The Pivotal Voter in the Bangkok Mayoral Election

6th in line at my district's ballot-voting tent, I waited patiently to vote for my preferred candidate. However, I accidentally voted for a different candidate.

Today, we have an election for Bangkok’s mayor. I can’t stop thinking about how much my vote will matter for the candidate I favor. This led me to calculate the probability of being the pivotal voter.

Given that there are 4,428,646 eligible voters for the Bangkok mayoralty, with 18 candidates who have varying levels of popular support from the NIDA poll. The poll uses multistage sampling, drawing more than 2,000 samples across 50 districts.

• 1st place: Chadchart Sittipunt → 67.3%
• 2nd place: Chaiwat Sathawornwichit → 8.2%
• 3rd place: Mallika Boonmeetrakool Mahasook → 7.3%
• 4th place: Anucha Burapachaisri → 3.10%
• 5th place: Undecided → 10.20%
• 6th place: M.L. Kornkasiwat Kasemsri → 1.55%

If the poll is already strongly in favor of the 1st candidate, one additional vote would not really matter to the election result.

I know you probably have the thought that if this is the case, and everyone has the same thought, wouldn’t it lead to a catastrophe with low turnout? And yes, that is a collective-rationality problem that would require another theorem and different assumptions.

Back to our current setting: the individual-rationality approach to calculating pivotal voting probability.

The setting where 1 vote matters can range from the minimum case, where all 18 candidates are tied, to the maximum case, where there are only 2 candidates: Chadchart and Chaiwat, and both are tied.

Minimum Case: All 18 Candidates Are Tied

We have:

$$4{,}428{,}646 \bmod 18 = 16$$

So, assume 16 people are missing. Then the number of voters left is:

$$4{,}428{,}646 - 16 = 4{,}428{,}630$$

Each candidate would receive:

$$\frac{4{,}428{,}630}{18} = 246{,}035$$

votes.

Using the multinomial distribution:

$$P(\text{18-way tie}) = \frac{T!}{(246{,}035!)^{18}} \left(\frac{1}{18}\right)^T$$

where:

$$T = 4{,}428{,}630$$

So:

$$P(\text{18-way tie}) = \frac{4{,}428{,}630!}{(246{,}035!)^{18}} \left(\frac{1}{18}\right)^{4{,}428{,}630}$$

This is approximately:

$$P(\text{18-way tie}) \approx \frac{1}{9.56 \times 10^{51}} \approx \frac{1}{10^{52}}$$

Maximum Case: Exactly 2 Candidates Are Tied

Now suppose only 2 candidates matter: Chadchart and Chaiwat.

Each voter independently chooses either Chadchart or Chaiwat with probability:

$$\frac{1}{2}$$

When:

$$N = 4{,}428{,}646$$

A tie occurs when each candidate receives exactly half of the votes:

$$\frac{4{,}428{,}646}{2} = 2{,}214{,}323$$

Using the binomial formula:

$$P(\text{tie}) = \binom{4{,}428{,}646}{2{,}214{,}323} \left(\frac{1}{2}\right)^{4{,}428{,}646}$$

This is approximately:

$$P(\text{tie}) \approx 0.000379$$

or:

$$0.0379\%$$

or about:

$$\frac{1}{2{,}638}$$

Even in the ideal case, it is still pretty low, isn’t it?

I think this simple model points out the real motive for why people turn out to vote: not just to change the result, but for something deeper.