# MPV/IRO and equal rankings

Mike Ositoff ntk at netcom.com
Fri Oct 16 23:10:33 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

That would greatly mitigate IRO's problem, and would bring
some Approval advantages to IRO. As you say farther down, that's
in keeping with what other rank-counts allow. But IRO advocates
aren't like other people. I've proposedd that method, with
names like "Approval IRO", and IRO advocates didn't accept it.
They're afraid it would violate 1-person-1-vote :-) IRO advocates,
you see, tend to be the people who haven't taken the trouble to
study single-winner methods at all--too busy promoting to find

vote more than 1 candidate per rank-position. It's too bad
that not only do they propose a bad method, but they resist
all attempts to mitigate the method's problem. Likewise,
for example, I've found that IRO advocates also oppose
IRO with the candidate withdrawal option.

That concludes this reply. I'd delete the following lines,
but the keyboard might lock-up first.

Mike

> 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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