Approval and LO2E

Michael A. Schoenfield maschoen at
Wed Nov 4 02:02:03 PST 1998

Isn't this an example of an Arrow problem as articulated by Ken Arrow and

Michael S.

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-----Original Message-----
From: Mike Ositoff <ntk at>
To: election-methods-list at <election-methods-list at>
Cc: ntk at <ntk at>
Date: Wednesday, November 04, 1998 7:55 AM
Subject: Re: Approval and LO2E

>Blake said:
>Anyway, my problem is this:  Consider an election with
>> a united but minority totalitarian movement and a fragmented
>> democratic movement.  A and B will be the democrats.  C will
>> be the totalitarian.  The public will be 55% democratic, 45%
>> totalitarian.  The totalitarians will likely all vote for C
>> alone.  Voter preferences are as follows:
>> 30 A B C
>> 25 B A C
>> 45 C A=B
>> Now if not enough A and B voters compromise the result could be
>> like this
>> 25 A
>> 10 A B
>> 20 B
>> 45 C
>> A 35
>> B 30
>> C 45
>> With C winning easily
>I believe that the best rank methods, by which I mean the VA
>methods, are better than Approval. Of course one can't expect
>Approval to have the full strategy advantages of the best rank
>This kind of example has been used before. Typically, either
>A or B would be more of a middle alternative, in which case
>it would be CW unless one extreme has a majority. That makes
>it easy for A & B voters to know whom to vote for. If B is
>middle, then its voters have no reason to vote for anything
>else, and A voters know that B is the best they can get, and
>so vote for B.
>Or, if neither A or B is more middle than the other, maybae
>one is known or strongly suspected to be significantly bigger
>than the other. Then, too, it wouldn't be hard for A & B voters
>to know which to vote for. The bigger one is the obvious place
>for A & B votes to come  together. Its voters vote for it only
>and the other alternative's voters vote for it too.
>If neither of those things is true, and A & B are equally
>distant from C, and seemingly equal in size? What would
>happen would depend of whether trust exists and is justified.
>Not satisfactory? You can't have everything with such a simple
>method. As an A or B voter, I'd vote for the other in that
>situation, without either of the above-described helpful
>conditions. I'd vote for both. Of course, if the other
>voters defected, then I'd do the same next time, and they'd
>know to expect that. That's why they'd be foolish to defect
>under those conditions. With neither A nor B seeming bigger
>or more middle than the other, it would be quite obvious that
>both groups of voters need to co-operate to beat C. Consider
>how it would make the A voters look if they defected and caused
>the election of the totalitarian. The fact that there then
>might not _be_ a next election would be all the more reason
>why defection would be out of the question.
>> On the other hand, one of the A or B groups could stare down the other
>> side, using its fear of C to force it to compromise.
>> 35 A B
>> 20 B
>> 45 C
>> A 35
>> B 55
>> C 45
>> Here B won because the C voters panicked.  This is what I mean by
>> government of the stubborn.
>Not plausible. If it looked as if A is likely to have 75% more
>voters than B, the A voters would have reason to expect the
>B voters to vote for A. The A voters would be in a better
>position to exxpect that. Whether they'd refuse to vote for
>B would depend on what degree of amity or enmity exists between
>A & B voters. If they get on very well, then of course both
>would vote {A,B}. If not, then the A voters would vote {A}
>and the B voters would know that, and would vote {A,B}, if
>we're assuming that they want to optimize the outcome.
>> If both A and B compromise the results could look like this
>> 50 A B
>> 3 A
>> 2 B
>> C 45
>> A 53
>> B 52
>> C 45
>> See how compromise can result in the election between A and B being
>> made by a few people who don't compromise.
>The A voters, in your above example, defected in smaller numbers,
>percentage-wise, but A still won. If A & B voters aren't
>on the best of terms, and defection by either side is considered
>possible, and A voters are determined not to let B voters steal
>the election, then the A voters would make it clear, before
>the election, that they were just voting for A, because A has
>probably about 75% more voters than B and is therefore the natural
>choice to beat C.
>By the way, if voters were using the mathematical strategy
>that I described the other day, then, if A is expected to
>be about 75% bigger than B, and it were known that both
>A & B votes know that, then it would be obvious that A
>has a higher probability of being the other frontrunner with
>C. The psycological strategic considerations would make
>it look like a virtually zero chance that B could be a frontrunner
>with C. A would win for sure. Just the fact that A is 2nd biggest
>and B is smallest would make A look more like a frontrunner,
>and then the consideration of what the other voters are
>doing would make it pretty much a sure than that B can't
>be a frontrunner.
>> This is in contrast to Smith//Condorcet, Tideman, Schulze (VA and
>> and AV/IRO, where the democratic voters can prevent the election of C
>> just by ranking C last.  And the C voters have no way to use insincere
>> voting to get C elected.
>I'll check that claim out.
>> This is of course related to my problems with Approval and Clones,
>> since A and B are clones as far as their sincere rankings in this
>> example.
>Clones are a special case. Approval doesn't _automatically_
>do the right thing in this clone example, but it doesn't
>have Margins' tendency to require complete abandonment of
>favorites as can happen in Margins if its poorly-deterred
>order-reversal happens.
>Rankings provide much opportunity for a method to screw up,
>which is why if rankings are used, it's important to count
>them in the best way.
>> ---
>> Blake
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