# [EM] Blake's margins arguments

AVnow avnow at fgn.net
Tue Feb 18 15:42:02 PST 2003

```Adam Tarr wrote:

> Tom McIntyre Wrote:
>
>>> MIKE OSSIPOFF wrote:
>>>
>>> 101: A
>>> 50: BAC
>>> 100: CBA
>>>
>>> About 60% of the voters have indicated that they'd rather elect
>>> B than A. And so margins elects A.
>>>
>>> WV counts, keeps, & honors the B>A majority. A has a majority defeat
>>> that wv doesn't lose or erase. With margins, what happens to that
>>> majority against A? Margins erases it.
>>
>>
>> And about 60% prefer A to C.  What about honoring *that* majority?
>> One of these majorities has to be lost.  Both WV and margins count,
>> keep, & honor one of them, and erase the other.
>
>
> The A over C defeat is not erased.  The only way you could consider
> that defeat erased, would be if C had won the election.  Take the
> extreme example: say there was a fourth candidate, "Adam Tarr", who
> received zero votes.  Does electing A, B, or C constitute "erasing a
> majority", because the other two candidates' defeats of mr. Tarr are
> not counted?  Of course not.  Such a standard would force us to
> declare every nontrivial multicandidate election as a tie.

So the question is, which of these majority wins matter, and which ones
don't?  "Adam Tarr" is not in the Schwartz set, so does not even make it
to our tiebreaker.  But notice, under both methods, that C's
"non-majority" defeat of B has mattered, to this point, just as much as
A's majority win over C and B's majority win over A, insofar as all
three candidates are still in the running.  We should not be able to say
at this point, without applying further criteria, whether C should or
should not be among the winners.  Yet it seems that you have already
decided C cannot be among the winners (therefore the A over C defeat is
not erased).  Can there be no tiebreaker under which C wins?  My own
intuition, looking at the numbers, says this is pretty close to a
three-way tie.  Perhaps the A over C majority has not been respected
after all?

Let's take another example:
101: A
1: BAC
101: CBA

In this case, B defeats A 102>101,  A defeats C 102>101, and C defeats B
101>1 (with 101 abstaining).  B>A and A>C are victories by majority, but
very weak victories.  C>B is a non-majority win, but a resounding
victory.  Just to get to a three-way tie, we must assume that all 101
abstaining votes really meant to choose B over C (and that's one
possibility out of 2^101 or over
2,000,000,000,000,000,000,000,000,000,000, if we force each one take a
preference).  The only possible choice that will reflect the will of the
electorate is that C wins.  C is the winner picked by margins, but WV
stubbornly insists on the sanctity of a majority of all voters and picks B.

I really think this insistence on victory by a majority only makes sense
in a two-option election, in which case there's really no other way to
go.  Even then, what about the large number of eligible voters who
usually stay home?  We must consider that at least some of them do so
because they have no preference.  If a third candidate were to enter the
race, some of these voters may choose to come and vote, selecting only
the third candidate and not specifying a preference between the initial
two.  By the counting method that says C>B is not a majority victory, we
now may be able to say that there is not a majority win between the
first two.  So even in a two-candidate election, what we really mean
when we say "majority" is "majority of the voters who bothered to
specify a preference".  That's the only way to define majority that
gives the same result under both these two and three candidate
scenarios.  And under this definition, C>B is a majority win.

I think the concept of "majority", in the sense that you mean it, is a
concept that sounds intuitive, but is ultimately not workable in the
same way that cyclic ties sound counterintuitive, but actually make
sense (which, BTW, I have a good geometrical argument as to why cyclic
ties make sense, if anyone cares to see it).  In fact, we can alter the
above examples to make C the Condorcet winner despite not having this
sort of majority in either pairing, and despite B>A being this sort of
majority.  That's because margins -- not WV -- is used when determining
the Condorcet winner.  Everyone seems to be happy with margins and with
non-majority wins when picking a Condorcet winner, so it doesn't make
sense to me that they should suddenly become inadequate halfway through
the counting process.  I'd rather toss the idea that a win involving a
majority of all those coming to the polls must never be defeated by a
win involving a majority of those specifying a preference at the polls.

Tom McIntyre

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