Proportional representation needs a rounding rule because legislatures contain whole seats.
By the end: Trace highest-average allocation and distinguish D’Hondt from unmodified Sainte-Laguë.
With five seats and vote shares 50%, 31%, 19%, exact proportional entitlements are 2.5, 1.55, and 0.95 seats. Independently rounding each number is not a general allocation algorithm because the rounded sum may differ from the available seats. A rule must allocate exactly the stated total.
List proportional representation allocates seats to lists. Closed-list rules let a party’s ordered list determine which candidates fill its seats. Open-list designs give voters some influence over that order, with details that vary. Party proportionality does not itself establish which individual candidates voters can choose.
Party proportionality compares a party's share of specified votes with its share of specified legislative seats. It is not a rule for selecting one officeholder. With votes fifty, thirty-one, and nineteen out of one hundred and five seats, multiply each vote fraction by five. The ideal entitlements are 2.5, 1.55, and 0.95 seats. They sum to five, but people cannot occupy fractional seats in this assembly. The allocation problem is to replace these entitlements with whole numbers that still sum to exactly five under a declared method.
Rounding each entitlement independently does not solve that problem in general. With three seats and votes forty, thirty-five, twenty-five, the ideals are 1.2, 1.05, and 0.75. Ordinary nearest-integer rounding gives one seat each and happens to work. With votes forty-nine, forty-nine, and two, the ideals are 1.47, 1.47, and 0.06. The same rounding produces one, one, zero, leaving a seat empty. The defect is not imprecise arithmetic. Separate rounding decisions have no step ensuring that their total equals the number of available seats.
| Votes A / B / C | 3-seat ideals | Nearest whole seats | Seats filled |
|---|---|---|---|
| 40 / 35 / 25 | 1.2 / 1.05 / 0.75 | 1 / 1 / 1 | 3 of 3 |
| 49 / 49 / 2 | 1.47 / 1.47 / 0.06 | 1 / 1 / 0 | 2 of 3 |
A highest-average method avoids that defect by making exactly one award at each step. Begin with zero seats for every eligible party. Compare the parties' current quotients, give one seat to the largest, update the winning party's divisor, and continue until all seats are awarded. This lesson's five-seat example has no eligibility threshold, geographic tiers, or exemptions. Those choices are held fixed so that the divisor sequence explains the difference between methods. Treating a party as ineligible would change the competition before the first award.
List allocation also leaves a separate question about which people occupy the seats. In a hypothetical closed list, suppose A orders its candidates Ada, Arun, and Asha. If A receives two seats, Ada and Arun fill them under that list-order rule. A different permitted order could replace Arun with Asha while leaving A's seat count unchanged. An open-list design instead gives personal votes some specified role in determining which listed candidates win. The phrase open list alone does not tell you the exact formula for that role.
Consider an explicit invented personal-vote rule. A has already received two party seats. Its listed candidates have personal totals Ada forty, Arun thirty-five, and Asha twenty-five. If the rule simply elects A's two highest personal totals, Ada and Arun win. Reverse Arun's and Asha's personal totals and Ada and Asha win, still with two seats for A. This isolates candidate selection from party allocation. It does not describe an additional feature of the allocation lab, which reports party seats only and has no personal-vote or list-order controls.
| Candidate-selection specification | Information used after A gets 2 seats | People elected |
|---|---|---|
| Closed list: Ada, Arun, Asha | First two names in fixed order | Ada and Arun |
| Closed list: Ada, Asha, Arun | First two names in changed order | Ada and Asha |
| Invented personal-vote rule | Ada 40, Arun 35, Asha 25; highest two | Ada and Arun |
| Same personal-vote rule, changed votes | Ada 40, Arun 25, Asha 35; highest two | Ada and Asha |
Neither proportional party totals nor personal-vote selection guarantees representation of particular social groups or policies. Candidate nomination determines who can compete, and later legislative behavior determines how representatives act. A seat formula cannot answer those questions without more information. The selected list-system discussion in IDEA's 2005 handbook explains the distinction between party allocation and candidate choice. Its country illustrations are historical. Use the invented list examples here to understand the mechanism, rather than treating that source or this model as a statement of any jurisdiction's present rules.
Highest-average methods compare each party’s votes divided by a divisor based on the seats it already holds. D’Hondt uses divisors 1,2,3,4 and so on. Unmodified Sainte-Laguë uses 1,3,5,7 and so on. Award the next seat to the highest quotient, update that party’s divisor, and repeat.
