**"Maximize Affirmed Majorities" (MAM) is resolvable
and reasonably deterministic.**

Revised: December 31, 2002

* Resolvability*: For all collections of
votes for which two or more alternatives

have a chance of being elected, there must exist some admissible vote such that,

if this vote were added to the collection of votes, one alternative would be elected

with certainty and the rest defeated with certainty. In particular, the one which

would be elected with certainty with this vote added must be one of those having

a non-zero chance of being elected without this vote added.

* Reasonable
Determinism*: For all collections of
votes such that no alternatives

tie pairwise and no majorities are the same size, there must exist an alternative

which is elected with certainty.

These two criteria are closely related, satisfaction implying the voting rule
will usually

leave nothing to chance when electing the winner. Compliance with these criteria

means that in large public elections the
outcome will nearly always be *deterministic*.

It would be undesirable to require complete determinism in all
scenarios. That would

require sacrificing either * neutrality* of the alternatives or * anonymity* of the
voters.

To illustrate this, suppose there are two alternatives *x*
and *y*, and half of the voters

rank *x* over *y* and the other half rank *y* over *x*.
*Neutrality* requires neither *x* nor *y*

be privileged over the other, and * anonymity* requires no voter be
privileged over

any other voters. Therefore in this example both alternatives must be given an equal

chance (½) of being elected. (If one of the alternatives is "continue the status quo,"

it would be
reasonable to violate *neutrality* by privileging the status quo for
the sake

of stability, rather than
resort to chance when breaking a tie.) If the number of voters

is large, it is unlikely
that exactly half will rank *x* over *y* (or *y* over *x*). What
matters

more is that the voting rule be * deterministic* in more likely scenarios, which raises

the question of how to express such a
criterion.

*Resolvability* is more widely recognized than *reasonable
determinism* by the political

science community as a desirable criterion. *Reasonable
determinism* is tailored in

favor of
"pairwise" voting procedures such as MAM, Simpson, Copeland,
etc. (but

not Borda nor Instant Runoff) and
thus should only be used when comparing pairwise

election procedures. *Reasonable determinism* is mentioned here because MAM

complies, in case anyone wonders and because understanding the proof
(provided

elsewhere) may augment one's understanding of MAM.

Two examples illustrate *resolvability* and *reasonable determinism*:

__Example 1__: Suppose there are 3
alternatives *x*, *y* and *z*. Suppose there are 15
voters

who rank the alternatives as follows:

4 |
5 |
6 |

x |
y |
z |

y |
z |
x |

z |
x |
y |

It can be seen that a majority (9 voters) rank *y* over *z*, and
a majority (10 voters) rank *x*

over *y*, and a majority (11 voters) rank *z* over *x*.
Thus no two majorities are the same size

and no two alternatives are tied pairwise, so *reasonable determinism* demands
that the

voting procedure elect one of the alternatives with certainty. Consider
the Copeland

procedure, which awards each alternative a point for each pairing it wins plus
a half point

for each pairing it ties, and then chooses the alternative(s) having the
highest score. (This

is the same procedure used to rank teams in many professional sports, such as
baseball.)

In this example, Copeland awards each alternative 1 point. (For instance,
since a majority

ranked *x* over *y* and a majority ranked *z* over *x*, *x*
wins the "*x* versus *y*" pairing and loses

the "*x* versus *z*" pairing; thus *x* has a score of
1 + 0 = 1.) Therefore Copeland chooses

all three alternatives, proving Copeland does not satisfy *reasonable
determinism*.

It can also be checked that there is no admissible vote (a ranking of the
alternatives)

which, if added to the 15 votes, would affect the three Copeland scores. Thus

Copeland also does not satisfy *resolvability*.

__Example 2__: (Illustrates satisfaction
of *resolvability*) Suppose there are 3 alternatives

*x*, *y* and *z*. Suppose there are 14 voters who rank
the alternatives as follows:

4 |
5 |
5 |

x |
y |
z |

y |
z |
x |

z |
x |
y |

It can be seen that a majority (9 voters) rank *y* over *z*, and a
majority (9 voters) rank *x*

over *y*, and a majority (10 voters) rank *z* over *x*.
Consider the operation of MAM in

this example: Only two of the three majorities can be affirmed since the
three
together

are not internally consistent. (In other words, they *cycle*.) First MAM affirms
"*z* over *x*"

since it has the largest majority. Since both of the other majorities are
the same size,

the one given greater precedence by MAM depends on a tiebreaking strict ranking

constructed using the Random Voter Hierarchy procedure. Thus there is a 4/14

chance that the tiebreaking ranking will be "*x* over *y* over *z*" and a 5/14
chance that

the tiebreaking ranking will be "*y* over *z* over *x*" and a 5/14 chance
that the
tiebreaking

ranking will be "*z* over *x* over *y*." The loser of the "*y* over *z*" majority is *z*, and
the loser

of the "*x* over *y*" majority is *y*. Since there is a 9/14 chance
that the tiebreaking ranking

ranks *y* over *z*,
there is a 9/14 chance that MAM gives greater precedence to "*y* over *z*"

than to
"*x* over *y.*" Thus "*x* over *y*"
has a 9/14 chance of not being affirmed, and "*y* over *z*"

has
a 5/14 chance of not being affirmed. Thus there is a 9/14 chance that MAM
elects *y*

and a 5/14 chance
that MAM elects *z*. There is no alternative that MAM elect with

certainty. Now suppose one extra vote that ranks "*z* over *x* over *y*"
were added to the

14 votes. Then, as in example 1, there would be a majority of 9 who rank *y* over *z*,

and a majority of 10 who rank *x* over *y*, and a majority of 11 who rank *z* over *x*.
This

means
MAM would affirm "*z* over *x*" and then affirm "*x* over *y*" and would not affirm

"*y*
over *z*." Thus, if that vote were added, MAM would elect *z* with certainty and

would defeat *x* and *y*
with certainty.

