[EM] A PR quality gauge for lotteries

Forest Simmons fsimmons at pcc.edu
Wed Dec 7 13:05:43 PST 2016


Dear Friends,

Given a collection beta of cardinal ratings ballots, where the ratings are
on a scale from zero to one, and a lottery L on the candidates, here's a
way to gauge the Proportional Representation quality Q of the lottery L
relative to the collection of ballots beta:

For x between zero and one, let G(x) be the fraction of ballots for which
the lottery expectation is greater than x.

The quality gauge Q is given by

the integral over the interval 0<x<1 of the integrand G(x)^(ln(1-x)/ln(x)).

Note that Q is always in the interval [0, 1], and attains the value of one
only if G(x) is identically one of the interval 0<x<1.

If the lottery is uniform on a sub-set of candidates, then Q gauges the PR
Quality of that slate of candidates.

In particular if that subset is a singleton, then Q is a gauge of the
quality of the single candidate as a representative of the electorate.

If all of the ballots are voted approval style, and the lottery is
restricted to a single candidate C, then the function G(x) will be
identically equal to the fraction of the ballots that approve C.  Therefore
the candidate with the greatest approval will get the greatest value of Q.
In other words, the approval winner will win in a single winner approval
style election.

It does not follow (and is emphatically not the case in general) that a
slate of more than one candidate wins solely on the basis of most total
approval.

Is it true that a candidate rated at 100% on fifty percent of the ballots
gets the same Q as a candidate rated at fifty percent on 100 percent of the
ballots?

No.  The latter gets a Q value of 50%, while the former gets the smaller
value of Q equal to the integral over 0<x<1 of the integrand

(1/2)^(ln(1-x)/ln(x))

which is less than 48 percent.

In general, if the lottery expectation is constant over all ballots, then
that constant value will be the value of Q.

Now try it on some of your own favorite lotteries.

In another email I'll explain where the exponent of the integrand comes
from.

My Best,

Forest
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