[EM] MCA strategy

[Kevin Venzke]

I put some thought into MCA strategy, and I think I have an observation
or two worth sharing.  Be warned, though, that I might have seriously
messed this up.  I invite corrections.

To calculate the expectation for placing two candidates in two different
ranks, I think you would use the following:

A as Favorite, B as Approved: W * Ma * Mb
A as Favorite, B as Unacceptable: (W * Ma * Mb) + (W * Aa * Ab)
A as Approved, B as Unacceptable: W * Aa * Ab

definitions:
W is A's worth minus B's worth.
Mx is the probability that X will win by having the largest majority of
"Favorite" rankings.
Ax is the probability that X will win by having the fewest "Unacceptable"
rankings.

These may suggest why it is not safe to approve a lousy candidate even
though you have named a favorite.  Unless both candidates have a shot
at getting a majority of Favorite rankings, you're giving your favorite
no edge at all.


For zero-info Approval, all you do is say that everyone's odds of winning
are equal.  In MCA this is complicated by there being two ways of winning
(get a majority of Favorite rankings, or have the fewest Unacceptable
rankings).  But this can probably be fixed, by estimating what the odds
are of the election being won by one method or the other.  It can be 50% if
one really doesn't know.

Let F be the probability that the method will be won by someone having
a majority of Favorite rankings.  We don't need to worry about specific
candidates' odds because they're all equal.  Zero-info expectations:

A as Favorite, B as Approved: W*F
A as Favorite, B as Unacceptable: W, or: W*F + W(1-F)
A as Approved, B as Unacceptable: W(1-F)

I started generating random elections, with trials to find the optimal
arrangement, and this is what I've been finding:
1. Adjusting "F" doesn't seem to do much.
2. The candidates you would Disapprove in zero-info Approval are 
usually the same ones you would Disapprove in zero-info MCA.
3. The "Approved" rank is almost never used.  It is used when a candidate's
worth happens to equal the mean worth.  Not sure why.

I'll write again if I come up with more.

Kevin Venzke
stepjak at yahoo.fr


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