Majority Rule in Social Networks

In popular culture, there is much talk of social networks. Indeed, Facebook, Twitter, Foursquare, and Linkedin are ubiquitous – how did we ever live without them?

One interesting property of these social networks or social graphs that isn’t talked about much is the concept of Majority Rule. While I haven’t seen much written about Majority Rule in social networks, we have all used the tools available to us: For Facebook, they have the Like button. For Twitter, there is the Retweet. For Digg, there’s the Digg button. And the same goes with every other tech company and their acquisitions.

One important note: votes can be explicit votes, such as the examples above. But, there are implicit votes also, such as how many times a YouTube video has been seen.

For Foursquare, the check-in can be both explicit and implicit. For example, when I take the kids to the park, my reason is to spend time with the kids. But, while I’m there I can choose to check-in.

But, there are cases where one chooses to go to a specific location for the sole purpose of checking-in. In that case, I consider that explicit. I imagine the commercial viability of Foursquare lies in the explicit check-in: purposely going to a location to check-in, then receive some reward from the place visited (coupon or whatever).

For the purposes of this post, we’re only talking about explicit votes.

So, how does Majority Rule work in a social system?

The Physical World: Majority Rule and Social Graph

In the physical world, majority rule systems appear in electing government representatives, board of directors approval of new policies, and other like instances.

One property that exists in the physical world, but not so much in virtual social graphs, is that the members of the group involved in some voting process are to support and sustain the majority rule.

For example, if a government representative wins from a democratic process and obtains the majority of the votes, he or she wins.  As a member of that group, I am to sustain and support that person, if I choose.  What I don’t have control over, however, is that I become a recipient of that person’s decisions, such as the legislation of a new law.

Indeed, majority rule in the physical world means that we are affected in very real ways.

The Virtual World: Majority Rule and Social Graph

Well, for virtual social graphs such as the Twitter or Facebook, if a majority rule is achieved, there is no contract to compel the members of that group to follow the majority rule. For example, as a member of the broad community of Facebook, if a fan page is Like’d 500 times, I have a choice to like or not to like. And, just because something is like’d a gazillion times, it still has no tangible affect on me.

What a majority rule does do, for me, is pique my interest. It creates curiosity and, if I choose to clickthrough to the thing that has achieved majority rule (majority like’d or majority retweeted), then I might receive some utility such as laughter, entertainment, or joy.

So, what is it exactly that makes Majority Rule work in social systems?

Properties of Majority Rule

There are three main properties of Majority Rule Systems:

1. Option Equality: The characteristics of choice A and choice B are the same.  If the same amount of people vote for choice A, the results will switch.
2. Voter Equality: Voter equality means that no person’s vote is more influential than that of another’s vote.
3. Sensitivity: Sensitivity means that if the voting is tied and one more person votes, then the option he votes for wins.

One major presupposition in the notion of majority rule is that of choice. Social scientists call the method of choosing a social choice function. A social choice function is a rule for choosing between options where some of the citizens prefer one option and others prefer the other.

The majority rule system chooses one option over another when more citizens prefer the one option than the other. From this we get the definition of a social choice function (SCF):

SCF: C is a social choice function for options a and b and citizens V iff a≠b, and V is finite and non-empty, and for all disjoint subsets A and B of V, C(A, B) equals {a} or {b} or {a, b}.

We also get the definition of the majority rule social choice function:

mrSCF: C is a majority rule social choice function for options a and b and citizens V iff C is a social choice function for options a and b and citizens V, and for all disjoint subsets A and B of V, C(A, B)={a, b} if A≈B, C(A, B)={a} if A>B, and C(A, B)={b} if B>A.

In addition to the definition of majority rule social choice function, we should formalize the definitions of the three properties of majority rule.

Option Equality (OE) can be defined thus:

OE: C satisfies option equality for options a and b and citizens V iff for all disjoint subsets A and B of V, C(A, B)={a} iff C(B, A)={b}.

Voter Equality (VE) can be defined as the following:

VE: C satisfies voter equality for options a and b and citizens V iff for all disjoint subsets A and B of V, if A≈B, then C(A, B)=C(B, A).

Sensitivity (S) can be formalized as follows:

S: C satisfies sensitivity for options a and b and citizens V iff for all disjoint subsets A and B of V, if C(A, B)={a, b}, then for every non-empty subset X of V-(A∪B), C(A∪X, B)={a}.

Now, let’s prove that Majority Rule satisfies the three properties above. In other words,

Theorem: C is the Majority Rule social choice function for options a and b and citizens V iff C is a social choice function that satisfies Option Equality, Voter Equality, and Sensitivity for options a and b and citizens V.

