[EM] Strategic voting in Condorcet & Range N-canddt elections

[Dave Ketchum]

I would go for a different strategy - whatever leads toward long-term  
strength.

Winning the current election is usually pretty good.

A third party, looking ahead, can think of what will make a stronger  
future for it, even though perhaps not having a chance to win the  
current election.

On Jun 9, 2009, at 10:18 PM, Warren Smith wrote:

>> WDS responds:
>>      really it is better to regard N as the cardinality of the subset
>> of candidates that the
>> public thinks have a chance of winning.
>>
>> Dave Ketchum:
>> I would add any candidates that a significant percentage of the  
>> voters
>> wish belonged above, even if their current odds were less.
>
> --no.  Strategy is based on who has chances of winning, Not who
> somebody wishes have chances of winning.
>
>
>> I can read below that Approval is better than Condorcet for N=3 - and
>> be puzzled.  Both can rank A=B while Condorcet can also rank A>B (or
>> B>A) when that is a voter's desire.
>
> --So? Approval can rank B>A too.  There are 8 possible approval  
> votes in a
> 3-candidate election and 6 possible Condorcet rank-order votes but  
> 13 possible
> if equalities allowed in the ranks.  For honest Condorcet voters, only
> the 6 ever arise
> because two random normal utilities are never exactly equal.  For  
> Approval
> only 6 of the 8 ever arise if we use mean-as-approval-threshold,  
> same reason.

Agreed you never get exact equality - but you can get close enough for  
voters to treat voting equality  to be good enough.
>
>
> Anyhow that all is irrelevant.  What is relevant is what is the
> Bayesian Regret of
> the two methods.   I reported the numbers for the random-normal  
> utilities case
> in 2,3,4,5-candidate elections, 200 voters, for honest basic Condorcet
> and mean-as-threshold approval voting, which both were what Venzke's
> question basically concerned.
> Venzke wanted to know what if the identites of the 2 frontrunners A &
> B were not known ahead of time, even though he conceded A & B existed.
> In that case of ignorance, Venzke I presume believed the strategic
> voters for Condoret would just be honest, while the strategic voters
> for range would use approval-mean-as-threshold style voting.
> These numbers I gave compare those two.
> They show, in that case, Approval is better than Condorcet for N=3,4
> (and approx same for N=4)  and worse for N=5.
>
> Your personal puzzlement perceptions do not affect the validity of
> those numbers.
>
> [Perhaps Venzke believes that in case of ignorance abotu all the  
> other voters,
> Condorcet voters to be strategic should NOT be honest. That may be.
> But if so, he has never said so and never explained what his amazing
> better-than-honesty strategy is.]
>
>>> So for the rest let us assume voters will cast votes with
>> A,B top &
>> bottom (or reverse)
>> only, due to them being strategically wise and also due to
>> them
>> wanting to simplify
>> their lives if they can (and they can, here).
>
>> Venzke:But the "strategically wise" characterization, as well as  
>> the assumption
>> that only A or B can win, go out the window as soon as neither A  
>> nor B
>> has an outright majority. This happens already in Plurality  
>> elections.
>> I am pretty sure that third party supporters will analyze this  
>> situation
>> and find that the odds of C>A>B votes creating a harmful cycle are  
>> small
>> compared to the odds of C being the (voted) Condorcet winner.
>
> --No.  Remember.  We agreed at an earlier stage of the argument, that
> strategic votes were going to be intended to influence who wins among
> {A,B}, for only
> TWO candidates A,B... because we could regard 3-way near-ties
> [where that your vote had a chance of electing any of THREE canddts]
> as a neglectibly unlikely scenario.

Often there is only one leader and method does not matter - even  
Plurality would do.

Sometimes there are two and Approval, Range, and Condorcet become of  
interest.

Agreed three is rare, but when it happens, method really matters and  
the possibility needs to be considered when choosing a method.

Cycles are a possibility for three or more with Condorcet.  These  
should not be used as an argument against Condorcet, but simply a rare  
possibility to prepare for, should they happen.

Dave Ketchum
>
> OK?  That is stipulated. Only A or B can win.   I do not give a damn
> about anybody possibly having an outright majority.  Or any cycle
> maybe existing in the minds of third parties.  It is unnecessary to
> let those concepts enter our minds.  Expunge them from consideration;
> they serve no purpose except to confuse.  When things are very simple
> don't make them complicated.  This argument needs very few axioms.
> Keep it that way.
> All we need to know is
> 1. only A or B can win.  Others neglectibly unlikely, so neglect them,
> dammit.  Be utterly convinced that no matter what, nobody besides A &
> B can win.
> 2. method is monotone.
> Period.
>
> Now.  In that case, in any monotone method, an A-top B-bottom vote
> is always maximally strategic, in the sense that if ANY vote you can
> cast will make A win (and remember, only A or B can win - we agreed on
> that earlier)  then a vote
> of this form will always do the job.
>
> OK? So, assuming strategic voters know that theorem, why would they do
> anything else?  They would not, unless they have an insane desire to
> complicate their lives
> (and maybe not even then).  I put it to you: 99% of people smart
> enough to know this theorem, are also smart enough not to have insane
> desires to complicate their lives and live on the edge.
>
> It's kind of like if we agreed a best way to drive was in your lane.
> Perhaps under some circumstances, it might be equal-best to swerve way
> the hell off the road thru a field
> for 5 loop-the-loops...   by why would anybody do that if they were
> sure that just
> staying in their lane could not possibly be a worse strategy, but
> could be better?
> -- 
> Warren D. Smith