Puzzle (Wisest allocation of vote-weights):
Suppose there are some very large number N of independent
voters who need to make
a yes/no decision.
We suppose this decision has a "right" and "wrong" answer.
Suppose voter K would have (independent) probability PK
of making the "right" decision on her own. (Hopefully all the PK≥1/2,
i.e. hopefully humans are better decision-makers than coin tosses.
The PK may
differ because some voters are wiser or better informed than others, but
we'll assume the PK-1/2 don't differ too tremendously.)
Suppose voter K gets to cast WK votes.
Question:
What (approximately)
is the best choice of the weights WK, maximizing the chance that
the collective decision will be the right one?
Answer: The probability society makes the right choice is approximately the same as the probability that a normal random deviate with expectation value E=∑1≤K≤N (PK-1/2)WK and variance V=∑1≤K≤N PK(1-PK)WK2 is positive. To maximize this, we want to maximize E/√V by choice of the WK subject to some convenient normalization constraint (without which the WK would only be determined up to scaling). The simplest is to maximize E subject to V=constant. "Lagrange multiplier technique" solves that problem and the result is
where λ is a Lagrange multiplier, i.e. constant normalization factor, which you could, e.g, choose to make ∑1≤K≤N WK = 1. (It doesn't matter what λ is so long as it is positive.)
If all PK≈1/2, then this formula simplifies to just
approximately, if we ignore quadratically-small terms. I.e, very simply, it is best if each voter has weight linearly proportional to his advantage over a coin toss if we are allowed to ignore quadratically-small terms. That could, for certain kinds of votes, be reckoned historically – give everybody a vote-weight proportional to their average historical "success rate" in making choices, where making the right choice counts as +1 and the wrong choice -1. [Or, perhaps better, use the more-exact formula which doesn't neglect quadratic terms with the Dirichlet estimate P=(S+1)/(S+F+2) for a voter with S successes and F failures.] But this all can only work if we have an objective way of retroactively reckoning success rate.