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

Ross Hyman rahyman at sbcglobal.net
Fri Mar 30 16:24:53 PDT 2018

```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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