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Bart Ingles | 1 May 2004 06:17
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Re: Some unfortunately too strong Defensive Strategy Criterion


Jobst Heitzig wrote:
> 
> In the last days, I thought about some form of strategy-proofness like
> the following:
> 
> Criterion:
> Suppose that, with all voters voting sincerely, the method elects A, but
> some voter prefers B to A and can get B elected by voting insincerely.
> Then those voters not preferring B to A must have a way of voting which
> ensures that A or some option C gets elected which the first voter ranks
> *below* A, so that either the sincere result can be guaranteed or the
> incentive to vote insincerely is removed.
> 
> However, I then came up with the following, very simple 3-by-3-example
> which seems to render those thoughts ridiculous...
> 
> Problematic Example:
> 
> Sincere preferences
> Voter 1: A>B>C
> Voter 2: B>C=A
> Voter 3: C>A>B
> 
[...]
> 
> I sincerely hope I missed some essential point in that example... Can
> anyone tell what this would look like with Approval (I mean, what is a
> sincere Approval vote in the first place?)?
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Bart Ingles | 1 May 2004 06:31
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Re: Some unfortunately too strong Defensive Strategy Criterion


Bart Ingles wrote:
> 
> So the likely outcome should be:
> 
> Voter 1:  AB
> Voter 2:  B
> Voter 3:  CA
> 
> Result:
> Tie between A and B.

So the overall social utility of the tie should be:
[(1/2 + 1/4) + (1/2) + (0 + 1/4)] / 3 = 1/2

For comparison, the overall social utility if one of the following was
the sole winner would be:
A wins:  1/2
B wins:  1/2
C wins:  1/3

So the tie between A and B seems oddly appropriate in terms of SU.
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Curt Siffert | 1 May 2004 07:27
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condorcet/borda


I'm wondering what the verdict is on using Borda as the tie breaker for 
deciding within the Schwartz Set.  Some of the strongest Condorcet 
advocates on this list feel that Borda is actually superior to 
Condorcet *if* it can be ensured that only sincere voting is going on.  
So if Borda is used only as a tiebreaker (which would only be rarely 
needed), then there wouldn't be an incentive to use insincere 
techniques that would normally work with Borda.

Or if not Borda, some other cardinal system.  You'd only look ordinally 
for Condorcet, and then use the cardinal calculations for the 
tiebreaker within the Schwartz Set.

Curt

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sdfgdfg sdfgdfs | 1 May 2004 08:36
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questions about SF's version of IRV

Hello, I am wondering what you think of the IRV voting
system being implimented in SF, specifically:

It seems that voters will be only allowed to rank
their top three choices, instead of ranking the whole
field? Does this make a difference? If so how would
this type of difference effect a race with 5
candidates and an incumbent, vs. 20 candidates in a
non incumbent race? Is it theoretically possible if
the 20 candidates formed 3 "alliances" which only
voted for each other, then a 51+ percent majority
could not be achieved? 

Is anyone aware if the SF system will allow people to
only choose one choice, I guess what is called
"truncation"?

Keep in mind that this would be a very first try for
IRV, so even with the possibility open for strategic
voting people would very likely not know what to do
anyways.

Thanks

	
		
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Bart Ingles | 1 May 2004 09:15
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Re: questions about SF's version of IRV


sdfgdfg sdfgdfs wrote:
> 
> Hello, I am wondering what you think of the IRV voting
> system being implimented in SF, specifically:
> 
> It seems that voters will be only allowed to rank
> their top three choices, instead of ranking the whole
> field? Does this make a difference? If so how would
> this type of difference effect a race with 5
> candidates and an incumbent, vs. 20 candidates in a
> non incumbent race? Is it theoretically possible if
> the 20 candidates formed 3 "alliances" which only
> voted for each other, then a 51+ percent majority
> could not be achieved?

That was my understanding.  Of course, the definition used for
"majority" under any form of IRV is pretty weak to begin with-- it
basically means that the Condorcet loser should be eliminated.  In other
words, if more than half of voters include in their rankings a vote for
the eventual winner, then you can say that a majority of voters
preferred the winner over at least one other candidate.  

So with a ten candidate field, the "majority" winner would be preferred
head-to-head over at least one other candidate (even though the
remaining eight candidates may have head-to-head majorities over the IRV
winner).

