[Election-Methods] RE : Re: Is the Condorcet winner always the best?

[Dave Ketchum]

On Thu, 13 Dec 2007 12:00:21 -0800  Jonathan Lundell wrote:
> On Dec 13, 2007, at 9:25 AM, Dave Ketchum wrote:
> 
>> On Thu, 13 Dec 2007 08:00:23 -0800 Jonathan Lundell wrote:
>>
>>> On Dec 11, 2007, at 6:17 PM, Dave Ketchum wrote:
>>>
>>>> A and C agree that B is better than their standard enemy.
>>>>
>>>> C voters will be happy to help install B, since this is better  
>>>> than  installing A.  A voters may be a bit unhappy, but they at  
>>>> least  avoided installing C.
>>>>
>>> That argument makes sense after the election, once the A or C  
>>> voters  know for certain that C or A, respectively, would have won  
>>> had it not  been for B. But the argument fails *before* the  
>>> election. Given the  implied utility function of this election,  both 
>>> A and C voters have a  strong incentive to bury B if they  think 
>>> their own candidate has a  good chance of winning outright.
>>
>>
>> Later in that same post of mine:
>>
>>>> Choices can be hard.  Get far enough from a tie and A or C will  
>>>> win.  If we manage a cycle we can debate the results of that.
>>>
>>
>> IF A or C expected a solid win, the same voting would have been  
>> appropriate, since it would not prevent the win from being recognized.
>>
>> Of course, there can be cycles - and hopefully the method will  handle 
>> them well - but this does not seem to be the place to debate  handling 
>> cycles.
> 
> 
> Cycles don't enter into it (and if A is guaranteed a solid win, then  of 
> course strategy is irrelevant).
> 
> My argument is about expected utility. Let's go back to Diego's  scenario:
> 
> 46: A >> B > C
> 5: B >> A > C
> 5:  B >> C > A
> 44: C >> B > A
> 
> Suppose the A voters' utility for {A, B, C} is {100, 10, 0}, and B's  
> likewise, mutatis mutandis; their estimated probability of A's (B's)  
> election must be low indeed before it's rational for them to approve B  
> (or to rank B in a Condorcet election).
> 
Lets assume:
      B voters are going to vote as described (avoids debating how they 
might change).
      There are 90 major party voters, desperately wishing to win, but 
willing to let B win on near ties, rather than risking opponents winning 
without earning full backing.
      No cycles (which require full 3-way competition, while this is 
mostly 2-way.

The pattern above works for this:
     A or C may get more than 50 votes - and win.
     Best of A or C may be less than 50 votes, letting B take a turn as 
winner.
     A or C can get 50 (of the 90) - making a tie in A>B or C>B - have fun 
unless this gets resolved.

Note that the pattern is neutral as to A vs C, and provides for hand-off 
to B on near A-C ties WITHOUT needing to estimate chances of a tie.

A and C actually using it makes sense to me for:
     Whichever is strong on election day wins.
     If neither is strong then, better B than the enemy.

I vote against "Random Ballot" for this task.
-- 
  davek at clarityconnect.com    people.clarityconnect.com/webpages3/davek
  Dave Ketchum   108 Halstead Ave, Owego, NY  13827-1708   607-687-5026
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