[EM] An improved Condorcet Hare hybrid and its generalization to STV

[Jameson Quinn]

OK, I get it now.

I have a hard time seeing how the way this method resolves a Condorcet
cycle makes any a priori normative sense. But that's a very weak objection,
because I also think that this method is relatively strategy-resistant for
a Condorcet method. So, good job.

My bigger objection is that this isn't a summable method, because of the
eliminations.

2018-03-31 10:22 GMT-04:00 Ross Hyman <rahyman at sbcglobal.net>:

> Scores for B and C do not change when A is cloned.
>
> The part of the score for B that depends on A is M_B,A V_A.
> When A is cloned it becomes
> M_B,A1 V_A1 + M_B,A2 V_A2.
> These are equal because M_B,A1 = M_B,A2 = M_B,A and V_A1 + V_A2 = V_A.
>
> On Mar 31, 2018 8:13 AM, Jameson Quinn <jameson.quinn at gmail.com> wrote:
>
> I'm not sure I buy your proof sketch of clone independence. Say that we
> have a Condorcet cycle A>B>C>A, and C is eliminated so that A wins. Now
> clone A a bunch of times; won't that increase the score of C and decrease
> that of B, so that B will be eliminated and C will win?
>
> 2018-03-30 19:24 GMT-04:00 Ross Hyman <rahyman at sbcglobal.net>:
>
> An improved Condorcet Hare hybrid and its generalization to STV.
> The traditional Condorcet Hare hybrid methods retains the Hare elimination
> method but changes the election criterion from majority top votes to
> Condorcet winner.
>
>
> Another way to construct a Condorcet Hare hybrid is to retain the Hare
> election criterion but change the elimination criteria to one that uses
> information from the Condorcet matrix. Such a method will be better at
> elimination in general and such a method is easier to generalize to STV.
>
>
> One such method that elects a Condorcet winner if there is one, is clone
> invariant, and satisfies proportional completion is the following:
>
> V_A>B is the number of ballots that rank A above B. V_A is the number of
> ballots that rank A at the top.
> M_A,B = (V_A>B - V_B>A)/(V_A>B + V_B>A)
> S_A = sum_B M_A,B V_B is the score for candidate A.
> So long as there is no majority winner, eliminate the candidate with
> lowest score.  Recalculate V_A's and S_A's.  (M_A,B’s do not need to be
> recalculated.)Repeat until there is a majority winner (that is one with
> more top votes than
> the quota = total votes/2.)
>
>
> M_A,B can be any antisymmetric, M_A,B + M_B,A =0, function of V_A>B and
> V_B>A that is positive if V_A>B > V_B>A.  The choice above satisfies
> proportional completion.
>
>
> Using matrix multiplication notation the score is S = MV where M is the
> antisymmetric matrix, V is the vector of top votes and S is the score
> vector. The above method uses information from the Condorcet matrix, M,
> and the top votes vector V in an equal way. I think it gives better
> results than the Condorcet Hare hybrid methods
> that only use information from V to eliminate.
>
>
> Proof that Condorcet winner will not be eliminated:If there is a Condorcet
> winner it will have a non-negative score.  The weighted average of S_A,
> sum_A (V_A/V_Tot) S_A =0 so that if there is a Condorcet winner it is
> guaranteed that there will be at least one other candidate with negative
> score so the Condorcet winner will not be eliminated.
>
> Proof of independence of clones:
> S_A does not change if one of the other candidates is cloned.  If A is
> cloned to A1,A2 etc. then the weighted average over the clones sum_i
> (V_Ai/V_A)S_Ai = S_A so some of the clones will have a higher score than
> the original A and some less (unless they all exactly equal S_A).This might
> mean that one of the clones of A would be eliminated before A would have
> been, but since other clones of A remain, and we are eliminating just one
> at a time, everything is ok.
>
>
> Example 6 A2, A1, B, C
> 1 A1, A2, B, C
> 5 B, C, A1, A2
> 4 C, A1, A2, B
> Ignore denominators for score calculation since they are all the same:
> S_A1 = (10 -6)V_A2 + (11 -5)V_B + (7-9)V_C
> S_A2 = (6-10)V_A1 +(11-5)V_B + (7-9)V_C
> S_B = (5-11)(V_A1 + V_A_2) + (12-4)V_C
> S_C = (9-7)(V_A1 + V_A2) + (4-12)V_B
> Initially V_A1 = 1, V_A2 =6, V_B = 5, V_C=4. No candidate’s votes exceed
> the quota of 8 so scores must be calculated and candidate with lowest score
> eliminated.
> S_A1 = 4*6 + 6*5 - 2*4 = 46
> S_A2 = -4*1 + 6*5 – 2*4 = 18
> S_B = -6*7 + 8*4 = -10
> S_C = 2*7 + -8*5 = -26
> C is eliminated.
> Now V_A1 = 6, VA2 = 5, V_B = 5.  No candidate exceeds quota.
> S_A1 = 4*6 + 6*5 = 54
> S_A2 = -4*5 + 6*5 = 10
> S_B = -6*11 = -66
> B is eliminated.  V_A1
> = 10, V_A2 =6. A1 is elected.
>
>
> Condorcet Hare hybrid eliminates A1 then C and elects A2. A1 is the more
> sensible winner since more voters prefer A1
> to A2 than A2 to A1.
>
>
> Generalization to STV.  I lack good notation to show this in complete
> generality.  For simplicity I will show it for electing 2 candidates.
> Elect candidates that exceed the quota as in your favorite flavor of STV.
> When no candidate exceeds thequota eliminate the candidate with the lowest
> score according to the following score formula:
> S_A = sum_{B,C} M_A,{B,C}  V’_BV’_C
> M_A,{B,C} + M_B,{A,C} + M_C,{A,B} =0
> sum_{BC} is the sum over candidate sets that, for determining the score
> for candidate A, does not include A, does not include any excluded
> candidate, and must include all elected candidates.
> V’_A = V_A/Q where Q is the quota.  This will be less than 1 for unelected
> candidates and 1 for elected candidates.  Note that the V’A’s inside M are
> not the same as the V’s outside.
>
>
> M_A,{BC} is determined in the following way:Eliminate all candidates
> except A, B and C.  Elect two candidates as in your favorite flavor of
> STV.  If A is one of those candidates, M_A,{BC}  = V’_A -1.  This will be
> positive because A will only be elected if V_A > Q. If B and C are elected,
> which means that A is not elected, M_A,{BC} = - M_B,{AC} – M_C,{AB) =
> 2-V’_B  - V’_C.   Just as in the one winner case, M_A,{BC} need only be
> calculated once.
>
>
> This Condorcet STV method improves on conventional STV by ensuring that a
> candidate that wins in every three-candidate election for the two seats
> cannot be eliminated.  It has the same clone independence as conventional
> STV.
> Example Elect 2.
> 9 T A2, A1, B, C
> 1.5 T A1, A2, B, C
> 7.5 T B, C, A1, A2
> 6 T C, A1, A2, B
> (The fractional vote totals are to make things work out nice.  You can
> multiply by 10 if you want)
> The quota is 24/3 = 8 votes.  V_T =24 and is elected. Using the Gregory
> method, the V_A’s for the other candidates are now: V_A1 = 1, V_A2 = 6, V_B
> = 5, V_C =4.
> I choose the definition of M so that it would give me the
> exact same values for the scores S_A as for the one candidate case, so C
> and B are eliminated and T and A1 are elected.  With Conventional STV, T
> and A2 are elected.
> ----
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>
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