An argument against committee monotonicity
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I don't think this is original but it does not appear to be present on this forum yet.
Consider the following election:
A 20
AB 29
BC 31
C 20If there is only one winner then clearly B is optimal. However, if there are two winners then {A,C} seems more logical than {B,C} since the former represents all voters while the latter only represents 80%, with 31% being doubly represented.
Optimal d'Hondt PAV gives {A,C} a score of 100 and {B,C} a score of 80 + 31/2 = 95.5, so {A,C} win. SeqPAV, of course, elects B first.
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@last19digitsofpi I think this is more illustrative of how Perfect Representation, being NP-hard to satisfy, is not always attainable with greedy (aka sequential) rules.
I agree with both of your statements that B is logical for k=1, and AC is better for k=2. However
- There are rules which are not committee monotone but will still not elect AC (e.g. MES)
- There are rules which are committee monotone which will elect AC (e.g. SAV)
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@andy-dienes said in An argument against committee monotonicity:
There are rules which are committee monotone which will elect AC (e.g. SAV)
That means that SAV elects C for k=1, which is definitely illogical. As for MES, part of that might be because it's a little unclear for which values of x the following generalization should elect AC instead of BC:
A (1-x)/2
AB x/2 - epsilon
BC x/2 + epsilon
C (1-x)/2All-at-once d'Hondt PAV elects AC for x < 2/3. I think my argument is more persuasive if x is closer to (but still more than) 1/2, and epsilon is small compared to x-1/2.
