[EM] Converting super-majorities to margins

[Blake Cretney]

In my previous post I comment on the possibility of combining some
sort of super-majority requirement with various single-winner methods.
 At one point I mention that I assume that a qualifying super-majority
should always have a greater margin of victory than a normal majority.
 It isn't obvious, however, that this would be the case.  For example,
one common super-majority requirement is a 2/3 vote.  However

12 to 8 -> margin of 4
has a greater margin than
6 to 3 -> margin of 3

But, 6 to 3 is a 2/3 vote, while 12 to 8 is not.

The general procedure I suggested was:

1.  Order the options using a method without regard to SQ.
2.  Pick the highest ordered option with a path of super-majorities
to the SQ.  For convenience, you can also use only super-majorities
from a restricted set, as I suggest for Tideman.

This will work for the 2/3 requirement, but the procedure is
complicated by the fact that two kinds of precedence are required.

As well, the method seems to have some internal conflict.  That is,
if on the one hand, when using Tideman's locking procedure, we claim
that 12-8 takes precedence over 6-3, on what basis do we decide that
for a different purpose, 6-3 takes precedence over 12-8.  Tideman is
based on the principle that we should be more confident in a majority
with a higher margin.  If we discard this principle, why do we use
Tideman?

It therefore seems desirable to convert the 2/3 requirement into a
margin.  Fortunately, this is easy to do.  We define the required
margin as the minimum margin necessary to ensure that an option does
not win without having a 2/3 vote.  A convenient formula, is
round(v/3), where "round" means round to nearest integer.

A side effect of this is that it means that contests with less than
full participation will require more than 2/3 to qualify.  For
example, 6-3 would not qualify for a vote of 21 ballots.  However,
this can be viewed as desirable, as our confidence might diminish as
participation drops.

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Another case is where a quorum is desired.  For example, it may be required
that at least 5 people participate in a vote for the SQ to be changed.  

Now, although quorums are often specified simply in terms of participation,
it makes little sense to allow a vote to pass because of the presence of
objections.  For example, with a quorum of 5, 
4 to 3 passes
4 to 0 fails

This absurdity is avoided because the 3 "no" voters would likely withdraw for
the purposes of the vote, if this allowed it to be defeated by lack of quorum.
 However, this kind of action is harder to manage in a ranked ballot, so I
will make it a provision that "no" votes should never be counted towards
making a quorum.

Clearly, "yes" votes should be considered in making up the quorum.  It is a
little more difficult to decide what should be done with neutral votes.  That
is, voters who rank two options as equal.  If the neutral voters really are
neutral, they don't care how their vote is tallied, but it could make a
difference to the outcome.  My choice is to count the neutral vote as half a
vote towards making quorum.  Not only does this seem natural for a vote in
between a "yes" and a "no", but it allows a quorum requirement to be easily
converted to a margin requirement.  The formula is that the minimum margin is
equal to two times the quorum, minus the total votes. m=2q-v

Let y, n, a be number of yes, no, and abstention votes
Let v be total number of votes
Let q be quorum required
Let m be minimum margin required

Clearly,
v=y+n+a
m=2q-v
Assume y-n>=m
   y-n>=2q-v
   y-n+v>=2q
   y-n+y+n+a>=2q
   2y+a>=2q
   y+a/2>=q, which is how I said I wanted votes towards quorum to be tallied
So, y-n>=m -> y+a/2>=q

The above argument can be reversed to show that y+a/2>q -> y-n>=m
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Here's an example.  40 ballots.  The vote will require a quorum of 23 and a
2/3 vote.

The 2/3 vote gives us the requirement round(v/3)=round(40/3)=13

The quorum required gives m=2q-v=46-40=6.  Since 13 is also required for a
super-majority, this doesn't have any effect.
Winner Loser Margin
A      SQ    16
B      A     15
B      SQ    12

Using Tideman, we 
lock A->SQ 16
lock B->A  15
lock B->SQ 12

Order B > A > SQ

Even though B does not have a 13 margin directly against SQ, it does through
a path.  B qualifies, and is therefore the winner.

---
Blake Cretney