[EM] Another kind of approval equilibrium
fsimmons at pcc.edu
Thu Dec 4 15:32:10 PST 2003
This is a second installment on this topic.
In the first post on this topic, I finished with a claim that there is a
way to get around the final random drawing step (i.e. the "enforcing" of
the equilibrium probabilities) that was a feature of my previous
The key idea that obviates the necessity of that drawing is the use of an
ensemble of statistically close ballot sets to find the probabilities.
Roughly speaking, suppose that E is a large (and randomly chosen)
collection of ballot sets which are close statistically to some ballot set
S of approval ballots, and suppose that candidate X is the approval winner
in 37 percent of the members of E. Then BEFORE tallying up the ballots of
collection S we could argue that X has about a 37 percent chance of being
the approval winner of ballot set S.
Furthermore, if the members of E are so close to S that it would be
difficult to distinguish them by any pre-election poll, then the 37
percent figure would be more reliable than any pre-election poll estimate.
There are many ways to generate a suitable ensemble E from a ballot set S,
but here is a simple method that is probably adequate:
Let N be the number of ballots in the ballot set S.
To get one member of E, randomly draw N ballots from S. Of course this
drawing must be "with replacement" in order to get anything different from
To get additional members for E, repeat the procedure (using different
seeds for your random number generator).
At the risk of straining your patience, I wish to point out that if this
simple method of generating E is used, then it isn't really necessary to
generate E after all. We can consider the ideal case of E consisting of
(one each of) all N^N possible results of such a drawing, and figure the
candidates' winning frequencies analytically from the statistics of the
ballot set S.
If S is a set of approval ballots, then the relevant statistics are the
sample means and covariances of the candidate approvals, a set of
M*(M+1)/2 numbers if there are M candidates. Furthermore these statistics
are summable over the ballots.
One might object that once we know the approval means for S we know the
winner, so it doesn't make sense to consider the calculated frequencies as
In reply I enjoin you to remember that these statistics are unavailable to
the voters at the time of voting, but if they did have access to these
frequencies through some oracle, their best strategy would be to use them
exactly as they would bonafide probabilities.
Now leaving behind these technical and philosophical details, suppose that
we start with a set of N cardinal ratings (CR) ballots which rate M
1. Initialize all winning probabilities P1, P2, ... as 1/M.
While no candidate has 100 percent probability of winning do
2. Use winning probabilities P1, P2, ... and the "above the mean"
strategy to generate a set S of approval ballots.
3. Use this set S to update winning probabilities P1, P2, ... as per our
Repeat these two steps until the updated probabilities are within some
preset tolerance of the previous estimates.
4. Declare as winner the candidate with the most approval in the ballot
set S, as last updated in step 2 before exiting the loop.
That's my idea in a nutshell. Of course there are many possible
variations, including versions based on ranked ballots.
One variation that doesn't involve iteration would be to use Rob LeGrand's
ballot-by-ballot method one hundred times on the original set of ranked
ballots to get a reasonable estimate of the equilibrium probabilities, and
then use Joe Weinstein's "weighted median" method on ranked ballots for
the final outcome if the same candidate didn't win all one hundred times.
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