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MendelSchmiedekamp comments on Arrow's Theorem is a Lie - Less Wrong
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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.
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?