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

Jonathan Lundell jlundell at pobox.com
Thu Dec 13 12:00:21 PST 2007

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  

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).

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