Theorem "** MAM is
resolvable.**" We provide two proofs; the second
demonstrates

MAM

in the document defining MAM.

Proof #1 that MAM is resolvable: Assume no alternative in * A* is
elected with certainty

by MAM given the votes

definition of MAM, at least one alternative always has a non-zero chance of being

elected by MAM. Relabel the alternatives if necessary so that

that has a non-zero chance of being elected by MAM given

σ ∈ L(

vote

certainty given

Maj =
Majorities(* A,R*).

For all

Aff = Affirmed(

Top = Top(

MAM = MAM(

Let
*r*
= *R*_{social}(*A**,**R**,*σ). By
theorem
"Properties of RSocial (1)", *r* is a strict ranking

of ** A**. (That is,

vote. By theorem "Properties of

(1.1) For all *(a,b)*
∈ Aff, *r* ranks *a* over *b*.

Let *P _{r}* denote {

pairs that "agree" with

By 1.1, Aff ⊆

Abbreviate * R'* =

Make the following abbreviations:

For all *(a,b)* ∈
Pairs(* A*), let Precede

Aff

Top

MAM

We aim to show MAM(* A,R',*σ

the following four propositions:

Proposition 1.2: Aff* '* ⊆

Proposition 1.3: Aff ⊆ Maj

Proposition 1.4: Aff ⊆ Aff

Proposition 1.5:

Proposition 1.6: For all

Proof of 1.2: Suppose the contrary.
This means Aff** '**\

"Precedence is a Strict Ordering" we can pick

in Aff

Thus #

Since

be inconsistent with Aff ∩ Precede(

Aff ⊆

and #

#

Since

Since

Since

exist

and

by theorem "Precedence is a Strict Ordering"

that precedes

so the contrary assumption cannot hold, establishing 1.2. QED

Proof of 1.3: Pick any *(a,b)*
∈ Aff. We must show *(a,b)* ∈
Maj** '**. By the

definition of Affirmed(),

By 1.1,

and #

Thus 1.3 has been established. QED

Proof of 1.4: Since Aff ⊆
*P _{r}* and

consistent with

By 1.3, Aff ⊆ Maj

consistent with Aff

Thus 1.4 has been established. QED

Proof of 1.5: Since *x*
tops *r*, *P _{r}* contains no pair in which

by 1.2 Aff

of Top() that

Proof of 1.6: Assume *y* ∈
* A*\{

Case 1.6.1: *y* ∈
Top. By the
definition of *R*_{social}(), MAM = *x*. Thus,

by the
definition of MAM *x* ∈
Top. Since {x,y} ⊆ Top, by theorem

"MAM is Reasonably Deterministic (1)" #** R**(

tops

and #

(1)"

Case 1.6.2: *y* ∉
Top. By the definition of Top(),
there
exists *a* ∈
**A**

such that *(a,y)* ∈
Aff. By 1.4 *(a,y)* ∈
Aff* '*. Therefore

Since *y* ∉
Top* '* in both cases, 1.6 has been established.
QED

By 1.4 and 1.5, Top* '* = {

L(

elect

Thus MAM is

Proof #2 that MAM is *resolvable*: We rely on theorem
"MAM_{2}
= MAM" and

show that MAM_{2} is *resolvable*. Assume *x* has a
non-zero chance of being elected

by MAM_{2}. This means there exists σ ∈
L(* R,A*) and

that

admissible vote. We will show UndominatedRankings(

all σ

and defeats all other alternatives with certainty given

Clearly both of the following statements hold:

(2.1) #*R** '*(

(2.2) #

Pick any σ* '* ∈

that

Agreed(*r*) = Agreed(*r, A,R*).

Agreed(

Agreed

Agreed

Distinct = DistinctAgreed(

Distinct

LDA = LargestDistinctAgreed(

LDA

Undom = UndominatedRankings(

Undom

Since *r* ∈ Undom,
the following statement must hold:

(2.3) If Distinct ≠
φ, then LDA ∈ Distinct
∩ Agreed(*r*).

Since *r* and *r'* are distinct strict
rankings of * A*, the following statement must hold:

(2.4) There exist *a,b* ∈ * A* such that

By 2.1 and 2.2, both of the following statements must hold:

(2.5) #*R** '*(

(2.6) #

By 2.3, 2.5 and 2.6, both of the following statements must hold:

(2.7) If Distinct ≠
φ, then LDA ∈ Distinct* '*
∩ Agreed

(2.8) Distinct

Since *r* ∈ Undom, *r'*
does not dominate *r* given *( R,*σ

Case 2.9: *r* does not
dominate *r'* given *( R,*σ

given

for all

By 2.5 and 2.6, both of the following statements must hold:

(2.9.1) For all

and

(2.9.2) For all

and

By 2.9.1 and 2.9.2, both of the following statements must hold:

(2.9.3) For all

and

(2.9.4) For all

and

By 2.4, 2.9.3 and 2.9.4, Distinct

This implies LDA

Case 2.10: *r* dominates *r'*
given *( R,*σ

LDA ∈ Agreed(

for all

given

Thus

Since *r* dominates *r'* given *( R',*σ

distinct from

UndominatedRankings(

MAM

given

MAM is