The Central Idea
Voting and apportionment begin with different kinds of data, but they share a common challenge:
\[ \boxed{\text{convert many inputs into one discrete collective outcome}} \]In voting, the inputs are preferences, approvals, rankings, or scores.
In apportionment, the inputs are population counts, membership totals, vote totals, or other measures of group size.
In both settings, the mathematical rule matters because different valid rules can sometimes produce different outcomes from the same underlying data.
Why These Problems Are Mathematically Interesting
A simple-looking rule such as choose the alternative with the most support becomes less simple when there are three or more alternatives.
Likewise, a proportional allocation becomes complicated when the exact shares are fractional but the available seats must be whole numbers.
These problems involve:
- counting;
- rankings and relations;
- ratios and proportions;
- rounding;
- optimization;
- fairness criteria;
- paradoxes caused by discrete rules.
Voting Is an Aggregation Problem
Suppose a group must choose among alternatives
\[ A,\qquad B,\qquad C. \]Each voter may have an individual preference order such as
\[ A\succ B\succ C, \]meaning the voter prefers A to B and B to C.
A voting method is a rule that combines many individual preference inputs into a group-level result.
Preference Ballots
A preference ballot ranks alternatives from most preferred to least preferred.
For three alternatives, one ballot might be
\[ B\succ C\succ A. \]This ballot says only how the voter orders the alternatives. It does not necessarily measure how strongly the voter prefers one alternative over another.
Preference Schedules
When many voters submit the same ranking, their ballots can be grouped in a preference schedule.
| Number of Voters | 18 | 14 | 10 | 8 |
|---|---|---|---|---|
| 1st | A | B | C | B |
| 2nd | B | C | A | A |
| 3rd | C | A | B | C |
The total number of voters is
\[ 18+14+10+8=\boxed{50}. \]Plurality Voting
Under the plurality method, only first-place votes are counted.
From the preference schedule:
- A receives \(18\) first-place votes;
- B receives \(14+8=22\) first-place votes;
- C receives \(10\) first-place votes.
Therefore, the plurality totals are
\[ A:18,\qquad B:22,\qquad C:10. \]B has the largest plurality total.
Plurality Is Not the Same as Majority
A majority means more than half of all votes.
With \(50\) voters, a majority requires more than
\[ \frac{50}{2}=25. \]Therefore, the smallest majority is
\[ \boxed{26}. \]B has \(22\) first-place votes, so B has a plurality but not a majority.
The Majority Criterion
The majority criterion says that if an alternative is ranked first by more than half of the voters, that alternative should win.
This criterion does not say what should happen when no alternative has a majority.
Different voting methods resolve that situation differently.
Runoff Voting
A runoff method narrows the field and conducts another comparison.
In a simple top-two runoff:
- count first-place votes;
- keep the two alternatives with the largest totals;
- compare those two using the voters' preferences.
From the earlier plurality totals,
\[ B=22,\qquad A=18,\qquad C=10. \]The runoff would therefore be between A and B.
Evaluating the A-versus-B Runoff
Return to the preference schedule:
| Number of Voters | 18 | 14 | 10 | 8 |
|---|---|---|---|---|
| 1st | A | B | C | B |
| 2nd | B | C | A | A |
| 3rd | C | A | B | C |
Voters preferring A over B:
\[ 18+10=\boxed{28}. \]Voters preferring B over A:
\[ 14+8=\boxed{22}. \]Thus A defeats B in the head-to-head comparison.
Instant-Runoff Voting
Instant-runoff voting uses ranked ballots and eliminates alternatives in rounds.
A common procedure is:
- count first-place votes;
- if one alternative has a majority, stop;
- otherwise eliminate the alternative with the fewest first-place votes;
- transfer those ballots to the highest-ranked remaining alternative;
- repeat until a winner is determined.
Instant-Runoff Example
The first-place totals are
\[ A=18,\qquad B=22,\qquad C=10. \]No alternative has a majority.
C has the fewest first-place votes and is eliminated.
The \(10\) voters who ranked C first have
\[ C\succ A\succ B. \]Their ballots transfer to A.
The new totals are
\[ A=18+10=28, \] \[ B=22. \]A now has a majority.
