# MPV/IRO and equal rankings

Blake Cretney bcretney at my-dejanews.com
Fri Oct 16 12:15:00 PDT 1998

```What does MPV/IRO do when candidates are ranked equally?  There seems
to me to be three possibilities.

1.  Give each of the alternatives a full vote.
This is certainly what the voter who is giving the equal rankings
would like.  After all, there is always the danger that by the
will be gone too.  If you rank them both equally, this means you
can help them both at the same time.

But what that means is that each voter must weigh their desire to
get their first choice, versus their desire to get one of their
most acceptable choice.

So, for example, A=B>C is less likely to elect A then A>B>C, but
more likely to elect one of A or B.  The voter must weigh these
considerations, as they would in a method like approval.  To me,
this seems out of place in a ranked method, but perhaps some MPV / IRO

2.  Give each of the alternatives an equal fraction of the vote.
So, for example, once A=B=C reaches the top of the ballot (through
elimination), each of A, B, and C will get 1/3 of a vote.  Once one
of them is eliminated, the each get 1/2.  And finally when two are
eliminated, 1.

This doesn't appear to have the problem I mentioned above, but it does
fail GITC.
Candidates are A and B, which are not twins, X and Y, which are.

42 A B X Y
30 B X Y A
27 X=Y=B A
32 X Y B A
31 Y X B A

A 42
B 39 - eliminated
X 41
Y 40

A 42 - eliminated
X 75.5
Y 44.5

X 117.5 - winner
Y 44.5 - eliminated

Now without Y

42 A B X
30 B X A
27 X=B A
63 X B A

A 42 - eliminated
B 43.5
X 76.5

B 85.5 - winner
X 76.5 - eliminated

So, having a twin caused X to win.  This is called the rich party problem
because it means that parties that can afford to run more candidates will

3. Just don't allow equal rankings, except by leaving candidates unranked.
This is the most obvious solution.  It is possible that the electorate
wouldn't understand, and use, equal rankings anyway.  And it passes
GITC.

Unfortunately, it passes GITC for the same kind of technical reasons
that make plurality pass GITC.  That is, because voters are forced
to distinguish between candidates randomly, even if they have no
preference, they will break up what based on their true preferences,
are twins.  However, the rich party problem remains.

Consider the above example.  Imagine that each voter randomly
distinguished between candidates they consider equal.  On average
each equally ranked candidate would get an equal share of the vote,
just like happened in the example above.  So, the banning solution
is equivalent in rich party effect to the fractional solution.
---
To me, this argument suggests that solution 1 is the best.  It may not
be perfect, but at least it doesn't have the rich party problem.  The
main danger is that voters won't understand how to equally rank
candidates, and that the results will be more like solutions 2 and 3.

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