[EM] Help naming a new method
Andy Jennings
elections at jenningsstory.com
Sat Apr 30 14:42:18 PDT 2011
"Chiastic". Great word. I love the image. Thanks for the suggestion. I
think the finalists right now are "chiastic average" and Jameson's "mutual
median".
On Fri, Apr 29, 2011 at 11:53 AM, <fsimmons at pcc.edu> wrote:
> This method can be generalized by replacing the line y=x with any monotonic
> graph that connects the corner points (0, 0) and (100, 100) of the square
> determined by these diagonally opposite corners.
>
Yes, in my dissertation (http://ajennings.net/dissertation.pdf) I call this
the grading function and I give a formula for the grading function that will
minimize the L_p-norm distance between the voter's grades and the societal
aggregated grade (for a given p and a given expected grade distribution).
The grading functions that turn out to have especially nice properties are
F(x)=x (this chiastic average) and F(x)=50 (the median).
A limiting case in which the line y=x is replaced by the lower and right
> sides
> of this square is particularly simple:
>
> Elect the candidate whose lowest rating (over all ballots) is greatest.
>
> In case of a tie at level r, among the tied candidates elect the one with
> the
> greatest number of ballots rating her above r.
>
> It follows from this tie breaking rule that if no candidate is rated above
> bottom on all ballots, the candidate with the greatest number of above
> bottom
> ratings wins.
>
> If instead of this deterministic tie breaking rule ties were broken by
> random
> ballot, then this limiting method would be a nice solution to Jobst's
> consensus
> method challenge.
>
Just speaking about compromise in general and not about Jobst's consensus
challenge for a moment:
I do think this chiastic average (or the "mutual median" or whatever we end
up calling it) would encourage compromise in some cases. If we have options
A and B:
55 people: A-100, B-0
45 people: A-0, B-100
So A has a chiastic average of 55 and B has a chiastic average of 45.
Suppose we come up with a compromise solution C that everyone is okay with.
If we can get all of the B voters to sign on to this solution (rate it
above 55), and we can get more than two-elevenths (18%) of the A voters to
rate it above 55, then C can win. But if the voters are all strategic (as
Jobst's consensus challenge specifies) then all the A voters will bury C and
it will never achieve a chiastic average above 45 and a majority can indeed
force their will upon the minority. But for sure if there is no majority, I
think the chiastic average will encourage compromise.
I thought a little bit last year about whether there was a grading function
that could solve Jobst's challenge, but was only considering deterministic
tie-breakers which would all end up choosing A. Forest's suggestion (choose
the option with the highest minimum grade) with a fallback to random ballot
is indeed a neat solution to the consensus challenge. For Jobst's specific
problem, you could choose any grading function which was equal to zero on
x=0 to 55, but it could increase above zero for x>55.
Choosing the option with the highest minimum grade can actually be a pretty
good system for making decisions in small groups. If everyone is forced to
accept the outcome, then minimizing the maximal inconvenience caused to
anyone in the group is, in some sense, a very fair way to make the decision.
It's how I decided on a date and time for my dissertation defense, because
I wanted to minimize the maximal inconvenience caused to any of my advisors.
Thanks for your thoughts, Forest!
Andy
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