Borda was a director for the French Academy of science. He proposed a Voting Mechanism for Voting Theory based on points:

- assign weights to all the candidates in the ranking
- do not consider just the most preferable

The Borda Rule:

- $k$ candidates from set $A$
- $N$ voters communicate their individual preferences $P_i$
- for each individual preference $P_i$ assign the points:
- the 1st candidate in $P_i$ gets $k$ points
- the 2nd candidate in $P_i$ gets $k - 1$ points
- the $j$th candidate in $P_i$ gets $k - j + 1$ points
- the $k$th candidate in $P_i$ gets 1 point
- so define the
*individual Borda score*$S_i(a)$ as the score that candidate $a$ receivers from a voter $i$ - $S_i(a) = k - \text{pos}_i (a)$ where $\text{pos}_i(a)$ is the position of $a$ in $R_i$

- define the
*Borda score*$B(c)$ for candidate $c$ as the sum of points from all $P_i$- $B(c) = \sum_j S_j(c)$

- the candidate with the highest Borda score wins the election

Individual preferences:

- $a > b > c$ - 11 votes
- $b > a > c$ - 8 votes
- $c > b > a$ - 2 votes

Scores:

- $B(a) = 3 \cdot 11 + 2 \cdot 8 + 1 \cdot 2 = 51$
- $B(b) = 3 \cdot 8 + 2 \cdot 11 + 2 \cdot 2 = 50$
- $B(c) = 1 \cdot 11 + 1 \cdot 8 + 2 \cdot 3 = 25$

$a$ gets elected

This method satisfies:

This method does not satisfy:

- let $A$ be the set of candidates: $A = \{a, b, c, ...\}$
- $N$ voters
- suppose $x$ improves its positions only for one voter $j$
- i.e. there's a new preference ranking $P'_j$ where the candidate $x$ has a better position than in the old preference ranking $P_j$
- if $a$ improved his ranking on $m$ positions, then $S'_j(x) = S_j(x) + m$ ($m > 0$)

Now let's analyze how it will affect the global score

- the score before: $B(x) = \sum_i S_i(x)$
- the score after: $B'(x) = \sum_i S'_i(x) = ...$
- $... \sum_{i \ne j} S'_x(x) + S'_j(x) = ...$
- the scores for $i \ne j$ has not changed, so can replace them by the old score
- $... \sum_{i \ne j} S_x(x) + S'_j(x) = ...$
- and $S'_j(x) = S_j(x) + m$
- $... \sum_{i \ne j} S_x(x) + S_j(x) + m = B(x) + m$

- so $B'(x) = B(x) + m \Rightarrow B'(x) > B(x)$
- and therefore the Monotonicity principle is respected

Separability is respected if

- we divide the region $\Omega$ into two regions $N$ and $\overline{N}$
- the global ranking is the same in $N$ and $\overline{N}$ and the same candidate $a$ wins
- then if considered the whole region $\Omega$, the global ranking should be the same and the same candidate $a$ should win

Suppose that

- $S_N(a) \geqslant S_N(b)$ and $S_\overline{N}(a) \geqslant S_\overline{N}(b)$
- and therefore $B_N(a) \geqslant B_N(b)$ and $B_\overline{N}(a) \geqslant B_\overline{N}(b)$ (1)
- i.e. in two regions the relation between candidates $a$ and $b$ is the same

- consider the whole region $\Omega$:
- $B(a) = \sum_j S_j(a) = ...$
- can split the sum into two parts: for $N$ and for $\overline{N}$
- $... \sum_{j \in N} S_j(a) + \sum_{j \not \in N} S_j(a) = B_N(a) + B_\overline{N}(a)$
- the same for $b$: $B(b) = B_N(b) + B_\overline{N}(b)$

- so because of (1) can say that $B(a) \geqslant B(b)$
- thus the Separability principle is respected

Consider this example:

- $N = 7, A = \{a, b, c, d\}$

Preferences:

- 3 voters: $c > b > a > d$
- 2 voters: $b > a > d > c$
- 2 voters: $a > d > c > b$
- The outcome is $a > b > c > d$

$d$ decides to withdraw - anyway he has no chance of winning

- but it has a strong effect on the result!
- Now the global ranking is $c > b > a$ - the complete opposite!

Another example:

- $N = 5, A = \{a, b, c\}$

Preferences:

- 2: $a > b > c$
- 1: $c > a > b$
- 2: $b > c > a$

Scores:

- $B(a) = 9, B(b) = 11, B(c) = 9$
- outcome: $b > a \geqslant c$

Now $d$ also decides to participate:

- 2: $a > d > b > c$
- 1: $c > a > d > b$
- 2: $b > c > a > d$

Scores are:

- $B(a) = 15 > B(b) = 14 > B(c) = 10 > B(d) = 9$
- note that even though $d$ is the last one in the global rating, adding him changed the winner!

So we see that by carefully choosing new candidates it's possible to manipulate the results.

Consider these individual rankings for $A = \{a, b, c\}, N = 5$

- 3: $a > b > c$
- 2: $b > c > a$

Borda Score:

- $S(a) = 3 \cdot 3 + 2 \cdot 1 = 11$
- $S(b) = 3 \cdot 2 + 2 \cdot 3 = 12$
- $S(c) = 2 \cdot 2 + 3 \cdot 1 = 7$
- $b$ wins the election

However in pair-wise comparison we see that

- $a > b$ for 3 voters
- $b > a$ for 2 voters

$\to$ the majority prefers $a$ over $b$ but $b$ wins the election

- the Condorcet Fairness criterion is not satisfied

Slightly different approach

- instead of assigning points to all $k$ candidates
- assign points to $n < k$ candidates
- i.e. assign $n$ points to 1st, $n-1$ to 2nd, ..., $1$ to $n$th and 0 to the rest

- Decision Engineering (ULB)
- Voting Fairness Criteria [1]