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

[Warren Smith]

> 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.

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.

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
http://RangeVoting.org  <-- add your endorsement (by clicking
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and
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