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PhilGoetz comments on Arrow's Theorem is a Lie - Less Wrong
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Comments (62)
I don't see how this can possibly be correct. Suppose we require that no person can give the same score to 2 different options. Your reasoning would indicate that we can then combine these scores in a way that produces a total score for every option that satisfies the four conditions. Conditions 1,2, and 4 are stated in ways that don't depend on whether we use rankings or scores; condition 3 explicitly assumes the use of a score. So if we take the output scores, and list them in order, we get an output ranking, which must still meet those conditions, hence violating Arrow's theorem.
"So if we take the output scores, and list them in order, we get an output ranking, which must still meet those conditions, hence violating Arrow's theorem."
If we only take into account rankings as data, and only output rankings, then of course; the point is that we don't, because it's an artificial constraint, and there's no reason for us to adhere to it.
You can take your input rankings, map them into scores, use your process on the numbers, and get output scores. These output scores satisfy the 4 criteria. You can then map these output scores back into rankings. Because the 4 criteria don't depend on whether you use rankings or scores, they must still be satisfied, except for determinism. Determinism might not be satisfied because the mapping of rankings into scores is one-to-many.
Does Arrow's theorem allow us to find solutions to the original ranking problem if we throw out the determinism criterion? I don't see how determinism is useful.
By Arrow's Theorem, any process by which rankings are mapped to scores must violate one of these criteria, once the scores are processed using my algorithm. This is not terribly surprising, as you lose all the information about relative importance in doing so.
Yes; but can we find a process which violates only determinism? Because we don't need determinism.
That's exactly what tommccabe's process does. It violates determinism in Arrow's sense of the word, but it satisfies the other conditions. Moreover, it violates Arrow-style determinism in a way that we arguably want anyway.
Note, according to the wikipedia article listed, Arrow's theorem is valid "if the decision-making body has at least two members and at least three options to decide among". This makes me suspect the Pareto-efficiency counter-example as this assumes we have only 2 options.
It doesn't matter if there are ten thousand other options. If you sum numbers A-1 through A-N, and you sum numbers B-1 through B-N, and A-X > B-X for all X, then A must be larger than B; it doesn't matter how many alternatives there are.
Fair enough. Although in considering the implications of more than two options for the other conditions, I noticed something else worrisome.
The solution you present weakens a social welfare function, after all if I have two voters, and they vote (10,0,5) and (0,10,5) the result is an ambiguous ordering, not a strict ordering as required by Arrow's theorem (which is really a property of very particular endomorphisms on permutation groups).
It seems like a classic algorithmic sacrifice of completeness for power. Was that your intent?