2nd example of manipulation in Condorcet's method

Mike Ossipoff dfb at bbs.cruzio.com
Thu Jun 6 02:50:14 PDT 1996


Bruce Anderson writes:
> 
> On May 11, I posted an example of an election in which Condorcet's method, as 
> "officially" defined on this list (EM-Condorcet), was vulnerable to several 

That was the 1st manipulation example that you were referring to? That
example where voters entered a party & told how they voted, and so
other voters went out and took advantage of them? 

Look, in case I didn't make it clear when I answered you before
about that, any method will hurt people who tell the others
how they've voted. Must I give an example for Copeland? In 
a 3-candidate race, some Copelands voters enter a party, telling
how they voted. Others, hearing that, realize that they can
create a circular tie when they go out to vote, and that, based
on what's been told to them, their favorite is Plurality winner,
and so will be picked by Regular-Champion.

I believe it was Riker who showed that, no matter what method
is being used, the Condorcet winner has an easy win if everyone
has complete knowledge of eachother's preference orderings.

The problem is that we don't have perfect information like that.

Methods can be judged according to how much voters depend on
good predictive information in order to get the best result
they can for themselves. Because Condorcet doesn't need
any defensive strategy unless perverse & risky offensive
strategy is attempted on a scale sufficient to change the
election result, and because, even then, it doesn't need
drastic defensive strategy, as does Copeland, Condorcet
is better by that standard, among others.

What a desperate attempt to find a strategy problem.


> kinds of manipulation.  This example seems to have been dismissed as applying 
> only to "open" elections, i.e., ones in which some voters can know, in advance, 
> how all the other voters have voted.  There might be good reasons for arguing 
> that this example is inconsequential; but saying that it only applies to such 
> open elections is not one of them.  To see this, take my May 11th example and 
> multiply (across the board) the numbers of voters involved by 1,000,000.  Then, 
> instead of 101 voters, there would be 101,000,000 voters participating in the 
> election.  Also, instead of the closest pairwise matchups being decided by 1 
> vote, every pairwise matchup would be decided by at least 1,000,000 votes.  
> Accordingly, if the (now) 2,000,000 voters with preference order B,C,A could 
> take (i.e., could pay to take) a poll that is known to be accurate to within 
> 1,000,000 votes, then those 2,000,000 voters could manipulate the election as 
> indicated in that example.

But you're still creating a problem where some voters have predictive
knowledge, and others don't. a) That's a problem with your
methods too; b) It isn't a likely situation that we need
to base the choice of a sw method on.

> 
> True, 1,000,000 divided by 101,000,000 is about 0.99%.  But, with enough money 
> to spend on the poll, is this level of accuracy necessarily impossible?
> 
> I don't understand how any voting system could be vulnerable to manipulation 
> under the restriction that each voter has absolutely no idea how any other voter 
 will vote.  So it seems to me to be just a question of how accurate a poll any 
 > one group of voters can take (and yet still have time to inform the voters in 
> that group of the analyzed results of that poll before those voters must cast 
> their ballots).
> 

As I said, if we all knew eachother's prefernce orderings, there'd
be no problem. The Condorcet winner would have an easy win. If
we know nothing of eachother's prefernce orderings, a group of
innocently-intended truncators can make their alternative win,
instead of the Condorcet winnenr, in Copeland, but not in
Condorcet.

In reality, of course, we have incomplete & not-entirely-reliable
polling information. Unlike your examples, that information is
equally available to all. But when the informtion is wrong, it's
your methods, and not Condorcet, that will screw up and defeat
a Condorcet winner as a result. 

Oh, you haven't seen a proof of that? Ok, this is a demonstration,
though I don't know what you mean by a "proof". If a proof has
to be written in formulese, then this isn't a proof:

Demonstration of Condorcet's Invulnerability to Mis-Estimate:

First I'll state the property:

With Condorcet's method, even if everyone mistakenly believes
a certain alternative to be the middle Condorcet winner,
and everyone includes it in their ranking, and no one includes
in their ranking any alternative that they like less than it,
that can't give the election away to that alternative if there's
a Condorcet winner ranked over it by a full majority of all
the voters.

