the Y faction)
15%: B (from 50% of X supporters)
10%: C (33% of X supporters)
5%: D (17% of X supporters)
45%: F (Z supporters)

Meanwhile, X's grades will be:
30%: A
12.5%: B
8.33%: C
4.17%: D
45%: F

With the numbers above, if the Y faction starts to use strategy, starting
with the strongest anti-X partisans among them, it would take a full 5% of
strategic voters to shift the result in MJ or CMJ.

Obviously, it is possible to create a scenario which is not quite so
perfectly monolithic as the original scenario, where MJ still fails. But
you can see from the above why I believe that in realistic situations,
strategy will NOT be as tempting for the marginal X or Y voter.

However, in this situation, it's not only the voters who can act
strategically. Consider the situation of candidate Y, deciding whether to
run an "honest" negative ad about Z, or a "backstabbing" negative ad about
X. While there is certainly a risk of it backfiring into voter disgust,
from a purely game-theoretic perspective the "backstabbing" ad is more
attractive. Starting from the original scenario, that holds true in all
three systems: Approval, Score, and MJ.

If voters are purely rational, they can use Strategic Fractional Rating
(Mike Ossipoff's SFR) to guard against a Z win. For instance, in this
situation, the polls say that Z has 45% support, +/- 2%. For safety's sake,
let's call it 47%. So X and Y together have 53%. If the X and Y factions
were equally matched (26.5% each), then under score, each faction would
have to give the other at least 23.5/26.5 = 89/100 on average to ensure
that at least one of the factions (the larger) would win. Similarly, under
approval, each faction would have to approve the other with at least 89%
probability. In my opinion, 89% is quite a high level of cooperation; the
kind of thing that even a few backstabbing ads would almost certainly
overcome.

However, in MJ, it only takes an 89% chance of giving the other faction's
candidate anything above an F. It's much easier to cooperate when
cooperating means giving your favorite an A and the other faction a D, than
when it means giving the other candidate an 89/100 or even a full approval
on the same level as your own candidate. In fact, the good thing about MJ
is that, as explained above, if everyone is honest, you already have up to
a 5% buffer against strategy — the maximum possible. And if voters want to
be strategic, but not to use any probabilistic strategies that would risk a
Z victory, they can simply give the other faction's candidate a D. (And
generally, those who are most inclined to be strategic will be exactly
those who would honestly rate the other faction's candidate lowest; as long
as this group is smaller than the half the combined margin of victory over
Z, their strategy will have little or no impact).

OK, that post is long enough for now. If anyone has any questions, I'd be
happy to clarify.

Jameson

¹"Like" in this case means "the same, except in cases which are vanishingly
improbable with realistic numbers of voters"

²"Like" in this case means "gives the same ordering of candidates, except
in some cases if the trimmed mean in question would include more than two
grades; which is to say, if there are more votes at the median than at the
grades next door; which is to say, if the cumulative grade distribution is
unusually bendy around the median."
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