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Two-Round Voting

voting-theory

Two-Round Voting

This a voting mechanism from Voting Theory. It is essentially the same as Plurality Voting, but run in two rounds

Given set $A$ of candidates

  • Round 1: Using plurality voting mechanism, choose two candidates $a, b \in A$
  • Round 2: Plurality voting only between the two $a$ and $b$

Example

  • here we assume that preferences are ‘‘stable’’: people don’t change their preferences between two rounds
  • Round 1:
    • $a > b > c$ - 11 votes
    • $b > a > c$ - 8 votes
    • $c > b > a$ - 2 votes
    • $a$ and $b$ win the 1st round
  • Round 2:
    • (we just remove $c$ from the previous rankings)
    • $a > b$ - 11 votes
    • $b > a$ - 8 + 2 votes

Criteria

This method satisfies:

This method does not satisfy:

Monotonicity

$N = 16$ and $A = {x, y, z}$

We have the following individual preferences:

  • 6 voters $x > y > z$
  • 5 voters $z > x > y$
  • 4 voters $y > z > x$
  • 2 voters $y > x > z$

Elections:

  • Round 1: $x$ and $y$ win ($x=6, z=5, y=6$)
  • Round 2: $x$ wins (6+5: $x > y$, 6: $y > x$)

But suppose that $x$ manages to also convince the last two voters that he is better:

  • 6 voters $x > y > z$
  • 5 voters $z > x > y$
  • 4 voters $y > z > x$
  • 2 voters $x > y > z$

Note that $x$ by improving his position should remain the winner

Elections:

  • Round 1: $x$ and $z$ ($x=6+2, z=5, y=4$)
  • Round 2: $z$ wins (not $x$!) (6+2: $x > y$, 9: $z > x$)   This counter-example shows that the Monotonicity principle is not respected by Two-Round Voting method.

Separability

Suppose we run an election in Belgium

  • there are 2 communes - 2 regions
  • we have 3 candidates: $A = {a, b, c}$
  • $N = 13$ for each region

| | Region I | Region II | preferences || |- 4: $a > b > c$

  • 3: $b > a > c$
  • 3: $c > a > b$
  • 3: $c > b > a$ | |- 4: $a > b > c$
  • 3: $c > a > b$
  • 3: $b > a > c$
  • 3: $b > a > c$ | Round 1 || ${\color{blue}{a: 4}}, b: 3, {\color{blue}{c: 6}}$ || ${\color{blue}{a: 4}}, {\color{blue}{b: 6}}, c: 3$ || Round 2 || ${\color{blue}{a: 7}}, c: 6$ || ${\color{blue}{a: 7}}, b: 6$ | In both regions $a$ wins.

But if we consider the global region, we’ll have different results:

  • Round 1: ${\color{red}{a: 8}}, {\color{blue}{b: 9, c: 9}}$ - note that $a$ loses and doesn’t go to the next round
  • Round 2: ${\color{blue}{b: 17}}, c: 9$
  • $c$ wins

So the separability principle is not satisfied in this example.

Condorcet Fairness Criterion

Is not satisfied

  • http://www.ctl.ua.edu/math103/voting/methodof.htm

Sources

✏️ Edit on GitHub