Here's another method that alas, didn't entirely work. Consider the
following "obvious" random approach to proportionality: - If the council is full, stop. Otherwise: - Choose a non-blank ballot at random. - Elect the candidate listed first on that ballot. If there's a tie for first, choose one at random. - Eliminate that candidate from every ballot and loop to the beginning. Since it's random, it's subject to the luck of the draw. So take the mode (or do the process above many many times and choose the council that appears most often). The result is monotone: raising A can only get him elected sooner. In what sense is the determinized method proportional? If everybody votes closed party list style, then the process is equivalent to drawing from a multinomial distribution as many times as there are seats. From theorem 2 of White et al (http://dx.doi.org/10.1016/j.spl.2009.09.013), one can start with a council of zero people from every party (which is a "mode" of the multinomial with n=0) and run the D'Hondt algorithm, and at every step still be in a mode of the multinomial with that many seats. So the D'Hondt apportionment is a mode of the multinomial. Thus (up to tiebreaks), the process above finds the D'Hondt apportionment if everybody votes closed party list style. On the other hand, if there's just one seat, then the system reduces to Plurality. Plurality fails the mutual majority criterion, and thus the derandomized method fails the DPC. So it's not proportional in general. I'm finding it a bit difficult to determine just "how proportional" it is. It seems to depend on both the size of the factions, and on the number of seats. (Obviously, if there are as many seats as there are candidates, and each candidate is ranked by at least one voter, then everybody gets elected.) I also suspect the disproportionality is in some way analogous to the disproportionality of Plurality itself. If so, some kind of pairwise comparison might rescue the method, like Condorcet does Plurality. Of a related note, using the G-test or chi-square test as an objective function instead of setting the mode directly results in a Sainte-LaguĂ« reduction, not a D'Hondt one; and those tests are based on likelihood-ratio tests, which compare two situations. So perhaps there's something that can be done by combining Condorcet logic (pairwise comparisons) and multinomial tests to get something that, in party list situations, reduce to Sainte-LaguĂ«. But perhaps I'm just reinventing my own "Statistical Condorcet" of 2014 :-) --- In any case, the derandomized method above can be directly calculated by observing it has the Markov property. The probability of some candidate being elected depends only on the candidates who have already been elected: the probability of candidate A being chosen is proportional to the number of (non-blank) ballots that rank A first after every candidate who's already elected has been eliminated. Each council cX then has a score equal to p(X_s=cX|X_0={}), and the winning council is the one that maximizes this probability score. In theory, one can find the score for every council by dynamic programming (or even more brute-force, by taking powers of the Markov matrix). In practice, that doesn't scale well at all for large councils. (A consequence of the Markov property is that, when electing a council of s candidates, the method only looks at the first s ranked candidates on the ballot.) To take the example from the Droop Judgment post: 58: A>C>B>D 18: B>C>A>D 21: C>A>B>D 32: D>B>A>C two to elect. p(X_1={A}|X_0={}) = 58/129 = 0.45 p(X_1={B}|X_0={}) = 18/129 = 0.14 p(X_1={C}|X_0={}) = 21/129 = 0.16 p(X_1={D}|X_0={}) = 32/129 = 0.25 p(X_2={AB}|X_1={A}) = 18/129 = 0.14 p(X_2={AC}|X_1={A}) = 79/129 = 0.61 p(X_2={AD}|X_1={A}) = 32/129 = 0.25 (everything else is zero) p(X_2={AB}|X_1={B}) = 58/129 = 0.45 p(X_2={BC}|X_1={B}) = 31/129 = 0.30 p(X_2={BD}|X_1={B}) = 32/129 = 0.25 p(X_2={AC}|X_1={C}) = 79/129 = 0.61 p(X_2={BC}|X_1={C}) = 18/129 = 0.14 p(X_2={CD}|X_1={C}) = 32/129 = 0.25 p(X_2={AD}|X_1={D}) = 58/129 = 0.45 p(X_2={BD}|X_1={D}) = 50/129 = 0.39 p(X_2={CD}|X_1={D}) = 21/129 = 0.16 Summing everything up: p(X_2={AB}|X_0={}) = first elect B then A (0.14 * 0.45) + first elect A then B (0.45 * 0.14) = 0.1260 p(X_2={AC}|X_0={}) = first C then A (0.16 * 0.61) + first A then C (0.45 * 0.61) = 0.3721 p(X_2={AD}|X_0={}) = first D then A (0.25 * 0.45) + first A then D (0.45 * 0.25) = 0.2250 p(X_2={BC}|X_0={}) = first C then B (0.16 * 0.14) + first B then C (0.14 * 0.30) = 0.0644 p(X_2={BD}|X_0={}) = first D then B (0.25 * 0.39) + first B then D (0.14 * 0.25) = 0.1325 p(X_2={CD}|X_0={}) = first D then C (0.25 * 0.16) + first C then B (0.16 * 0.25) = 0.0800. And the winning council is AC, which in this particular example respects the Droop constraints (two candidates should come from {ABC} and one should be A). ---- Election-Methods mailing list - see https://electorama.com/em for list info |
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