But if you limit the choices to three, then even this limited definition
only works for four or fewer candidates.
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Dave Ketchum | 1 May 2004 11:00

Re: questions about SF's version of IRV

On Sat, 01 May 2004 00:15:12 -0700 Bart Ingles wrote:

> 
> sdfgdfg sdfgdfs wrote:
> 
>>Hello, I am wondering what you think of the IRV voting
>>system being implimented in SF, specifically:
>>
>>It seems that voters will be only allowed to rank
>>their top three choices, instead of ranking the whole
>>field? Does this make a difference? If so how would
>>this type of difference effect a race with 5
>>candidates and an incumbent, vs. 20 candidates in a
>>non incumbent race? Is it theoretically possible if
>>the 20 candidates formed 3 "alliances" which only
>>voted for each other, then a 51+ percent majority
>>could not be achieved?
>>

To demand ranking all the candidates has to be UNACCEPTABLE - think of the 
CA recall election and how many candidates they had for governor.  Also 
think of the many voters who are not going to do useful thinking beyond 
bullet voting much as they would do in Plurality.

Actually, as a Condorcet backer, I would welcome IRV being so stupid, but 
I want IRV to propose this right for least static in defining Condorcet rules.

Thus, when IRV promoters talk of "majority" they are NOT thinking of 51+ 
percent - they are using this label for their winner, while not talking of 
percents.
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Markus Schulze | 1 May 2004 12:35
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Re: condorcet/borda

Dear Curt,

Schwartz//Borda violates independence of clones.

Markus Schulze
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Markus Schulze | 1 May 2004 12:35
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Re: Some unfortunately too strong Defensive Strategy Criterion

Dear Jobst,

you wrote (1 May 2004):
> Criterion:
> Suppose that, with all voters voting sincerely, the method elects A, but
> some voter prefers B to A and can get B elected by voting insincerely.
> Then those voters not preferring B to A must have a way of voting which
> ensures that A or some option C gets elected which the first voter ranks
> *below* A, so that either the sincere result can be guaranteed or the
> incentive to vote insincerely is removed.

In the scientific literature, election methods with this property
are called "strategyproof with counterstrategies". Especially
Prasanta K. Pattanaik wrote many papers about this topic.

Markus Schulze
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Bart Ingles | 1 May 2004 18:50
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Re: questions about SF's version of IRV


Dave Ketchum wrote:
> 
> To demand ranking all the candidates has to be UNACCEPTABLE - think of the
> CA recall election and how many candidates they had for governor.  Also
> think of the many voters who are not going to do useful thinking beyond
> bullet voting much as they would do in Plurality.

But that's exactly what's required for some Australian elections.  So
it's not an unreasonable question.  Although the original was also about
voters being *permitted* to rank more than three choices.

> Thus, when IRV promoters talk of "majority" they are NOT thinking of 51+
> percent - they are using this label for their winner, while not talking of
> percents.

True-- it really means "majority of voters expressing a preference for
one of the last two candidates standing."

> Assume 10 voters bullet vote:  4A, 3B, & 3C;  A wins, and I believe they
> claim their winner ALWAYS has a majority.

At least in that case, they can claim (with partial validity) that the
voters with exhausted ballots voluntarily abstained, and therefore
shouldn't be counted as part of the denominator.  But by limiting the
number of candidates that can be ranked, they forfeit even that claim.

> > So with a ten candidate field, the "majority" winner would be preferred
> > head-to-head over at least one other candidate (even though the
> > remaining eight candidates may have head-to-head majorities over the IRV
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Jobst Heitzig | 1 May 2004 20:09
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Examples with 4 options for immune methods

This is a comprehensive (at least I hope so) list of examples with 4
options of which more than one is immune, with all defeats of different
strength.

An option is *immune* here when each of its defeats is countered by a
chain of stronger defeats leading back. When there are at least two
immune options, then there is no Condorcet winner or loser, exactly 2
options are immune, and the options can be renamed so that the defeats
are A>B>C>D>A,A>C,B>D without loss of generality.

Interesting sets:
Smith = {ABCD}
Undominated = Banks = {ABD}
Copeland = {AB}

The immune set varies and is denoted by {..} in the list.

The strength of defeat X>Y is abbreviated by XY in the list.

The examples are ordered by largest defeat, 2nd largest defeat, etc.,
considering AB before AC, BC, BD, CD, and DA.
Those of the 6!=720 orders which have only one immune option are not
listed. Note that, for the remaining examples, the immune set is thus
determined by the 3 largest defeats!

The winners of the three major immune methods and of Plain Condorcet are
listed beneath.

COMPREHENSIVE (?) LIST OF 104 EXAMPLES:
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