Some Math for Repairing a Hacked Democracy

(Article originally appeared here)

There are cases in which the democracy can be hacked to serve the interest of a single political party. For example, party A controls the opinion of 51% of the population either through clientelistic promises (e.g. vote for me and you get a job in the public administration) or through blackmailing (e.g. if you do not vote for me, this or that will happen to you or your family). Let us say that the 49% of people that are not under the control of party A are also against the party A and its clientelistic/blackmailing practices. Then, if party A wins, how many people are unhappy? Likely more than 49%, as the blackmailed are certainly not happy, such that the number of unhappy may be well over 50%.

Majority voting seems to be the natural criterion for winning a democratic election. Yet, as elaborated in Chapter 17 of the (brilliant!) book “How not to be wrong” by J. Ellenberg, majority voting is a clean winning solution only when there are two options to vote for. For example, there is political party A and political party B, and, say, 57% of the people vote for A means automatically that 57% of the people voted against B. In general, any party that got more than 50% of the votes implies that more than half of the population does not support the other party.

Now, let us check the case with three parties A, B, C. The voting is, say, 40% for A, 30% for B and 30% for C. The party A is a majority winner. However, if we ask the voters that supported B and C it may happen that all of them are against the party A. In this example, majority voting makes 60% of the people unhappy. As explained in the book by J. Ellenberg, there is no ideal solution to this problem.

But can this system of a “hacked democracy” be repaired, at least partially? Let us take a voting system with the following rules:

1. Each voter must give one vote “for” one political party X and one vote “against” one political party Y. The options X and Y can be any parties from the list, but X and Y cannot be identical, i.e. it does not make sense to vote for and against the same party.
2. For each party (or, more generally, political option so as to include various political civic initiatives) one calculates the difference between the votes for (VF) and votes against (VA) to get the score of the party as S=VF-VA. Clearly, S can also be negative for highly unpopular political options.
3. The party with the highest score wins.

Let us say that party A controls 51% of the voters. It is still possible for A to win the election, even if all the remaining 49% of the voters are against A. For example, see the following table:

party | A | B | C |

VF | 51 | 9 | 40 |

VA | 49 | 10 | 41 |

However, the catch is that party A needs to properly steer its controlled population of 51% how to distribute their “against” votes. The problem of party A is that it does not have a control over the people that vote for party B or C and cannot tell them how to distribute their “for” votes. Sure, it can have an estimate of the number of die-hard supporters of B and C, but there may be a large undecided population for which it is difficult to predict how they will vote. In fact, the party A faces some form of betting, where it has to predict the outcome of the votes for the other parties. This is certainly more difficult than only controlling 51% of the voters. For example, the following situation could occur:

party | A | B | C |

VF | 51 | 20 | 29 |

VA | 49 | 10 | 41 |

The party A can have the following strategy. It knows that in the worst case it will have 49% “against” votes, such that its lowest possible score is 2%. By instructing its supporters to put 10% votes against B, the party A guesses that less than 12% will vote for B, i.e. it guesses that the score of B must be less than VF-VA=12–10=2%, such that A would surely win. Similarly, A guesses that C will not get more than 43% of the votes. The Table above shows an example where the guesses of party A are wrong. Note that the outcome in this example is not ideal, as the winner is party B, although it has the least number of “for” votes. However, it becomes more acceptable knowing that many of the “for” votes for party A are forced ones.

This model could also be used as a tool against duopoly, in which two political parties are shifting power to each other at the elections, without giving an opportunity to a third party C or a civic political option. For example, 40% are members of A, 35% are members of B and 25% are undecided or support C. If A or B wins, there are 60% or 65% unhappy voters. It is often that most of the members of A are unhappy if B wins, but are more indifferent if C or another option wins. As A and B are direct and bitter rivals, it follows directly that VA for A is at least 35% and VA for B is at least 40%. Let us say that the undecided voters are tired of the power shift between A and B and they spread their “against”votes, say, 15% to A and 10% to B. These percentages are added to the VA that is already given to A and B, making VA of 50% in total for A and VA of 50% for B. Hence, the score of A is S(A)=VF(A)-VA(A)=40–50=-10%, while the score of B is S(B)=35–50=-15%. The score of C is S(C)=25–0=25%, a clear winner.

It can be shown that this voting model has the following features:

1. If party A controls more than 2/3 (66.67%) of the votes, then it surely wins.
2. The winning party cannot have a negative score, i.e. it cannot have more people against it than people for it. (proof below)

This is not an ideal system, as there are degenerate examples in which a party with very few “for” votes can win the election. However, the merit of this model is that it makes it much more difficult for a single political party to control the election outcome through a clientelism or blackmailing. Instead, a party that aspires to win should also try to minimize the number of people in the society it makes unhappy.

Appendix

Proof of Feature 2:

We do it by contradiction. Let the winning party X1 have a negative score. Let the score of party X1 be S(X1)=VF(X1)-VA(X1). Then all parties must have a negative score as well, i. e. less than the winning party, that is S(X2)<S(X1), S(X3)<S(X1), etc.. Note that the sum of all votes for VF is 100 and the sum of all votes against VA is 100. This means that the total sum of scores must be zero:

S(X1)+S(X2)+S(X3)+…=0

However, this cannot happen if all of the scores are negative. Hence the contradiction and therefore the winning party X1 cannot have a negative score.