A runoff can use a new ballot or transfer ranked ballots. The denominator matters.
By the end: Trace instant-runoff voting and distinguish continuing votes from all original voters.
A new election is not the same as a transfer
A two-round system can hold a later ballot between selected finalists. Voters then choose again, and turnout, information, or alliances may change. Exact qualification rules vary. A simple top-two runoff is one design, not the definition of every two-round system.
Instant-runoff voting, also called the alternative vote in some contexts, uses ranked ballots in one election. In our model, count each ballot for its highest-ranked remaining candidate. If nobody has more than half of continuing votes, eliminate the lowest candidate and transfer affected ballots to their next remaining choices.
What a later choice can establish
A runoff addresses the choice of one officeholder. It does not distribute shares of that office among candidates. Start with one hundred valid first choices, forty for A, thirty-five for B, and twenty-five for C. A leads, but no candidate exceeds fifty. A top-two procedure keeps A and B for a later contest. That selection uses first-round support, not a claim that C is nobody's preferred compromise. Which candidates qualify and which votes count toward qualification must belong to the stated rule.
There are sixty first-round votes for candidates other than A. That fact does not identify a sixty-person coalition for B. Some C voters might prefer A to B, while others prefer B to A. A later ballot asks the finalists' question directly of those who participate then. Ranked ballots can instead record a conditional answer in advance. These are different ways of obtaining information. Neither permits an analyst to read the missing preferences out of the original three totals alone, even when those totals are exact.
Suppose, on paper, all twenty-five C voters prefer B to A and everybody participates again without changing that preference. B then receives thirty-five plus twenty-five, giving sixty against A's forty. Reverse only the C voters' comparison, and A receives sixty-five against B's thirty-five. Both continuations fit the same first-round totals. It is information that a one-choice ballot did not collect. A prediction between these outcomes therefore needs evidence beyond the first count and the identity of the finalists.
The same first round permits opposite runoff results
Additional assumption
A in A/B contest
B in A/B contest
All 25 C voters prefer B to A
40
35 + 25 = 60
All 25 C voters prefer A to B
40 + 25 = 65
35
Invented fixed-participation continuations. First choices are A 40, B 35, C 25 in both rows. Later preferences are additional information.
The complete ranked profile supplies the missing information explicitly. Forty people rank A above C above B, thirty-five rank B above C above A, and twenty-five rank C above B above A. If only A and B remain, the last group contributes to B because B is its highest remaining choice. Calling these conditional choices does not assume voters admire B or regard A as unacceptable. A ranking states an order. It supplies neither an approval boundary nor numerical distances between the candidates for any voter.
A separate election can differ from this conditional count even if its finalists match. People can abstain, previously absent people can vote, and information can change preferences. For example, a later result of A forty and B forty-five among eighty-five voters would give B a majority of that later turnout. It would not reproduce B's stipulated sixty votes in the complete ranked profile. This is a comparison of two invented processes, not evidence that a second ballot generally increases or decreases any candidate's support or the electorate's participation.
Before following the player, write down what each comparison holds fixed. The ranked count keeps the submitted papers and their listed orders fixed while changing the set of remaining candidates. The hypothetical second election permits a new collection of ballots, so it needs a participation assumption as well as preferences. Ask whether the proposed conclusion concerns a counting mechanism, a forecast of choices, or the desirability of requiring another election. Those questions require different support. Only the first is settled by the deterministic transfer trace that follows.
New papers and fixed papers answer different questions
Process
Paper collection
Illustrative final count
Transfer complete compromise rankings
Same 100 submitted papers
A 40; B 60
Separate later ballot, stipulated result
85 new papers at later turnout
A 40; B 45
The second row is an independent paper scenario, not a forecast. B has 45/85, about 52.94% of that later turnout; this is not 60/100.
Trace the same voters
Our complete profile has forty A>C>B ballots, thirty-five B>C>A ballots, and twenty-five C>B>A ballots. First choices are 40, 35, 25. C is eliminated. Its twenty-five ballots next prefer B, so B wins with sixty continuing votes against A’s forty.