The Borda Count
The Borda count assigns points according to position in each ranking.
With three alternatives, one common scoring convention is:
- 2 points for first place;
- 1 point for second place;
- 0 points for third place.
The alternative with the greatest total score is selected.
Borda Count Example
Using the same preference schedule, calculate A's score.
A is:
- first for \(18\) voters;
- third for \(14\) voters;
- second for \(10\) voters;
- second for \(8\) voters.
Therefore,
\[ S_A=2(18)+0(14)+1(10)+1(8)=54. \]For B,
\[ S_B=1(18)+2(14)+0(10)+2(8)=62. \]For C,
\[ S_C=0(18)+1(14)+2(10)+0(8)=34. \]Under this Borda scoring rule, B has the largest total:
\[ \boxed{62}. \]Borda Scores with \(m\) Alternatives
If there are \(m\) alternatives, a common Borda convention assigns
\[ m-1,m-2,\ldots,1,0 \]points from first place through last place.
Different equivalent score offsets may be used, such as
\[ m,m-1,\ldots,2,1. \]Adding the same constant to every position on every ballot changes total scores but does not change their ordering.
Pairwise Comparison
A pairwise comparison asks which alternative is preferred by more voters when two alternatives are considered at a time.
With alternatives A, B, and C, the pairwise contests are:
\[ A\text{ vs. }B, \] \[ A\text{ vs. }C, \]and
\[ B\text{ vs. }C. \]How Many Pairwise Comparisons?
With \(n\) alternatives, every unordered pair can be compared.
The number of pairwise comparisons is
\[ \boxed{\binom{n}{2}=\frac{n(n-1)}{2}}. \]For \(5\) alternatives,
\[ \binom{5}{2} =\frac{5\cdot4}{2} =\boxed{10}. \]The Condorcet Criterion
A Condorcet winner is an alternative that defeats every other alternative in head-to-head majority comparisons.
The Condorcet criterion says that if such an alternative exists, it should be selected.
A Condorcet winner does not always exist.
Pairwise Cycles
Group preferences can be cyclic even when every individual voter has a consistent ranking.
It is possible for the group to prefer
\[ A\succ B, \] \[ B\succ C, \]and yet
\[ C\succ A. \]This creates a cycle:
\[ \boxed{A\succ B\succ C\succ A}. \]The cycle shows that majority preference need not form a transitive ordering.
A Simple Condorcet Cycle
Suppose there are three equally sized groups of voters:
| Group 1 | Group 2 | Group 3 |
|---|---|---|
| \(A\succ B\succ C\) | \(B\succ C\succ A\) | \(C\succ A\succ B\) |
Two groups prefer A to B.
Two groups prefer B to C.
Two groups prefer C to A.
Therefore, no alternative defeats both others.
Approval Voting
Under approval voting, each voter may approve any number of alternatives.
Each approved alternative receives one approval from that voter.
The alternative with the largest number of approvals is selected.
Approval voting records acceptable alternatives rather than requiring a complete ranking.
Voting Methods Use Different Information
Different methods use different parts of a ballot.
| Method | Information Used |
|---|---|
| Plurality | First-place choice |
| Instant runoff | Ranked preferences used across rounds |
| Borda count | Every rank position |
| Pairwise comparison | Relative preference within each pair |
| Approval voting | Approved versus not approved |
Because the information is aggregated differently, the methods need not produce the same result.
Fairness Criteria
Voting systems can be studied using formal criteria.
Common criteria include:
- majority criterion;
- Condorcet criterion;
- monotonicity criterion;
- independence of irrelevant alternatives;
- consistency-related properties.
A criterion is a mathematical test for a particular kind of behavior. It is not itself a complete voting method.
Monotonicity
Informally, a voting method is monotonic when moving a winning alternative upward on some ballots, while making no other changes that hurt it, does not cause that alternative to lose.
This criterion captures a natural expectation:
\[ \boxed{\text{more favorable ranking should not by itself make a winner lose}} \]Some voting procedures satisfy this criterion and some can fail it.
Independence of Irrelevant Alternatives
One version of the independence of irrelevant alternatives idea asks whether the social comparison between A and B should depend only on how voters rank A relative to B.
The criterion becomes important when adding or removing another alternative can change the result between alternatives that remain.
