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

Dave Ketchum davek at clarityconnect.com
Thu Dec 13 18:52:03 PST 2007

```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
Do to no one what you would not want done to you.
If you want peace, work for justice.

```