Ranking a second candidate does not give the ballot a simultaneous second full vote in this single-winner count. Its current contribution goes to one remaining candidate. A transfer changes where that contribution is counted. Equal-lowest ties in this teaching model eliminate the alphabetically last candidate; actual tie rules differ.
Track one contribution through elimination
Think of each complete ranked paper as an instruction for locating its one current contribution. While A, B, and C remain, its first name supplies the destination. After an elimination, scan its order from the beginning and use the first name still present. This rule also handles a paper whose next written name has already left. It does not always mean use the next numbered line.
For the forty A>C>B papers, A remains available throughout the compromise count, so their destination never changes. The thirty-five B>C>A papers likewise stay with B. Only the twenty-five C>B>A papers need a new destination when C leaves. They move to B, not to both B and A. The final active totals are therefore forty and sixty. Keeping an active C total of twenty-five as well would manufacture an extra contribution for each transferred paper and give one hundred twenty-five active votes from one hundred papers.
The majority test happens before another elimination. Initially forty, thirty-five, and twenty-five are all at most fifty, so the count removes C. After recounting, sixty exceeds half of one hundred, so the model declares B the winner and stops. It does not eliminate A merely to produce a unanimous last column. That stopping point matters for interpretation. The result establishes B's majority against the remaining alternative under these submitted orders, not that every voter chose B first or that the original A voters changed their minds.
Only the eliminated group's contributions change destination
Schematic paper routes, not distance or time scales. Each strip denotes the stated group of whole unit-value papers. After C leaves, B holds 35 + 25 = 60; A holds 40. No active value remains with C.
Exact ties need an instruction too. Consider a separate paper example with forty A>B>C papers, thirty B>A>C papers, and thirty C>B>A papers. B and C tie for lowest first-choice support. The teaching model removes the alphabetically last tied candidate, C. Its thirty contributions then join B, giving B sixty against A forty. If a different declared rule removed B instead, those thirty B papers would join A and produce A seventy against C thirty. A tie policy can therefore affect the officeholder, not merely the presentation of the count.
That tied example is not another preset in the player. Reproduce it on paper with two rows, one for the initial count and one for the chosen elimination. The point is to distinguish a missing rule from random numerical noise. With exact counts, thirty equals thirty; neither candidate has less support. Alphabetical priority is a reproducibility convention here, not a recommendation for actual election administration. A statutory procedure might resolve ties differently, and determining its operation would require checking the applicable legal specification rather than importing this example's convention.
A tied elimination needs a declared rule
First choices A/B/C
Tie decision
Final count
40 / 30 / 30
Model removes C
A 40; B 60
40 / 30 / 30
Alternative removes B
A 70; C 30
Papers are 40 A>B>C, 30 B>A>C, 30 C>B>A. This paper-only contrast changes the tie policy, not the papers. The model removes the alphabetically last tied candidate.
These are single-winner transfers of whole current contributions. Do not carry the instruction unchanged into a multi-seat transferable count. In the later bounded STV example, a winner retains value for an elected seat and transfers only its surplus at a fractional value. Here no candidate retains a quota for a filled seat while another seat remains vacant, because there is only one office. Counting papers, continuing contributions, and retained value are different accounting tasks. For this complete single-winner profile, the first two totals happen to remain equal at one hundred.
Work through the mechanism
Move a ballot to its next remaining choice
Presets replace the whole ballot profile and reset playback. The original count remains on the left.
Stipulated ballots, not inferred from first-choice totals
Voters
Ranking, best first
40
A > C > B
35
B > C > A
25
C > B > A
Count 1: Eliminate C; move each affected ballot to its next active choice, or exhaust it. Continuing 100; exhausted 0.
Step 0 / 1
Complete count, available offline
Count
A / B / C
Exhausted
Decision
1
40 / 35 / 25
0
Eliminate C; move each affected ballot to its next active choice, or exhaust it.
2
40 / 60 / 0
0
B has more than half of continuing votes.
Active votes plus exhausted ballots always total 100. The bars stay on the original 0–100-vote scale even when the continuing-majority denominator changes. Exact elimination ties remove the alphabetically last tied candidate in this teaching model.