No Single Rule Captures Every Desirable Property
Voting theory shows that desirable fairness properties can conflict.
As a result, choosing a voting method is partly about deciding which mathematical properties the system is intended to emphasize.
The role of mathematics is to make those properties and tradeoffs explicit.
Apportionment Problems
An apportionment problem distributes a fixed number of indivisible units among groups according to their relative sizes.
Examples include allocating:
- representative seats among regions;
- committee positions among departments;
- delegates among organizations;
- machines among production units;
- resources among service areas.
The Standard Divisor
Suppose a total population \(P\) must receive \(H\) seats.
The standard divisor is
\[ \boxed{D=\frac{P}{H}}. \]It represents the average population per seat.
Standard Quotas
If group \(i\) has population \(p_i\), its standard quota is
\[ \boxed{q_i=\frac{p_i}{D}}. \]Equivalently,
\[ q_i=\frac{p_iH}{P}. \]The standard quota usually is not an integer.
A First Apportionment Example
Suppose three regions have populations:
| Region | Population |
|---|---|
| A | 48,000 |
| B | 32,000 |
| C | 20,000 |
Total population:
\[ P=48000+32000+20000=\boxed{100000}. \]Suppose \(10\) seats must be allocated.
The standard divisor is
\[ D=\frac{100000}{10}=\boxed{10000}. \]Quotas in the First Example
Region A:
\[ q_A=\frac{48000}{10000}=4.8. \]Region B:
\[ q_B=\frac{32000}{10000}=3.2. \]Region C:
\[ q_C=\frac{20000}{10000}=2.0. \]The quotas sum to
\[ 4.8+3.2+2.0=\boxed{10}. \]Lower and Upper Quotas
The lower quota is the standard quota rounded down:
\[ \boxed{\lfloor q_i\rfloor}. \]The upper quota is the standard quota rounded up:
\[ \boxed{\lceil q_i\rceil}. \]If
\[ q_i=4.8, \]then
\[ \lfloor q_i\rfloor=4 \]and
\[ \lceil q_i\rceil=5. \]The Quota Rule
An apportionment satisfies the quota rule when each group receives either its lower quota or its upper quota.
Symbolically, if \(a_i\) is the assigned number of seats,
\[ \boxed{ a_i\in\{\lfloor q_i\rfloor,\lceil q_i\rceil\} }. \]Not every apportionment method always satisfies this rule.
Hamilton's Method
Hamilton's method, also called the largest-remainder method, begins with the standard quotas.
The procedure is:
- calculate the standard divisor;
- calculate each standard quota;
- assign each group its lower quota;
- count the remaining seats;
- give those seats to the groups with the largest fractional remainders.
Hamilton Example
Using the quotas
\[ 4.8,\qquad3.2,\qquad2.0, \]the lower quotas are
\[ 4,\qquad3,\qquad2. \]These allocate
\[ 4+3+2=9 \]of the \(10\) seats.
One seat remains.
The fractional remainders are
\[ 0.8,\qquad0.2,\qquad0. \]The largest remainder belongs to A, so the final allocation is
\[ \boxed{A=5,\qquad B=3,\qquad C=2}. \]Why Ordinary Rounding Can Fail
Suppose quotas are
\[ 2.6,\qquad2.6,\qquad1.8. \]Their sum is
\[ 7. \]Ordinary rounding gives
\[ 3,\qquad3,\qquad2, \]whose sum is
\[ 8. \]The number of assigned seats no longer matches the required house size.
Apportionment methods exist because independent rounding does not reliably preserve the total.
Divisor Methods
A divisor method adjusts the divisor until a specified rounding rule produces exactly the required number of seats.
The general structure is:
\[ \text{modified quota} = \frac{\text{group population}}{\text{modified divisor}}. \]Different divisor methods use different rounding rules.
Jefferson's Method
Jefferson's method uses a modified divisor and rounds each modified quota down.
The divisor is adjusted until the rounded-down allocations sum to the required number of seats.
Symbolically,
\[ a_i=\left\lfloor\frac{p_i}{d}\right\rfloor, \]where \(d\) is selected so that
\[ \sum_i a_i=H. \]Adams's Method
Adams's method uses a modified divisor and rounds each modified quota up.