For votes 50,31,19 and five seats, D’Hondt gives 3,1,1 seats while Sainte-Laguë gives 2,2,1. The difference comes from the divisor sequence. Some jurisdictions modify the first divisor or add other stages; the lab implements only the two explicitly named unmodified sequences.
For D'Hondt, a party with no seats uses divisor one. After its first award it uses two, after its second it uses three, and so on. The quotient is a vote count divided by the next divisor, not a percentage of seats and not a number of votes being transferred. With votes fifty, thirty-one, and nineteen, the initial quotients equal those vote totals. A receives the first seat. Its next quotient becomes twenty-five, while B stays at thirty-one and C at nineteen. B therefore receives the second seat.
After B's award, the next D'Hondt quotients are A twenty-five, B 15.5, and C nineteen. A receives seat three. Its next quotient becomes fifty divided by three, about 16.67. B stays at 15.5 and C at nineteen, so C receives seat four. The counts are now two, one, one. C's next quotient falls to 9.5. For seat five, A's fifty divided by three exceeds B's thirty-one divided by two, so A wins. The completed allocation is three, one, one, exactly five seats.
| Seat | Next A quotient | Next B quotient | Next C quotient | Award; seats A/B/C |
|---|---|---|---|---|
| 1 | 50 / 1 | 31 / 1 | 19 / 1 | A; 1 / 0 / 0 |
| 2 | 50 / 2 | 31 / 1 | 19 / 1 | B; 1 / 1 / 0 |
| 3 | 50 / 2 | 31 / 2 | 19 / 1 | A; 2 / 1 / 0 |
| 4 | 50 / 3 | 31 / 2 | 19 / 1 | C; 2 / 1 / 1 |
| 5 | 50 / 3 | 31 / 2 | 19 / 2 | A; 3 / 1 / 1 |
Unmodified Sainte-Laguë uses a different response to each award. Its divisors are one, three, five, seven, and so on. After the first seat goes to A, A's next quotient falls to fifty divided by three, about 16.67. B's thirty-one wins the second seat. B's next quotient then falls to thirty-one divided by three, about 10.33. C's nineteen wins the third seat. A's 16.67 then wins the fourth. The counts again reach two, one, one, but the path and next divisors differ from D'Hondt.
At that shared two, one, one position, Sainte-Laguë compares fifty divided by five, thirty-one divided by three, and nineteen divided by three. Their values are ten, about 10.33, and about 6.33. B receives seat five and the final allocation is two, two, one. A's seat share is forty percent rather than sixty percent under D'Hondt. This difference is twenty percentage points of a five-seat assembly, or exactly one seat. The voters and seat total did not change. The divisor sequence changed which party received the final award.
Displayed decimals help readers inspect the sequence, but exact comparisons should decide close awards. To compare fifty divided by three with thirty-one divided by two, multiply across the positive denominators. Compare one hundred with ninety-three, so A's quotient is larger without rounding. For Sainte-Laguë's fifth seat, compare thirty-one times five with fifty times three. One hundred fifty-five exceeds one hundred fifty, so B beats A. The bounded lab uses integer cross-products for awards. It rounds only the displayed quotient, not the decision that the table records.
| Seat | Next A quotient | Next B quotient | Next C quotient | Award; seats A/B/C |
|---|---|---|---|---|
| 1 | 50 / 1 | 31 / 1 | 19 / 1 | A; 1 / 0 / 0 |
| 2 | 50 / 3 | 31 / 1 | 19 / 1 | B; 1 / 1 / 0 |
| 3 | 50 / 3 | 31 / 3 | 19 / 1 | C; 1 / 1 / 1 |
| 4 | 50 / 3 | 31 / 3 | 19 / 3 | A; 2 / 1 / 1 |
| 5 | 50 / 5 | 31 / 3 | 19 / 3 | B; 2 / 2 / 1 |
Another view lists each party's successive claims in advance. D'Hondt gives A claims fifty, twenty-five, fifty-thirds, and so on, while B has thirty-one, 15.5, and so on. Selecting the five highest claims gives the same awards because each party's later claim cannot exceed its earlier one. Sainte-Laguë changes those lists to odd divisors. The values do not consume or divide physical ballots among representatives. They are comparison numbers used to allocate whole seats. Every party's original vote count stays unchanged throughout the trace.
vᵢ is party i’s vote count; sᵢ is its seats already awarded. The next seat goes to the highest quotient. Only that party’s seat count then increases.
At zero, one, and two awarded seats, the divisors are one, three, and five. These quotients compare votes per divisor; they are not vote percentages.
Votes stay A 50, B 31, C 19. Changing the rule replaces the selected allocation; D’Hondt stays visible as the reference.