C ballot ranking
After C leaves
Winner
C then B then A
A 40; B 60; exhausted 0
B
Only C
A 40; B 35; exhausted 25
A with 40/75 continuing votes
The first-choice totals are identical. Later preferences determine what transfers.
Predict each preset before advancing
Select the compromise preset and pause or reset the player if it has already advanced automatically. Read the listed rankings before using Next step. Predict which papers will need a new destination, then predict the entire next count rather than only its winner. The left panel shows original first choices and the right panel shows current active votes. After the transfer, the correct right-hand totals are A forty, B sixty, C zero. The left-hand C bar stays at twenty-five because it records the original count, not another active allocation.
Use the cycle preset next. Its papers are thirty-five A>B>C, thirty-three B>C>A, and thirty-two C>A>B. No first-choice total exceeds fifty, and C is lowest with thirty-two. Once C leaves, all thirty-two C-first papers find A before B. A therefore rises to sixty-seven while B remains thirty-three. This is a different ballot profile from the compromise case, not another stage of the same election. The preset replaces all listed rankings and resets playback, so inspect the new data before comparing its winner with the reference profile.
The cycle preset transfers C papers to A
Each rectangle is one ranked paper group, not one person. Routes are schematic. Complete rankings keep all 100 contributions active; this single-winner count stops at A's 67–33 majority.
Now choose C voters rank only C. The first-choice bars return to forty, thirty-five, and twenty-five, just as in the compromise profile. Before stepping, predict whether B's total will rise. It will not. None of the twenty-five C-only papers names A or B, so each lacks a remaining destination. A stays forty, B stays thirty-five, and twenty-five papers exhaust. Active contributions total seventy-five. The winning result changes even though the initial totals do not, because the papers differ in information that the initial bars cannot display.
Compare the final bars with the numerical status rather than estimating a majority from bar lengths. The bars keep their original zero-to-one-hundred scale when papers exhaust. A's forty-unit bar does not grow beyond the midpoint of that original scale, but forty is more than half of the seventy-five continuing contributions. A majority line for that round would lie at thirty-seven and a half on the original scale. A fixed visual scale supports honest count comparisons; the changing denominator must come from the explicit continuing total, not a hidden rescaling.
The complete count reveal supplies the same stages without requiring you to observe the animation at the right moment. For each row, add the three candidate totals and the exhausted count. The sum must remain one hundred in all three presets. Then identify which comparison justified stopping. B's sixty exceeds fifty in the compromise profile, A's sixty-seven exceeds fifty in the cycle profile, and A's forty exceeds thirty-seven and a half in the exhaustion profile.
Audit all three final counts
Preset
Final A / B / C
Exhausted
Majority comparison
Compromise
40 / 60 / 0
0
60 > 100/2
Cycle
67 / 33 / 0
0
67 > 100/2
C voters rank only C
40 / 35 / 0
25
40 > 75/2
Counts are whole active contributions from 100 original papers. Candidate totals plus exhausted papers equal 100 in each row; the majority base includes continuing votes only.
Finish the comparison by explaining one changed quantity and one unchanged quantity between compromise and exhaustion. B's final support changes from sixty to thirty-five, while A's forty stays fixed. The original count of papers also stays fixed, but the final continuing count changes. This identifies the mechanical effect of replacing the C group's complete orders with C-only orders. It does not establish why real people would omit names, whether omissions were intentional, or how a different validity requirement would change participation.
State the majority denominator
If incomplete rankings are permitted, a ballot can exhaust when none of its listed candidates remain. If the twenty-five C ballots list only C, they exhaust after C’s elimination. A then has forty of seventy-five continuing votes, more than half, although A still has only forty of the original hundred.
This is a hypothetical optional-ranking rule. It is not an instruction for completing an Australian House ballot or any other jurisdiction’s ballot. A real claim that a winner has a majority must say whether the denominator is eligible people, turnout, formal ballots, or continuing ballots at that round.
Attach every majority to its base
A majority means strictly more than half of a named total. For one hundred continuing votes, fifty is not enough and fifty-one is the smallest whole number that qualifies. For seventy-five continuing votes, the boundary is thirty-seven and a half, so thirty-eight is the smallest qualifying integer. This explains why forty can fail the first test and pass the second without any increase in its numerator.
