Condorcet Truncation Example

[Rob Lanphier]

On Tue, 6 May 1997, Markus Schulze wrote:
> But the way you have defined "truncation resistant" and
> Smith//Condorcet [EM], the following problems will occur:
> 
> (1)Upward truncation [i.e., the voter gives his best preference
>    to more than one candidate] becomes a usefull strategy,
>    especially if there is the danger that more than one of the
>    most favoured candidates comes into the Smith set and one of
>    the most favoured candidates in the Smith set would have his
>    highest defeat against another of the most favoured candidates
>    in the Smith set.

I see your point, and I think this is a problem that needs to be
addressed.  At the worst extreme, this causes Smith//Condorcet[EM] to
mimic Approval, only having solved the truncation problem.  This isn't an
altogether bad thing, really, but I agree that it is suboptimal.

Is this a problem that can be addressed very easily?  For those of you
that don't quite understand the implications of this, here is an example
(stolen and modified from some writing Bruce Anderson did a while back): 

Candidates A,B,C,D
# Voters   Votes
33         A > B > C > D
31         B > C > A > D
2          D > A > B > C
2          D > B > C > A
32         D > C > A > B

Results:
[A] 67 votes
[B] 33 votes

[A] 35 votes    [B] 68 votes
[C] 65 votes    [C] 32 votes
                
[A] 64 votes    [B] 64 votes     [C] 64 votes 
[D] 36 votes    [D] 36 votes     [D] 36 votes 

In this race, candidate D loses all pairwise elections, and so is knocked
out of the running by the Smith criterion.  This leaves A, B and C.  Of
these, candidate A loses by the smallest total votes against (to candidate
C, who gets 65 votes).

However, let's say that the 31 voters who vote B > C > A > D modify their
vote to B = C > A > D.  This leaves:

[A] 67 votes
[B] 33 votes

[A] 35 votes    [B] 37 votes
[C] 65 votes    [C] 32 votes
                [B = C] 31 votes
                
[A] 64 votes    [B] 64 votes     [C] 64 votes 
[D] 36 votes    [D] 36 votes     [D] 36 votes 

Now, candidate C doesn't have near as many votes against (only 37) and so
easily wins the tiebreaker.

> Again: Even if none of the other voters votes tactically,
> every Condorcet Criterion method punishes those voters, who
> give full information about their true opinion. This means, that 
> they won't give full information about their true opinion.
> This means, that it won't be possible to determine the 
> Condorcet Criterion winner due to the opinion of the voters
> even if no voter votes tactically. This means that the Condorcet
> Criterion method won't work.

Whoa, wait a second.  Your proof reads:

1.  X method has a problem
2.  X method isn't perfect
3.  X method won't work

I think Arrow did show that there's no such thing as the "perfect" method,
so we should stop trying for that.  Though I find the above scenario
troubling, it doesn't change my support of Smith//Condorcet[EM] for three
reasons:

*  It doesn't encourage order reversal
*  One can't seem to upward truncate and cause someone who wasn't in the
Smith set to enter the Smith set and win the election (this is gut
instinct; I could be wrong, but I'd like to see a counter-example).
*  Any other proposal I've seen has worse problems

It seems as though, in the worst case, Smith//Condorcet[EM] is reduced to
a version of Approval voting, with the advantage of no longer having a
downward truncation problem. 

Is there a way to tweak Smith//Condorcet[EM] to solve this problem without
breaking its downward truncation deterrance?  One way of doing this would
be to not allow tied preferences at any level (other than the implied ties
past the bottom of the ballot).  This seems a bit extreme, but perhaps
there is another way of fixing it.

Rob Lanphier
robla at eskimo.com
http://www.eskimo.com/~robla