IRO fails plurality CWs too.

[Bart Ingles]

I haven't had a chance to respond to anything in a few days, but let's
see what we have:

Mike Ositoff wrote:
> 
> Hi--
> 
> Before I get to the example, let me mention that IRO's proponents
> often say that it gets rid of the lesser-of-2-evils problem. They're
> talking about a particular special situation, like:
> 
>  47  45  8
>   A   B  C
>   B      B
> 
> The more extreme candidates are smaller--candidate size tapers
> toward the extremes. But the extreme candidate is still big enough

Questionable statement.  The middle may have "larger" candidates, or it
may have a lot of smaller candidates.  I'll go along with the former for
now, just to see where it goes.

> to tip the scales between the more middle candidates when its votes
> transfer inward.
> 
> Ok, then let's keep those assumptions, but say that there are
> more candidates. For example:
> 
>  60 70 100 83 75
>   A  B   C  D  E
>   B            D
> 
> (I've only listed the 2nd choices that affect the result. Only
> the extremes need get eliminated to establish C as a loser.
> 
> Note that in this example not only does IRO dump a CW, but
> it dumps a  CW who has a plurality. So much for IRO advocates'
> favoriteness standard.

So since when do we care about the plurality winner?

Actually, what is clear from the example is that there are three main
factions; namely B, C, and D.  A and E are fringe groups that have no
other choice but to join with B and D.  So in effect you have:

  130  100  158
   B    C    D

This is a fairly typical IRO example.  C loses, but it has more than
enough votes to tilt the election from B to D, depending on what C's
voters prefer.

Would Approval or pairwise have different results?  Under those methods
the default strategy for B and D would be to truncate, assuming both
believe that they have a good chance of winning.  A and E would still be
eliminated in favor of B and D, and again C would potentially determine
the outcome between B and D.

If B believes a loss to D is likely, and thinks that C is a reasonable
second choice, then B is capable of throwing enough support to C to
ensure C's win.  Under IRO, 30 of B's voters could switch and rank C
over B.  This should be enough to make sure C is a finalist even if D
tries to use the pushover strategy.  Assuming another 29 or so of the
remaining B voters rank C as second choice, C's win is guaranteed.  So
the worst-case final result would be:

  100   30  100  130   28
   B    C    C    D    B
   C    B              D

Under Approval, a similar strategy is possible.  If B believes a loss to
D is likely, then 59 or more of B's voters can also give approval to C. 
Actually more than 59 would probably be needed, since we don't know how
many of C's backers also gave approval to D; because of this it may
actually be _more_difficult_to guarantee a Condorcet winner under
Approval than it was under IRO.  Of course, D's backers could have given
some votes to C, but I am assuming D's voters know that it is in their
interest to truncate.

  130   29   71  158
   B    C    C    D
   C         D

In this example, all of B's supporters (including those originally
voting for A) are required to ensure a Condorcet winner if 71 of C's
supporters give approval to D.  Any more than 71 and the CW loses.

I would give a pairwise example, but tactical voting would probably
result in a circular tie, and I'm not sure I even understand what the
results mean at that point -- I suppose you can devise a tiebreaker to
do whatever you want.  I assume that's where Votes Against comes in.

Bart

> 
> And that isn't a rare special example. Though the middle candidate
> of 3 will sometimes be the smallest, a voter distribution that
> gives tapering support toward the extremes is especially plausible
> & likely and in keeping with a normal distribution of voters
> on the political spectrum.
> 
> So IRO not only fails the Condorcet Criterion & Monotonicity,
> but it also fails the main standard that its proponents defend
> it with: favoriteness.
> 
> Mike