In the exhaustion example, A has forty of seventy-five continuing votes, about 53.33 percent. Against the original one hundred valid papers, A has forty percent. Both calculations describe the same final count. The exhausted papers remain part of the historical turnout and original valid-ballot totals even though they no longer contribute to a continuing candidate. They have not become votes for A. They also have not become invalid retrospectively under this hypothetical optional-ranking rule. Exhaustion describes the absence of a remaining listed destination at that counting stage.
Add a separate invented participation context of one hundred twenty-five eligible people, of whom one hundred submit these valid papers. A's forty then represents thirty-two percent of eligible people. That does not contradict the continuing majority. It answers a different question with another base. If a resident population also includes ineligible people, a population percentage would need its own count and justification. The player supplies ballots, not an electoral register or census, so this eligibility example is an additional paper assumption rather than a result you can read from the interface.
Forty votes, three explicitly different denominators
Bars share a scale of four SVG units per person or contribution. The extra 25 eligible nonvoters are a paper assumption, not player data. Exhausted papers do not become A votes. A has about 53.33% of continuing votes, 40% of original papers, and 32% of eligible people.
A claim about representation in a legislature requires another step. Electing one officeholder by continuing majority gives that person the single seat. It does not give the runner-up a fraction of it. If many districts each elect one representative, the chamber's party composition comes from adding district winners, not pooling every final-round majority into a proportional national allocation. A rule can require a majority within each district and still produce party seat shares that differ from aggregate vote shares. The local majority test and legislative proportionality evaluate different objects.
To audit a public claim, ask for the numerator, denominator, stage, and applicable ballot rule. A statement that A received forty votes after C's elimination can be checked against the trace. A statement that A won a majority of all original voters cannot. A statement that the optional-ranking design is desirable requires explicit objectives, such as freedom to omit candidates or breadth of continuing participation. Arithmetic can show the consequences of a specified omission. It cannot decide how to weigh those objectives or establish the current law of a jurisdiction.
The selected IDEA passages describe preferential elimination and the possibility of requiring more complete rankings. The selected AEC material concerns an official counting context, not permission to submit the C-only paper in that context. Read source scope together with the model's hypothetical validity assumptions. For transfer practice, change the paper example to forty-one A, thirty-four B, and twenty-five C-only ballots. After C leaves, A has forty-one of seventy-five, about 54.67 percent, but forty-one percent of original papers. The same denominator distinction survives a changed numerator and remains independently checkable.
Strictly above half, not at half
Continuing votes
Halfway boundary
Smallest whole majority
Example
100
50
51
40 fails
75
37.5
38
40 succeeds
75, changed paper example
37.5
38
41/75 ≈ 54.67%
The changed example has A 41, B 34, C-only 25. A remains at 41 when C exhausts. A has 41% of original papers, not 54.67% of them.
Exercises
Transfer, do not duplicate
After C is eliminated from the complete profile, what are the totals and where did the twenty-five C votes go?
Hint
Each transferred ballot contributes to one remaining candidate.
Worked answer
Candidate
Before
After
A
40
40
B
35
60
C
25
0
Total active
100
100
Transfers conserve active value when no ballot exhausts.
A remains at forty. B rises from thirty-five to sixty. C no longer has an active total. The hundred ballots still contribute one hundred votes in all. Adding B’s transferred votes while retaining C’s active twenty-five would double-count them.
A continuing majority
If the twenty-five C ballots exhaust, calculate A’s share of continuing votes and original ballots.
Hint
The continuing total is seventy-five, not one hundred.
Worked answer
Denominator
A share
75 continuing votes
53.3%
100 original ballots
40%
Always attach a denominator to a majority claim.
A has 40/75, about 53.3% of continuing votes, and 40/100 = 40% of original ballots. These are both correct with different denominators. A statement that A won majority support from all original voters would not follow.
Sources and further reading
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.
Official explanation of preference distribution and counting. The lesson’s exhausted-ballot scenario is a hypothetical optional-ranking model, not the Australian House ballot-validity rules.
Selected explanations of quota, surplus transfer value, exclusions, and exhausted votes. Our example covers one initial equal-value surplus, not every statutory transfer or rounding rule.