# Supermajority pairwise methods?

Steve Eppley seppley at alumni.caltech.edu
Mon Dec 23 10:29:13 PST 1996

```Contents:
When None may win
Supermajority Methods

WHEN NONE MAY WIN
-----------------
In a pairwise preference vote among alternatives where None may win,
None is also a choice on the ballot.
The ballot format may vary:
1. NOTR (None of the Rest) or None as an explicitly rankable alternative
2. dividing lines between approved, neutral, and disapproved alternatives
3. explicit pairings of alternatives with None using Yes/Nos
etc.

Using ballot format #1, for instance, a voter might vote:
C > A=B > E=NOTR > D
The same vote using a format with #2:
C(AB)/E/D     <--- The parentheses are Tom Round's notation for '='
The same vote using a format with #3:
C(Yes) > A(Yes)=B(Yes) > E() > D(No)

The third format allows for intransitive preferences, and thus might
be considered to allow irrational votes such as A(No) > B(Yes).  The
following discussion assumes all votes adhere to the transitivity
assumption.

SUPERMAJORITY METHODS
---------------------
In some cases, the group might want to require a supermajority in
order to pass certain alternatives.  Some possible methods come to mind:

1a) Any alternative which needs supermajority but fails to beat None
pairwise by the required amount is defeated and eliminated from
the pairwise matrix.

1b) Any alternative which needs supermajority but fails to beat None
pairwise by the required supermajority amount is defeated but *not*
eliminated from the matrix.  (It may knock out other alternatives.)

2a) For every alternative X which needs supermajority, pairwise
preferences for None over X (p[None,X]) are multiplied by the
appropriate factor Sm.  Then the matrix is tallied normally
(such as with Condorcet's method).

2b) For every alternative X which needs supermajority, pairwise
preferences for X over None (p[X,None]) are divided by the
appropriate divisor Sd.  Then the matrix is tallied normally.

2c) A combination of 2a and 2b, with Sm=Sd for symmetry.

(Any other supermajority pairwise methods I've overlookd?)

In 1a and 1b, the "required supermajority" could take different
forms depending on whether indifferent voters (q[X,None]) and nonvoters
are neglected, treated as opposed, or something in between.
(The group might demand the supermajority be a percentage of the
entire set of voters, for example, though I don't know if this demand
would be sensible.)
For a given "indifference factor" k (0 <= k <= 1)
an alternative X passes the supermajority threshold S (.5 <= S <= 1)
if p[X,None]  >=  S * (p[X,None] + p[None,X] + kq[X,None])

(Theoretically, there could be rational reasons to sometimes use S < .5,
which would allow a minority to pass a proposal opposed by a majority.
One example is a group which wants its operating rules to evolve
quickly, so only rules which are preferred by a supermajority will
endure.  This could be chaotic, but possibly of value while the rules
are young.  A group could also schedule periods of fast and slow
evolution, a form of punctuated equilibrium, by varying S.)

As with methods 1ab, methods 2abc might also make use of an
"indifference factor" k, to swell p[None,X] by adding kq[X,None]
to it.

Methods 2abc would sometimes allow an alternative X which fails
to super-beat None to pass anyway, because of the possibility
of circular ties:

Example:  method 2a-Condorcet, Sm = 1.1,  k = 0
46:  Y > X=NOTR
20:  X > Y=NOTR
34:  NOTR > X > Y

X        Y      None
X               54L      20
Y      46                46L             P[] matrix
None   34*1.1L  34*1.1
-------  -------  -------
37.4     54       46       <--- Condorcet scores

X > Y > superNone > X
X passes, even though it was below supermajority vs. None.
Y fails, even though it exceeded supermajority vs. None.

Any opinions on the relative merits of the different methods?

It looks to me like 2a is better than 2b or 2c if Condorcet's method
is used, since Condorcet's method counts opposition and the point of
supermajority is to privilege opposition.

Methods 1ab give more special privilege to the None choice than
methods 2abc.  Perhaps there is some philosophical or rational
justification for doing so, especially when the group has already
chosen to give None some special privilege by setting S>.5; why not
give it more?  Or maybe most people would simply irrationally
prefer 1ab more than 2abc.  On the other hand, at first glance
methods 1ab appear to have significantly more misrepresentation
strategy problems than 2abc.

---Steve     (Steve Eppley    seppley at alumni.caltech.edu)

```