[Election-Methods] IRV-Tournament

Dave Ketchum davek at clarityconnect.com
Wed Jul 16 14:40:24 PDT 2008


On  Wed, 16 Jul 2008 18:31:04 -0000 Bruce R. Gilson wrote:
> --- In RangeVoting at yahoogroups.com, Dave Ketchum <davek at ...> wrote:
> 
>>On Tue, 15 Jul 2008 14:38:32 -0000
>>Bruce R. Gilson wrote:
>>[...]
>>
>>>As soon as you put in some cycle-resolving system, you will 
>>>downgrade the preferences of some of these 6 groups -- obviously 
>>>you have to, because that's the only way to break the cycle. And 
>>>the people in those groups will feel that the election mechanism 
>>>is disregarding their preferences (or at least weighting them 
>>>less than others' preferences). And that will be the scandal. 
>>>
>>
>>Saying it more clearly, for A+B+C as the simplest cycle:
>>
>>Given six equal sized groups of voters:
>>      A>B, B>C, and C>A can do an A>B>C>A cycle with 2/3 of the 
>>strength of each group pushing the cycle forward and 1/3 acting as 
>>a brake.
>>      A>C, C>B, and B>A can do a similar A>C>B>A cycle.
>>      Combine the two cycles and you still have a three member 
>>cycle, again with a tie.
>>
>>Stray far enough from equality and the weakest candidate has no 
>>effect, leaving a two candidate race,
>>
>>In between you have a headache for which the rules BETTER be 
>>decided on before the election.  This does not require favoring any 
>>groups - vote counts are the proper basis for the decisions.
> 
> 
> It does not require favoring any groups -- certainly that is so, but 
> whatever system is decided upon, there are going to be people who 
> PERCEIVE that they have been disfavored by the rules. Suppose you 
> have an A>B>C>A cycle and the cycle-breaking rules elect A. Then the 
> people who voted BCA and those who voted CBA are going to scream 
> bloody murder, because they see that their least-favored candidate 
> was declared the winner, despite the fact that it vis clear that C 
> (who was preferred by the first group and STRONGLY preferred by the 
> second) could have beaten him. 
> 
> Any Condorcet cycle-breaking algorithm has to lead to the fact that 
> some votes count more than others, despite the fact that there was no 
> INTENT to do so. If it didn't, it would not break the cycle, because 
> the cycle is inherent in the way Condorcet scores are computed.
> 
Your "some votes count more than others" seems to deserve more 
thought.  In most any election method the vote counts can be near a 
boundary, encouraging second thoughts by those who voted on both 
sides, while counts far from a boundary could not have been affected 
by one or two voters voting differently.

Condorcet has some fixed boundaries, for which the above applies:
      A beats each other candidate, winning the election.
      In a cycle A, B, and C beat all others while each beats one of 
the other two.  Similar cycles are composed of more than three candidates.

How to resolve cycles is a topic for debate.  Rules for this BETTER 
get resolved before the election to decide what shall be considered fair.
-- 
  davek at clarityconnect.com    people.clarityconnect.com/webpages3/davek
  Dave Ketchum   108 Halstead Ave, Owego, NY  13827-1708   607-687-5026
            Do to no one what you would not want done to you.
                  If you want peace, work for justice.






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