Thus,
\[ a_i=\left\lceil\frac{p_i}{d}\right\rceil. \]The divisor \(d\) is adjusted until the total allocation equals the required house size.
Webster's Method
Webster's method uses a modified divisor and rounds each modified quota to the nearest integer.
Conceptually,
\[ a_i=\operatorname{round}\left(\frac{p_i}{d}\right), \]with the divisor chosen so that
\[ \sum_i a_i=H. \]Halfway cases require a specified tie convention.
Huntington-Hill Rounding
The Huntington-Hill method uses a geometric-mean threshold between consecutive integers.
To decide whether a quota between \(n\) and \(n+1\) rounds down or up, compare it with
\[ \boxed{\sqrt{n(n+1)}}. \]If the quota is below the threshold, round to \(n\).
If it is above the threshold, round to \(n+1\).
Arithmetic and Geometric Rounding Thresholds
Ordinary nearest-integer rounding uses the arithmetic midpoint
\[ n+\frac12. \]Huntington-Hill rounding uses the geometric midpoint
\[ \sqrt{n(n+1)}. \]These thresholds are close but not identical.
That difference can change an allocation near a rounding boundary.
Divisor Direction
If the initial rounded allocations produce too few seats, the divisor generally must be decreased.
Why?
Decreasing the divisor makes the modified quotas larger:
\[ \frac{p_i}{d_{\text{smaller}}} > \frac{p_i}{d_{\text{larger}}}. \]Larger quotas tend to produce more seats after rounding.
Conversely, increasing the divisor tends to reduce the seat total.
Apportionment as Constrained Rounding
At its core, apportionment is a constrained rounding problem.
The ideal shares are fractional:
\[ q_1,q_2,\ldots,q_n. \]The final assignments must be integers:
\[ a_1,a_2,\ldots,a_n\in\mathbb Z. \]And the total must remain fixed:
\[ \boxed{a_1+a_2+\cdots+a_n=H}. \]The Alabama Paradox
The Alabama paradox occurs when increasing the total number of seats causes a group to lose a seat.
This can occur under Hamilton's method.
The paradox is surprising because one might expect that increasing the number of available seats could not make any group worse off.
It illustrates how remainder rankings can change when the house size changes.
The Population Paradox
A population paradox occurs when one group grows faster than another but nevertheless loses a seat to the slower-growing group after reapportionment.
The issue arises from the interaction between changed populations and the discrete allocation rule.
The New-States Paradox
A new-states paradox occurs when adding a new group and an appropriate number of additional seats changes the allocation among the original groups.
The paradox shows that an allocation rule can create interactions among groups even when their own populations have not changed.
Apportionment Methods Have Different Properties
Different methods emphasize different mathematical behaviors.
Questions include:
- Does the method always satisfy quota?
- Can increasing the house size hurt a group?
- Can relative population growth produce a counterintuitive transfer?
- Does the method systematically favor larger or smaller groups?
No method should be evaluated only by a single example.
Relative Representation
One way to inspect an allocation is to compare population per seat.
If group \(i\) has population \(p_i\) and receives \(a_i\) seats, its average population per seat is
\[ \boxed{\frac{p_i}{a_i}}. \]Large differences among these ratios may indicate unequal representation, although the discreteness of seats makes exact equality generally impossible.
Representation Ratios
Another perspective compares actual seats with ideal quota.
One simple ratio is
\[ \boxed{r_i=\frac{a_i}{q_i}}. \]If \(r_i=1\), the actual allocation exactly equals the ideal fractional quota.
Values above \(1\) indicate representation above quota, while values below \(1\) indicate representation below quota.
Voting and Apportionment Are Not the Same Problem
Voting chooses or ranks alternatives.
Apportionment distributes a fixed number of indivisible units.
However, both are collective-decision problems because they use mathematical procedures to transform many individual or group-level inputs into one public result.
Common Error: Confusing Plurality and Majority
The alternative with the most first-place votes does not necessarily have more than half of the votes.
With
\[ A=42,\qquad B=35,\qquad C=23, \]A has a plurality of \(42\%\), not a majority.
Common Error: Counting All Rankings as First-Place Votes
In plurality voting, only first-place choices matter.