Majority Rule – Proof

To prove this theorem, I must proceed in two phases. The first phase will contain the three properties of majority rule. The second phase will contain the three conditions of majority rule social choice function.

Phase I

1.1 (OE) C(A, B)={a} iff C(B, A)={b}

1.2 (VE) If A≈B, then C(A, B)=C(B, A)

1.3 (S) If C(A, B)={a, b}, then for every non-empty subset X of V-(A∪B), C(A∪X, B)={a}

Phase II

2.1 C(A, B)={a, b} if A≈B

2.2 C(A, B)={a} if A>B

2.3 C(A, B)={b} if B>A

The framework I’ve set-up allows me to show how the elements in Phase I satisfy the elements in Phase II. To do this, I will assume the elements in Phase II and prove the elements in Phase I.

Similarly, I need to show how the elements in Phase II satisfy those in Phase I. To do this, I will assume the elements in Phase I and prove the elements in Phase II.

I can prove the theorem this way because it is a bi-conditional statement (principle of transitivity). That is, if I assume the left-hand side, the right-hand side should follow. Similarly, if I assume the right-hand side, the left-hand side should also follow.

By showing how Phase I and II prove each other, I will ultimately show that the majority rule social choice function does satisfy option equality, voter equality, and sensitivity and hence demonstrate fairness and effectiveness.

Phase I : 1.1

Show that C(A, B)={a} iff C(B, A)={b}.

Proof: Assume that C(A, B)={a}; show C(B, A)={b}. If A≈B, then C(A, B)={a, b}. If B>A, then C(A, B)={b}. By process of elimination and cases, since C(A, B)={a}, A>B. So, C(B, A)={b}. Now, assume the other side of the bi-conditional, i.e., C(B, A)={b}; show C(A, B)={a}. If A≈B, then C(A, B)={a, b}. If A>B, then C(A, B)={a}. By cases and process of elimination, B>A. So, C(A, B)={a}. Q.E.D.

Phase I : 1.2

If A≈B, then C(A, B)=C(B, A).

Proof: Assume A≈B; show C(A, B)=C(B, A). So, C(A, B)={a, b}. Also, C(B, A)={a, b}. Hence, C(A, B)=C(B, A). Q.E.D.

Phase I : 1.3

If C(A, B)={a, b}, then for every non-empty subset x of V-(A∪B), C(A∪X, B)={a}.

Proof: Assume C(A, B)={a, b}; show that for every non-empty subset X of V-(A∪B), C(A∪X, B)={a}. Since X≠∅ and x⊆(V-(A∪B)), A∪X>A. If A>B, then C(A, B)={a}.

If B>A, then C(A, B)={b}. So, by the process of elimination and cases, since C(A, B)={a, b}, A≈B. Hence, A∪X>B. Consequently, C(A∪X, B)={a}. Q.E.D.

Thus, by proving Option Equality, Voter Equality, and Sensitivity, I have shown that they are derived from the notion of majority social choice function. But I am only halfway finished.

I now need to prove the elements in Phase II. Remember, that in doing so, I will be assuming the elements in Phase I.

Phase II : 2.1

If A≈B, then C(A, B)={a, b}.

Proof: Assume A≈B; show C(A, B)={a, b}. Since A≈B, by voter equality I get C(A, B)=C(B, A). Also, bear in mind the definition of a social choice function, namely, that C(A, B) must equal {a} or {b} or {a, b}. By option equality, C(A, B)={a} iff C(B, A)={b}. Yet, {a}≠{b}. So, C(A, b)={a, b}. Q.E.D.

Phase II : 2.2

If A>B, then C(A, B)={a}.

Proof: Assume A>B; show C(A, B)={a}. Sensitivity claims that if C(A, B)={a, b}, then for every non-empty subset X of V-(A∪B), C(A∪X, B)={a}. Let A’ be a subset of A such that A’≈B. Accordingly, by Phase 2.1, I get C(A’, B)={a, b}. So, C(A, B)={a}. Q.E.D.

Phase II : 2.3

If B>A, then C(A, B)={b}.

Proof: Assume B>A; show C(A, B)={b}. Sensitivity dictates that if C(A, B)={a, b}, then for every non-empty subset X of V-(A∪B), C(A∪X, B)={a}. Let B’ be a subset of B such that A≈B’. By Phase 2.1, I get C(B’, A)={a, b}. But this means that C(B, A)={a}. So, by option equality, C(A, B)={b}. Q.E.D.

Conclusion

We have proven both phases. The Majority Rule Social Choice function satisfies option equality, voter equality, and sensitivity. By satisfying these three properties, the majority rule proves to be both fair and effective.