A / B / C: 2 / 2 / 1. Total: 5 seats.
| Seat | Party | Divisor | Winning quotient | Seats A/B/C |
|---|---|---|---|---|
| 1 | A | 1 | 50 | 1 / 0 / 0 |
| 2 | B | 1 | 31 | 1 / 1 / 0 |
| 3 | C | 1 | 19 | 1 / 1 / 1 |
| 4 | A | 3 | 16.67 | 2 / 1 / 1 |
| 5 | B | 3 | 10.33 | 2 / 2 / 1 |
| Party | Votes out of 100 | D’Hondt, 5 seats | Sainte-Laguë, 5 seats |
|---|---|---|---|
| A | 50 | 3 | 2 |
| B | 31 | 1 | 2 |
| C | 19 | 1 | 1 |
Select D'Hondt in the lab and check that the reference and selected bars both show three, one, one seats. Then select Sainte-Laguë and predict which party changes before reading the result. Open the complete award table and follow the selected sequence A, B, C, A, B. Notice that Sainte-Laguë gives C its first seat one step earlier than D'Hondt, yet both methods give C one seat at the end. Intermediate order explains the mechanism, while final counts describe the resulting legislature. Neither should be substituted for the other.
The lab keeps the vote vector and five-seat total fixed. Its buttons change only the divisor rule for the selected panel, and the D'Hondt reference remains visible. There is no vote slider, district selector, threshold switch, or animation to reset in this view. Explore other vote totals with a written quotient table. The bars measure seats on the same zero-to-five scale. They do not show proportional shares of votes, even though the unchanged vote counts explain how the seats were assigned.
Test scale invariance on paper by multiplying the votes by ten, to five hundred, three hundred ten, and one hundred ninety. Every quotient becomes ten times its previous value, so every strict comparison and exact tie stays the same. Both allocations are unchanged. Vote totals are inputs whose common scaling does not affect this procedure. Adding ten votes to every party is different. That operation changes their relative support and need not preserve the result. Confusing multiplication with equal additions would hide an important assumption in the invariance claim.
For an exact tie example, use votes sixty and twenty with two seats under unmodified Sainte-Laguë. A takes the first seat with sixty. The next comparison is A sixty divided by three against B twenty divided by one, exactly twenty each. In the teaching model, the earlier input party wins a tied quotient. With parties supplied alphabetically, A receives the second seat, giving two and zero. A different declared tie procedure could award B that seat. Rounded labels are not the reason for the equality in this constructed case.
Change B's votes to twenty-one while keeping A at sixty. After A's first award, its next quotient remains twenty and B's is twenty-one. B now wins the second seat, giving one each. The combined valid party vote total rose from eighty to eighty-one. B's share rose from twenty-five percent to about 25.93 percent, while its seat share jumped from zero to fifty percent. Whole-seat allocation produces discontinuous outcomes at boundaries. That local jump is a property of the example, not evidence of a sudden change in voter behavior or government quality.
| Valid party votes A / B | First award | Second comparison | Final seats A / B |
|---|---|---|---|
| 60 / 20; total 80 | A | 60 / 3 = 20 / 1 | 2 / 0, A wins exact tie |
| 60 / 21; total 81 | A | 60 / 3 < 21 / 1 | 1 / 1, B wins outright |
You can check each completed run with two separate tests. Add the party seat counts to confirm that all available seats were awarded exactly once. Then reproduce the award comparisons to confirm that the stated method chose those parties. The first test alone cannot distinguish three, one, one from two, two, one because both sum to five. The second test needs exact votes, divisors, eligibility, and tie priority. This is why a correct-looking seat chart is insufficient evidence that a counting rule was implemented as specified.
Small assemblies cannot represent every vote share exactly. D’Hondt generally favors larger lists relative to some other proportional formulas, but the effect depends on the votes and district magnitude. One example does not establish the size of a systematic effect across elections.
Every exact tie also needs a rule. The lab awards tied quotients to the alphabetically earlier party for reproducibility. A real rule might use a lot, another count, or a legal procedure. Changing a divisor or tie policy is changing the specification, not merely improving the same calculation.
Compare seat shares with the original valid party vote shares using named denominators. Under D'Hondt, A has three of five seats, or sixty percent, against fifty percent of the votes. B has twenty percent of seats against thirty-one percent of votes, and C twenty percent against nineteen percent. The gaps are plus ten, minus eleven, and plus one percentage points. Under Sainte-Laguë, the corresponding gaps are minus ten, plus nine, and plus one. These calculations describe this allocation. They do not turn either method into exact fractional proportionality.