Second- and third-place rankings are not added to the plurality total.
Those lower rankings may matter under other systems such as Borda count or instant runoff.
Common Error: Forgetting Ballot Multiplicity
In a preference schedule, one column may represent many voters.
If \(18\) voters share a ranking, every score or pairwise comparison from that column must be multiplied by \(18\).
Treating the column as one ballot produces incorrect totals.
Common Error: Pairwise Comparison by First Place Only
In a head-to-head comparison between A and B, every voter contributes according to whether A is ranked above B or B is ranked above A.
The voter's first-place alternative could be C and the ballot still matters in the A-versus-B comparison.
Common Error: Ordinary Rounding in Apportionment
Rounding every quota independently can produce the wrong total number of seats.
The allocation must satisfy
\[ \boxed{\sum_i a_i=H}. \]An apportionment method coordinates the rounding decisions so that this total is preserved.
Common Error: Using the Wrong Divisor
The standard divisor is based on total population and total seats:
\[ D=\frac{P}{H}. \]Dividing seats by population reverses the ratio and gives the wrong units.
The divisor has units of population per seat.
Common Error: Forgetting to Check the Seat Total
After any apportionment calculation, add the assigned seats.
The result must equal the required house size:
\[ a_1+a_2+\cdots+a_n=H. \]This is one of the fastest checks for an arithmetic or rounding error.
Common Error: Confusing Quota with Allocation
A quota may be fractional.
An allocation is the final whole-number seat count.
For example,
\[ q_A=6.37 \]is an ideal share, not an instruction to assign \(6.37\) seats.
A Reliable Procedure for Preference Voting Problems
- Identify the alternatives.
- Record the preference schedule accurately.
- Confirm the total number of voters.
- Identify the requested voting method.
- Use only the ballot information required by that method.
- Multiply each ranking pattern by its number of voters.
- Calculate totals carefully.
- Apply any elimination, transfer, or pairwise steps in the correct order.
- Check ties and tie-breaking assumptions.
- State the result together with the method used.
A Reliable Procedure for Hamilton's Method
- Add the group populations to obtain \(P\).
- Divide \(P\) by the house size \(H\) to obtain the standard divisor.
- Divide each group population by the divisor to obtain its quota.
- Assign every group its lower quota.
- Add the lower quotas.
- Determine how many seats remain.
- Rank the fractional remainders from largest to smallest.
- Assign the remaining seats in that order.
- Check that the final seat total equals \(H\).
- Check the quota rule.
A Reliable Procedure for Divisor Methods
- Calculate the standard divisor as a starting point.
- Calculate the resulting quotas.
- Apply the method's required rounding rule.
- Add the rounded seat counts.
- If there are too few seats, decrease the divisor.
- If there are too many seats, increase the divisor.
- Repeat until the seat total equals the required house size.
- Record the final modified divisor and allocation.
- Check the total and any relevant fairness properties.
Example: Full Plurality and Borda Workflow
Consider:
| Number of Voters | 12 | 9 | 7 | 6 |
|---|---|---|---|---|
| 1st | A | B | C | B |
| 2nd | C | A | B | C |
| 3rd | B | C | A | A |
Plurality totals:
\[ A=12, \] \[ B=9+6=15, \] \[ C=7. \]Under plurality, B has the largest first-place total.
Using Borda scores \(2,1,0\):
\[ S_A=2(12)+1(9)+0(7)+0(6)=33, \] \[ S_B=0(12)+2(9)+1(7)+2(6)=37, \] \[ S_C=1(12)+0(9)+2(7)+1(6)=32. \]Under this Borda rule, B also has the largest score.
Example: Full Hamilton Workflow
Suppose four regions have populations:
\[ 42000,\qquad31000,\qquad17000,\qquad10000. \]Total population:
\[ P=100000. \]Suppose \(H=12\) seats are available.
The standard divisor is
\[ D=\frac{100000}{12}\approx8333.333. \]The standard quotas are:
\[ q_A=\frac{42000}{8333.333}\approx5.04, \] \[ q_B\approx3.72, \] \[ q_C\approx2.04, \] \[ q_D\approx1.20. \]Lower quotas:
\[ 5,\qquad3,\qquad2,\qquad1. \]These sum to
\[ 11. \]One seat remains.
