[EM] How I figured my ballot

[Richard Moore]

While Mike is figuring out how to count the ballots, here's a bit of
info on
the strategy I used to come up with my approval votes in the poll to
select
a topic for the second poll. Naturally, I followed a zero-information
strategy,
but as we were discussing a few weeks back we don't really know what
the best ZI strategy is when there are few voters. At that time I made a

conjecture about a first-order approximation to a good ZI strategy for
such
cases, and that is what I used. Of course, I also had no way of even
knowing how many voters there would be, so that part is guesswork.

First, I rated every topic on a scale of 0 to 100, as sincerely as I
could.

Second, I ranked the topics from highest rating to lowest rating.

Third, I put this information in a spreadsheet. I programmed the
spreadsheet
to evaluate the first-order approximation I previously cited for the
utility
threshold. That approximation is

M + (S-M) * (n-1) / (N+n-1)

where n is the ranking of the candidate we are evaluating, N is the
number
of approval marks that are expected to be made (all other voters and all

candidates), M is the mean utility of the other candidates, and S is the

full-scale utility (100 in this case); the low end of the scale is zero.

Now the fun part. I tried different values of N to see the effect on the

number of candidates that I should approve.

For N > 240, the top 6 candidates pass (same as above-the-mean).
If 17 <= N <= 240, then only the top 5 candidates passed.
If 6 <= N <= 16, then only 4 candidates passed.
If 3 <= N <= 5, then only 3 candidates passed.
If 1 <= N <= 2, then only 2 candidates passed.
If N = 0, the spreadsheet blows up (divide-by-zero error), but it's
obvious you would only vote for your favorite candidate in this case.

So my conclusion for this particular case is that the first-order
approximation converges fairly quickly with the above-the-mean
strategy. Remember, if each voter only approves 3 candidates
(and in a field of 13, I would expect a higher number), then with
anywhere from 6 to 80 voters other than myself, I should approve
5 candidates. So, I felt fairly confident in approving my top 5
candidates in this poll.

Richard