Form a matrix M whose entry in row i and column j is the percentage of
ballots on which alternative j is the highest ranked alternative that is not
majority defeated by alternative i.
[By definition no alternative is majority defeated by itself, so if every
alternative ranked higher than alternative i is majority defeated by i, then
alternative i itself is the highest ranked alternative that is not majority
defeated by i.]
Each column of M sums to 100 percent, so M is a stochastic matrix. In fact, M
is the transition matrix for the following Markov process:
Given an alternative A(k), let A(k+1) be the highest ranked alternative that
isn't majority defeated by A(k) on a randomly chosen (and replaced) ballot.
To each initial probability distribution vector V there corresponds a steady
state distribution vector V' = MV which can be approximated by pre-multiplying
V by a high power of M. In the limit of infinitely many iterations (of V
replaced by M times previous V) we get V' .
For our proposed lottery method we take the initial state vector V to consist
of the random ballot probabilities for the respective alternatives.
Then the lottery determined by the 100th power of M times V is equivalent to
choosing A(0) in the (above mentioned Markov process) by random ballot, and
letting A(100) be the winner.
If I am not mistaken, increasing the rank of alternative i relative to the
other alternatives cannot decrease the steady state probability of alternative
i.
Also, if alternative i is replaced by a clone set {x,y,z}, then the steady
state probabilities of the other alternatives are not affected. And,
furthermore, the steady state probabilities of x, y, and z will be in the same
relative proportion as they would be if all of the other alternatives were
deleted from the ballots.
Now let's leave the realm of non-determinism for awhile and consider an idea
for a related deterministic method:
Let v1, v2, v3, etc. be the initial state vectors consist of all zeros except
for a one in the respective positions 1, 2, 3, etc.
Let v1', v2', v3', etc. be the corresponding steady state vectors.
The alternative i that loses the least probability in going from vi to vi' is
the winner.
I don't have time to post any examples right now, but I will later.
Forest
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