Why that's true:

I'll call that mistaken Condorcet winner "X". By assumption, X
has a majority against it, because a majority have ranked the
Condorcet winner over it. Truncation can never cause a Condorcet
winner to have a majority ranking something else over it, but
can only reduce the number ranking it over something else. 
So though truncation can make the CW beaten by another alternative,
it can't create a majority against the CW. Since the mistaken CW
has a majority against it, and the CW doesn't, the mistaken CW
can't win.

***

So, then, as I said, it's not Condorcet that has a problem when
polling information is wrong. It's Copeland & Regular-Champion,
which don't have the Invulnerability to Mis-Estimate property.

***
> Further, while 0.99% accuracy is good enough to guarantee manipulative 
> capabilities in that example, other examples can be constructed in which poorer 
> accuracy can still guarantee vulnerability to manipulation.  Here's one:
> 
> Suppose that 99,000 voters will vote in an upcoming election, that A, B, and C 
> are the only candidates under consideration, and that 93,000 of the 99,000 will 
> cast ballots as follows:
> 
>  26,000:  A,(B&C)
>   2,000:  (A&B),C
>  20,000:  B,A,C
>  14,000:  (B&C),A
>  31,000:  C,(A&B)
> -------
>  93,000:  subtotal.
> 
> Suppose that the remaining 6,000 voters belong to a well organized group, and 
> that each of these voters has a strong preference for B over C, a strong 
> preference for C over A, and a very strong preference for B over A; i.e.,
> 
>  93,000:  subtotal
>   6,000:  B,C,A
> -------
>  99,000:  total # of voters.
> 
> Suppose that the voting method being used in the election is the version of 
> Condorcet's method as "officially" defined on this list (EM-Condorcet).  
> Accordingly, ranked ballots are used, with ties and truncations allowed.
> 
> Suppose that this highly organized group of 6,000 voters is planning to take a 
> very accurate poll of the general electorate in time to advise the members of 
> the group on how they should cast their ballots.  How much accuracy is needed?  
> If the poll were perfect (and if I didn't make a computational error), the 
> possibilities are as follows:
> 
> If those 6,000 voters vote honestly, then their second choice, C, will win.
> 
> If those 6,000 voters do not vote at all in that election (e.g., they "stay 
> home"), then their first choice, B, will win.
> 
> If they decide to vote, but to truncate instead of voting honestly (i.e., to 
> vote:
>   2,000:  B,(A&C)
> instead of:
>   2,000:  B,C,A,
> even though they strongly prefer C over A), then B wins again.
>   
> Of course, if they truncate, they would be the only voters who don't express a 
> preference between A and C, and so they might decide to reverse C and A on their 
> ballots instead of voting honestly (i.e., to vote:
>   2,000:  B,A,C,
> instead of:
>   2,000:  B,C,A,
> even though they strongly prefer C over A).  B wins in this case also.
> 
> In short, according to the EM-Condorcet voting method, if those 6,000 voters 
> vote honestly, then their second choice, C, wins; but if they don't vote, they 
> truncate, or they pair-reverse C and A, then their first choice, B, will win.  
> For comparison (assuming I did the calculations correctly):
> 
> Voting                   Stay             Pair-
> Method           Honest  Home  Truncate  Reverse 
> ---------------  ------  ----  --------  -------
> EM-Condorcet:       C     [B]     [B]      [B]  
> Condorcet(1\2):     C      C     [B&C]     [B]
> Borda:              B      C       B        B
> Hare:               C      C       C        C
> Kemeny:             C      C     [B&C]     [B]
> Nanson:             C      C       C      [B&C]
> Plurality:          C      C       C        C
> Pl-runoff:          C      C       C        C
> Reg.-Champ.:        C      C       C        C
> 
> Candidates in brackets, i.e., [B] or [B&C], indicate places where strategic 
> advantage is gained by manipulation.
> 
> Further, every pairwise matchup, and every "votes-against" result used by the 
> EM-Condorcet voting method, is decided by at least 3,000 votes here.  Thus, a 
> polling accuracy of 3,000 divided by 99,000, or about 3.03%, would seem to be 
> sufficient to allow those 6,000 voters to manipulate this election.

