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:
$N = 16$ 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.
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.
Is not satisfied
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