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

Ross Hyman rahyman at sbcglobal.net
Sat Mar 31 12:11:46 PDT 2018

Below is my motivation for the method.  You can decide if it makes any normative sense.  And, I agree it is not summable because it uses top ranked votes.  But it is meant to be a modification to Hare and STV which also uses top ranked votes.

I first developed the method in 2013 for the single winner case as a clone independent version of Baldwin. 


The idea of eliminating the candidate with lowest Borda score seems like a good elimination method to me except that Borda is not clone invariant and it has bad burial properties.  A good thing about it is that I like the visual of interpreting the Borda score as the average placement on the ballot. It makes sense to eliminate the candidate that's average placement is lowest.  The reason for both clone and burial failures is that each candidate takes the same amount of length on the ballot.  But if each candidate's length on the ballot was proportional to its top ranked votes, then cloning candidates would keep their total length the same so that problem is fixed.  Also if there are just two factions, one that top ranks A and one that top ranks B, it won't effect the scores for A and B no matter where The A voters place B or where the B voters place A.  The resulting Borda-like method without elimination isn't totally clone invariant because a clone can be pushed up beyond its score value without clones.  But if you use it as an elimination method it is both fully clone invariant and Condorcet.  I think the method could satisfy some IRV proponents because it can be interpreted as a minor modification of IRV and STV that preserves their good properties.    


On Saturday, March 31, 2018 9:31 AM, Jameson Quinn <jameson.quinn at gmail.com> wrote:

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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