[Fwd: MPV/IRO and equal rankings]

Mike Ositoff ntk at netcom.com
Sun Nov 15 00:12:33 PST 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
> time your first choice is eliminated, your compromise second choice
> 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
> advocates could comment.

You're quite right; the need to make that choice is out of place
in a ranked method. The problem that Approval IRO mitigates
is a problem unworthy of a rank method. But, if certain promoters
insist on IRO, with the problem that it shares with FPP, then
Approval IRO would be an improvement.

You speak of equal votes, but, as I said some days ago, the
rule could be written more briefly & simply, without mention
of votes:

Repeatedly eliminate from the rankings the candidate who
occupies or shares highest position in fewest rankings.


Yes, that rule doesn't have the majority stopping rule that
conventional IRO has. That rule has no effect in IRO, serving
only to reduce count labor. But such a stopping rule would
have an effect on Approval IRO. Of course, the majority stopping
rule in that method would simply say to stop the count if
1 or more candidates occupy or share highest position in a
majority of the rankings, and to elect the one who occupies
or shares highest position in the most rankings.

I don't recommend a majority stopping rule with Approval IRO,
because I don't have much regard for "majorities" created by
IRO. Better to let the method proceed till there's only 1
uneliminated candidate; that way, IRO, which tends to fail
to count preferences, will at least count as many as it
can. Besides, the rule is much briefer without a majority
stopping rule, consisting only of 1 sentence.

This concludes this reply to this message.

Mike Ossipoff

> 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 
> have an unfair advantage.
> 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
> 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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