A legislature also makes decisions after seats are assigned. Assume, only for this illustration, that all five representatives vote, each party votes together, and passing a proposal requires more than half of all five seats. Three votes suffice. D'Hondt's three, one, one allocation lets A pass a proposal alone under those assumptions. B and C together have only two votes. Sainte-Laguë's two, two, one allocation gives no party three alone. A with C, B with C, or A with B can each reach the stipulated majority threshold.
Those possible seat combinations are arithmetic coalitions, not predicted governments. A and C may disagree, individual legislators may dissent, and institutions may use different rules for particular decisions. Executive appointment, confidence requirements, abstentions, and constitutional voting thresholds can change what a seat combination permits. Even under the five-seat illustration, knowing that A and C have three seats does not tell you whether they would negotiate successfully. The counting rule supplies seats. A behavioral or institutional account is needed to connect those seats to actual agreements and policy decisions.
One way to summarize the vote-seat mismatch is to add the absolute percentage-point gaps. For D'Hondt here, ten plus eleven plus one equals twenty-two. For Sainte-Laguë, ten plus nine plus one equals twenty. Halving these sums gives eleven and ten percentage points, respectively. This index makes Sainte-Laguë slightly closer here. It measures a specified distance between party vote and seat shares. It does not measure personal accountability, social-group representation, ease of casting a valid ballot, or how much voters value any of those objectives.
D'Hondt can advantage larger lists relative to other proportional formulas. That does not mean it always gives the largest party an extra seat. Some vote vectors produce identical allocations under both methods. With votes fifty-four, twenty-nine, and seventeen and five seats, both give three, one, one. D'Hondt's winning claims are fifty-four, twenty-nine, twenty-seven, eighteen, and seventeen. Sainte-Laguë's are fifty-four, twenty-nine, eighteen, seventeen, and 10.8. The methods disagree on some inputs, but a difference in divisor sequences does not force a different final allocation on every input.
A proposal to change formulas therefore supports several distinct questions. Recounting fixed votes establishes a mechanical counterfactual. Predicting nominations, alliances, or turnout requires evidence about responses to incentives. Claiming that the change causes better governance needs a defined outcome and a credible causal comparison. Preferring closer party proportionality to another goal is a normative judgment. Current legal compliance requires the current complete rule. Selected IDEA and ACE discussions help identify these design components, but their general or historical explanations cannot settle those further questions for a present election.
| Party | Votes / 100 | D’Hondt seats / 5; gap | Sainte-Laguë seats / 5; gap |
|---|---|---|---|
| A | 50% | 3 / 5 = 60%; +10 pp | 2 / 5 = 40%; −10 pp |
| B | 31% | 1 / 5 = 20%; −11 pp | 2 / 5 = 40%; +9 pp |
| C | 19% | 1 / 5 = 20%; +1 pp | 1 / 5 = 20%; +1 pp |
| Half the absolute-gap sum | Not a vote count | (10 + 11 + 1) / 2 = 11 pp | (10 + 9 + 1) / 2 = 10 pp |
After four awards for votes 50,31,19, parties hold 2,1,1 seats. What quotients compete for the fifth D’Hondt seat?
Divide each vote total by its current seat count plus one.
| Party | Current seats | Next quotient |
|---|---|---|
| A | 2 | 16.67 |
| B | 1 | 15.5 |
| C | 1 | 9.5 |
A has 50/3 ≈ 16.67, B has 31/2 = 15.5, and C has 19/2 = 9.5. A wins the fifth seat, producing 3,1,1. Using total seats awarded as every party’s divisor would be a different and incorrect procedure.
At the same 2,1,1 allocation, which party receives the next Sainte-Laguë seat?
The next divisors are five, three, and three.
| Party | Next divisor | Quotient |
|---|---|---|
| A | 5 | 10 |
| B | 3 | 10.33 |
| C | 3 | 6.33 |
A has 50/5 = 10, B has 31/3 ≈ 10.33, and C has 19/3 ≈ 6.33. B receives the seat, producing 2,2,1. The method preserves the same total seats but changes how competing rounding claims are resolved.
The explanations, counts, and worked answers are complete without these optional links. Reference annotations identify the selected material reviewed. Historical country examples are not statements of current law, and the bounded models are not certified election-counting software. Mechanical results, behavioral evidence, and judgments about desirable representation remain separate.
Selected explanations of system families, district magnitude, thresholds, proportional formulas, mixed systems, and design objectives. Country examples and legal thresholds are historical, not statements of current law.
Overview of how magnitude and other design choices affect the translation of votes into seats. Not evidence that one institutional objective is universally preferable.