Bukit Timah Tutor Mathematics

A connected Mathematics learning system from school foundations to examinations, applications and advanced study. Use the Mathematics Hub to move between levels, concepts, diagnosis, examinations, applications and world routes.

Real-World Mathematics: Elections, Voting, Apportionment and Ranking

Three students in school uniforms work through open books at a classroom table, with textbooks and stationery nearby and study notes on the whiteboard behind them.

Application of Mathematics in Real-World Usage · Guide 27 · BTT Mathematics Hub

A candidate can win the most votes without winning a majority. A party can gain votes but lose vote share if turnout rises faster. A ranked-choice winner can depend on how lower preferences transfer. An apportionment method can allocate a different final seat because of rounding. Election Mathematics is therefore about counting rules as much as counting ballots.

This is a politically neutral Mathematics lesson. Every candidate, party, district and ballot below is fictional and labelled generically. It does not advocate any electoral system, party or political position. Real election rules differ by jurisdiction; official law and election authorities determine the actual procedure.

Vote share is votes divided by the declared vote total

Candidate A receives 4,200 votes, B receives 3,800 and C receives 2,000. Total valid votes are 10,000. Shares are 42%, 38% and 20%.

The shares sum to 100% because they use the same valid-vote denominator. If rejected or blank ballots are included in another statistic, its denominator must be stated separately.

Plurality is not majority

A has the largest share at 42%, so A has a plurality. A does not have a majority because 42% is below 50%.

“Most votes” and “more than half of votes” are different mathematical conditions.

Turnout depends on the chosen denominator

If 10,000 ballots are cast from 12,500 registered voters, turnout on the registered-voter base is 80%.

A turnout calculation based on eligible population, registered voters or voting-age population can produce different percentages. The denominator is part of the statistic.

Vote count can rise while vote share falls

A receives 4,000 of 8,000 votes in one fictional election, a 50% share. Later A receives 4,500 of 10,000, a 45% share.

A gained 500 votes but lost five percentage points of share because the total grew faster.

Margin can be expressed in votes or percentage points

If A has 4,200 and B 3,800, the vote margin is 400. Their share margin is 42%−38%=4 percentage points.

A four-percentage-point margin is not the same as a 4% relative increase. Relative to B’s vote count, 400/3,800≈10.53%.

Threshold wording is an inequality

Suppose a fictional rule elects a candidate directly only if share exceeds 50%. Exactly 50% does not satisfy “exceeds 50%.” If the rule says “at least 50%,” the equality case changes.

Mathematics cannot replace the legal wording. It can only apply the rule once the rule is known.

Ranked ballots contain more information than first choices

Suppose 40 voters rank A>B>C, 35 rank B>C>A and 25 rank C>B>A.

First-choice totals are A 40, B 35 and C 25. No candidate has a majority.

An elimination-and-transfer rule can change the winner

Under a simple fictional instant-runoff rule, C is eliminated and its 25 ballots transfer to their next preference, B. Final totals become A 40 and B 60.

A had the initial plurality, while B wins under this ranked-transfer rule. The ballots stayed the same; the aggregation rule changed which information mattered at each stage.

Pairwise comparisons ask a different question

Compare B with A. The 40 A>B>C voters prefer A, while the other 60 prefer B. B wins 60–40.

Compare B with C. The first 40 plus the 35 B-first voters prefer B, while 25 prefer C. B wins 75–25.

In this constructed profile B defeats each alternative head-to-head.

Borda-style scores depend on the scoring vector

Assign 2 points for first, 1 for second and 0 for third. A scores 40×2=80. B scores 40×1+35×2+25×1=135. C scores 35×1+25×2=85.

B wins under this scoring rule. A different point vector can change the ranking because lower preferences receive different weights.

Aggregation can create cycles

Consider three equal groups: one ranks A>B>C, one B>C>A, and one C>A>B.

By pairwise majority, A beats B, B beats C, and C beats A. Collective preference is cyclic even though every individual ranking is internally ordered.

This demonstrates an important social-choice idea: aggregating individually consistent preferences need not produce a single consistent group ranking.

Seat apportionment begins with exact quotas

Suppose three regions with populations 450,000, 300,000 and 250,000 share 10 seats. Total population is 1,000,000, so the standard divisor is 100,000 people per seat.