But, as I said, you're still talking about a situation where
some voters know how the others will vote, and that those others
are going to vote sincerely, as polled, while the strategizers
will base their strategy on knowledge of that sincere voting. 
That's nonsense. The same polling informtion is available to all
large segments of the poplulation. It wouldn't be possible to
give it to a significant number and keep it from the rest. That
makes nonsense out of your example.

Besides, as I discussed earlier in this reply, Regular Champion
can be equally vulnerable to that impossible unequal polling 
information. 

And the only reason why the B voters in your example can gain
hy truncation (if they can--I haven't checked that yet) is because
there's no Condorcet winner. This is a difference between Condorcet
& Regular-Champion: Truncation can steal an election from a Condorcet
winner in Regular-Champion (or any other Copeland version), but
not in Condorcet.

If B had a majority against him, and there were another alternative
that didn't, the B couldn't win in Condorcet, but could win in
Regular-Champion, or Copeland in general.

If B can win in your example, it's either because B doesn't have
a majority against him, or because everyone has a majority against
him. Either way, majority rule isn't being avoiodably violated.
But majority rule will often be seriously, avoidably, violated 
in Copeland & Regular-Champion.




> 
> As before, this example certainly does not mean that the EM-Condorcet voting 
> method is necessarily worse than any other particular voting method.  To me, 
> what it means is as follows.  It is not good enough just to state that EM-
> Condorcet satisfies some particular criterion and just to state that this 
> criterion is obviously important to satisfy.  Instead, EM-Condorcet should be 
> treated like any other voting method.  A claim that any particular voting method 
> satisfies any particular criterion needs, at a minimum, 1) a precise definition 
> of the voting method, 2) a precise definition of the criterion, and 3) a 
> valid proof that the voting method satisfies the criterion.  Without such 
> proof, the "claim" should either be dropped or be explicitly labeled as a 
> conjecture which may or may not be true.  Of course, examples that illustrate 
> the voting method, and a good discussion of the importance of the criterion 
> along with suitably illustrative examples, can be extremely helpful and are 
> invariably quite desirable.  But examples, by themselves, are insufficient to 
> justify the validity of such a claim.

I've given a precise definition of my proposed version of Condorcet's
method.

I've precisely defined Invulneerability to Mis-Estimate, and
my Generalized Majority Criterion.

But aside from that, I've pointed out that Condorcet's votes-against
count will obviously make it impossible for anyone to win with
a majority against him when there's someone who doesn't have a
majority against him. Do you need a proof of that?

And I never said that standards of mine are "obviously" the right
ones. I've pointed out that getting rid of the lesser-of-2-e-evils
problem is the sw goal of electoral reformers. It's the reason
why electoral reformers want a better sw method. That problem
may not be important to Bruce, but it's important to us, because
it's got millions of voters too cowed to express what they want.

Also, when I re-stated Bruce's MW criterion in a form that
isn't uniformly failed by all methods, plain Condorcet
& Smith//Condorcet won't be failing it, but Regular-Champion
& Copeland will regularly fail it.

Does Bruce also consider majority rule unimportant? Is that why
he proposes a method that can elect an alternative with a majority
against it, even when no other alternative has a majority against
it?

As I said, in plain Condorcet or Smith//Condorcet, an alternative
with a majority against it can't win unless every alternative in
the set from which Condorcet's choice rule is to choose has a majority
against it.


> 
> Bruce
> .-
> 


-- 



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