Assessing the efficiency of a single vote in a multiparty presidential election

Question

In a country there is a voting system where all parties get represented in parliament if they meet a bar of $n$ percent.

Suppose that the parties are grouped into two groups of red $R_1, \dots R_k $ and blue parties $B_1, , B_l$ and they poll percentages $r_1, \dots r_k, b_1 \dots b_l$ with standard errors $sr_1 , \dots sr_k, sb_1, \dots sb_l$ (e.g. $R_1 \sim \mathcal{N}(r_1, sr_1)$ etc. ). There are two presidential candidates a red and a blue. The presidential candidate with the largest coalition in parliament, e.g. the largest sum $\sum_{i=1}^k R_k 1_{R_k > n}$ or $\sum_{i=1}^l B_l 1_{B_l > n}$ wins the election.

Now, suppose that a voter only cares about which presidential candidate wins. I.e. a voter preferring a red candidate would vote for one of the parties $R_1, \dots R_k$ (with abuse of notation).

It is clear that if $r_j$ and $sr_j$ are both much much smaller than $n$ then the vote of the voter effectively does not matter. Whereas if $r_j$ and $sr_j$ are both much much larger than $n$ then the vote effectively count as one. But what happens when $r_j$ is very close to $n$. How do I define and compute how much a vote on $r_j$ counts compared to in the limit where $r_j$ is large?

You can say that the total amount of votes is fixed $N$ a very large number.

I guess one way to define this efficiency is the probability that the vote tips the election and of course there is a brute force way to simulate this looking at how often the single vote tips the election. But that seems computationally infeasible to me. Is there a nice way to go about this problem?

Answer 1

The effect of a vote on party $R_i$ on the total sum of votes is either

• $0$ when $r_i \leq n-2$
• $n$ when $r_i = n-1$
• $1$ when $r_i \geq n$

In terms of the expectation value you can compute it with the probability for the conditions (in this case you need to discretize the normal distribution).

Approximately you will get

$$n \cdot P(r_i = n-1) + 1 \cdot P(r_i \geq n) \approx n \frac{1}{sr_i} \phi\left(\frac{n-r_i}{sr_i} \right) + 1 - \Phi\left(\frac{n-r_i}{sr_i} \right)$$

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However, one may wonder whether the additional $1$ vote in the case of $r_i > n$ is so much relevant. The question is whether this single vote is gonna tip the outcome. It is the value of that which is important. Therefore, you would also need to include the probability that the outcome is gonna be tipped when the totals change by a single vote. A related question is here:

Do not vote, one vote will not reverse election results. What is wrong with this reasoning?

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Possibly you could use something as the following.

The effect of a vote on party $R_i$ is either

• $0$ when $r_i \leq n-2$
• $1$ when $r_i = n-1$ and $0 \leq \sum b_i - \sum r_i \leq n $
• $1$ when $r_i \geq n \hphantom{-1}$ and $0 \leq \sum b_i - \sum r_i \leq 1 $

The case $\sum b_i - \sum r_i = 1$ is difficult because in that case there is a difference in the value of rasing the result by $1$ (makes a draw) or by $n$ (makes a win).

It is also trick how you will compute the conditions with $\sum b_i - \sum r_i$ with your means and standard deviations you could use a normal distribution approximation but as the linked question shows the results will be different for different type of distributions and in your case you also have to deal with possibilities that the amounts are below $n$.