Exact quotas are 4.5, 3.0 and 2.5 seats. They sum to 10, but two quotas are fractional.

Largest remainder separates lower quotas from remainders

Lower quotas are 4, 3 and 2, allocating nine seats. Remainders are 0.5, 0 and 0.5. One seat remains.

The two largest remainders tie, so a real system needs a specified tie-break. Mathematics exposes the tie; it does not invent the legal resolution.

Divisor methods use a common divisor and rounding rule

Divisor methods adjust a common divisor until rounded quotas sum to the required seat total. Different rounding conventions can allocate the final seats differently from the same population data.

The method therefore belongs to the definition of the result, just as a statistical estimator belongs to the definition of an estimate.

Seat share can differ from vote share

A fictional party receives 40% of votes and 45 of 100 seats. Seat share is 45%, five percentage points above vote share.

The ratio seat share/vote share is 45/40=1.125. This descriptive ratio does not by itself explain why the difference occurred.

District magnitude changes the resolution of representation

A single-member district allocates one seat. A five-member district allocates five. The mapping from votes to seats therefore has a different granularity.

Formula, thresholds, district magnitude and geographic distribution can all matter in converting votes into seats.

Polling is sampling; a complete ballot count is not

A poll samples a subset and estimates a wider population preference, so sampling uncertainty applies. A complete count of all valid ballots is not a sample of those ballots.

Other errors can still exist in real systems, but they are conceptually different from survey sampling error.

A simple poll standard error depends on sample size

For a simple random-sample teaching model with p=0.50 and n=1,000, SE=√[0.5×0.5/1000]≈0.01581, or 1.581 percentage points.

A rough 95% normal interval is ±1.96SE≈±3.10 percentage points. Real polling may need weighting, design effects, nonresponse models and other adjustments.

A complete election calculation states the rule first

State which ballots are in the denominator, the exact majority threshold, the ranked-transfer rule, the apportionment formula, the tie-break and any qualifying threshold before calculating an outcome.

Election Mathematics is not merely arithmetic after votes arrive. The rule is itself a mathematical transformation from preferences to an outcome.

Practice: twenty election Mathematics questions

  1. Votes are 4,200, 3,800 and 2,000. Find total.
  2. Find each vote share.
  3. Does 42% constitute a majority?
  4. 10,000 ballots from 12,500 registered voters: find turnout.
  5. A rises from 4,000/8,000 to 4,500/10,000. Find old and new shares.
  6. Find vote margin between 4,200 and 3,800.
  7. Find percentage-point margin.
  8. Rank profile: 40 A>B>C, 35 B>C>A, 25 C>B>A. Find first-choice totals.
  9. Under elimination of C with transfers, find final A and B totals.
  10. Who wins pairwise B versus A?
  11. Who wins pairwise B versus C?
  12. Using Borda 2-1-0, find A’s score.
  13. Find B’s score.
  14. Find C’s score.
  15. Populations 450k, 300k and 250k share 10 seats. Find standard divisor.
  16. Find exact quotas.
  17. Find lower-quota seat total.
  18. A party has 40% votes and 45% seats. Find seat-share difference in percentage points.
  19. For p=.5 and n=1000, find the simple-random-sample standard error.
  20. Why can an electoral method change the winner without changing any ballot?

Worked answers

  1. 10,000.
  2. 42%, 38%, 20%.
  3. No.
  4. 80%.
  5. 50% then 45%.
  6. 400 votes.
  7. 4 percentage points.
  8. A 40, B 35, C 25.
  9. A 40, B 60.
  10. B, 60–40.
  11. B, 75–25.
  12. 80.
  13. 135.
  14. 85.
  15. 100,000 people per seat.
  16. 4.5, 3.0, 2.5.
  17. 9 seats.
  18. +5 percentage points.
  19. About 1.581 percentage points.
  20. Because the counting rule chooses how the same preference information is aggregated.

Sources and connected applications

For neutral background on voting and apportionment Mathematics, see LibreTexts: Voting Theory and LibreTexts: Apportionment. All candidates, parties and elections on this page are fictional.

Continue with Insurance, Risk, Premiums, Expected Loss and Pooling; Healthcare Screening, Diagnostic Tests, Risk and Rates; and Queues, Waiting Times, Capacity and Service Systems. Return to the BTT Mathematics Hub.