[EM] MJ: Worse Chicken Dilemma than Approval or Score, elaborate bylaws, computation-intensive count.

Michael Ossipoff email9648742 at gmail.com
Thu Sep 6 07:28:01 PDT 2012

On Thu, Sep 6, 2012 at 9:49 AM, Jameson Quinn <jameson.quinn at gmail.com> wrote:

> MJ's chicken dilemma is incontrovertibly less serious than Score's, and
> arguably less than Approval's.

Maybe that depends on one's arbitrary choice among the sets of elaborate bylaws.

But let's take an obvious and natural interpretation, and try it in
the original Approval bad-example:

Suppose a majority rate A at 0, and the rest rate A at s100. What's
A's median score? Well, if the right number of those zero-raters had
been a little more generous, and had given A a millionth, and one had
given A 1/2 of a millionth, you could establish A's median at 1/2 a

Therefore, if a majority of the voters rate A at an extreme, then it's
obviously fair and right to call that extreme hir median.

What if a not quite a majority rate B at zero, and a sub-majority rate
B at max, and the rest rate B at N?

An argument similar to that above shows that B's median should be taken as N.

Now, let's try that in the original, standard Chicken Dilemma:

Sincere preferences:

27: A>B
24: B>A
49: C

Actual MJ ratings:

27: A100, BN, C0
24: B100, A0, C0
40: C100, A0, B0

What are the candidates' MJ scores, by the above interpretation? Who wins?

MJ scores:

A: 0
B: N
C: 0

B wins. The B voters' defection has worked. The B voters have easily
taken advantage of the A voters' co-operativeness.

With Score, the A voters, by giving to B some fractional rating, could
have at least tried to give B just enough to win if B does better than
A, but not enough to win if B doesn't do as well as A. It's a
difficult estimate, but the A voters could make thereby make
successful defection more difficult. I call that Strategic Fractional
Rating (SFR).

With Approval, the A voters could do SFR probabilistically.

>> For computations in the count: I'd argue that it's actually easier to carry
> out in practice than Score. Even more so if you consider CMJ.

With Score, you add each ballot's rating of X to X's total.

With MJ, if one or two newly-counted ballots rate X above hir current
median, then you must raise X's MJ score to hir rating on the ballot
with the lowest X-rating above X's median (or maybe to the mean of two
such ballots?).

That means you have to go through the ballots again, to find the one
with the lowest X-rating above X's median.   ...unless you've sorted
all of the ballots, by their ratings, for each candidate.

You don't think that's a lot more computation-intensive than Score? (see above).

Michael Ossipoff

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