The largest fractional remainder is \(0.72\), belonging to B.
Therefore the Hamilton allocation is
\[ \boxed{A=5,\qquad B=4,\qquad C=2,\qquad D=1}. \]Practice 1: Majority Threshold
An election has \(83\) voters.
What is the smallest number of votes needed for a majority?
Practice 2: Plurality
First-place totals are
\[ A=31,\qquad B=28,\qquad C=19. \]Which alternative has the plurality?
Practice 3: Pairwise Comparison Count
How many pairwise comparisons are required for \(6\) alternatives?
Practice 4: Borda Points
With four alternatives, use Borda scores
\[ 3,\ 2,\ 1,\ 0. \]If \(15\) voters rank
\[ A\succ C\succ D\succ B, \]how many Borda points does A receive from this group?
Practice 5: Standard Divisor
A total population of \(240,000\) must be apportioned among \(16\) seats.
Find the standard divisor.
Practice 6: Standard Quota
Using the divisor from Practice 5, find the standard quota for a region with population \(52,500\).
Practice 7: Lower and Upper Quotas
A group has standard quota
\[ 7.36. \]Find its lower and upper quotas.
Practice 8: Hamilton Allocation
Suppose three groups have quotas
\[ 3.45,\qquad2.35,\qquad2.20 \]and \(8\) seats must be assigned.
Use Hamilton's method.
Practice 9: Seat-Total Check
An apportionment of \(20\) seats produces
\[ 7,\qquad6,\qquad4,\qquad2. \]Is the allocation complete?
Practice 10: Divisor Direction
A divisor method currently produces only \(14\) seats, but the required house size is \(15\).
Should the divisor generally be increased or decreased?
Practice Answers
Practice 1
Half of \(83\) is
\[ 41.5. \]A majority requires more than half, so the minimum is
\[ \boxed{42}. \]Practice 2
The largest first-place total is
\[ \boxed{31}, \]so A has the plurality.
Practice 3
\[ \binom{6}{2} =\frac{6\cdot5}{2} =\boxed{15}. \]Practice 4
A is ranked first and earns \(3\) points per ballot.
\[ 15\cdot3=\boxed{45}. \]Practice 5
\[ D=\frac{240000}{16}=\boxed{15000}. \]Practice 6
\[ q=\frac{52500}{15000}=\boxed{3.5}. \]Practice 7
\[ \boxed{\lfloor7.36\rfloor=7} \]and
\[ \boxed{\lceil7.36\rceil=8}. \]Practice 8
Lower quotas:
\[ 3,\qquad2,\qquad2. \]These use \(7\) seats, so one remains.
The largest remainder is
\[ 0.45. \]Therefore the allocation is
\[ \boxed{4,\qquad2,\qquad2}. \]Practice 9
The assigned seats total
\[ 7+6+4+2=19. \]Since the required total is \(20\), the allocation is not complete.
Practice 10
The divisor should generally be decreased.
A smaller divisor creates larger modified quotas, which tends to increase the rounded seat total.
What to Remember
Voting methods aggregate preferences. Apportionment methods convert fractional proportional shares into whole-number allocations.
- plurality counts first-place votes;
- a majority means more than half;
- runoff methods narrow or eliminate alternatives in stages;
- Borda count assigns points by rank;
- pairwise methods compare alternatives two at a time;
- a Condorcet winner defeats every other alternative head to head;
- group preferences can form cycles;
- the standard divisor is \(P/H\);
- standard quotas are population divided by the divisor;
- Hamilton assigns lower quotas first and then largest remainders;
- divisor methods adjust the divisor and apply a specified rounding rule;
- apportionment must preserve the required total number of seats.
The central habit is: identify the aggregation rule before calculating the collective result, because different rules process the same underlying data in different ways.
Where We Go Next
Voting and apportionment show how discrete mathematics organizes collective choices and proportional allocations. The next applied problem family moves from collective decisions to connected systems, where vertices and links represent transportation, information, infrastructure, or organizational relationships.
The next lesson, Networks and Communication, examines how graph structures model connectivity, transmission, paths, bottlenecks, reachability, and the movement of information through networks.