The single transferable vote fills several seats by retaining quotas, transferring surpluses, and eliminating candidates.
By the end: Trace one fractional surplus and distinguish transferable vote value from a count of physical ballot papers.
A multi-seat count has another objective
Single transferable vote, or STV, uses ranked candidate ballots to fill multiple seats. A quota sets a vote total sufficient for election under the chosen procedure. The integer Droop quota is the floor of valid ballots divided by seats plus one, then add one. For one hundred ballots and two seats it is thirty-four.
Three candidates cannot each retain thirty-four of a hundred units, because that would require one hundred and two. This explains the quota’s exclusion property. It does not mean every elected candidate must always finish at quota; remaining-candidate provisions and exhausted ballots matter in real counts.
Explain what the quota limits
A multi-member transferable count divides representation among candidates rather than choosing one majority winner. Our district elects two members from one hundred complete ranked ballots. A quota of 34 is sufficient to elect a candidate under the stated procedure, even though 34 is below a majority of one hundred. That difference follows from the institution being filled. Two seats permit more than one body of voters to obtain representation; they do not require one candidate to defeat every rival for a single office.
Calculate the integer Droop quota in stages. Divide the initial valid ballot count by the number of seats plus one, giving one hundred divided by three. Discard the fractional remainder to obtain 33, then add one to obtain 34. Discarding the remainder differs from rounding to the nearest integer. The denominator here is three because there are two seats plus one, not because three candidates happen to stand. Adding a fourth candidate would not change this quota with the same ballots and seats.
The quota has an exclusion property. Three separate retained quotas would require 102 value units, more than the initial hundred. At most two candidates can each retain 34 without duplicating value. This argument explains why the threshold fits the available vacancies, but it does not give a complete counting rule. We still need rules for moving a surplus, eliminating candidates, and ending the count. In a more general procedure, exhaustion or remaining-candidate provisions can matter, so the arithmetic does not prove that every elected member in every STV election finishes with a quota.
Three quotas cannot fit into one hundred units
Each outlined compartment holds 34 value units on the same linear scale. Two quotas use 68; three require 102. The 100-unit budget ends two units before the third quota ends. This exclusion property does not itself specify transfers or elimination.
First preferences put 50 ballots with A, 26 with B, and 24 with C. A exceeds 34 and takes the first seat. The account now separates the value needed for A's election from the excess that may support another candidate. A retains 34 units; its surplus is sixteen. Neither B nor C has yet reached quota. We cannot determine the second member from these first preferences alone, because the destination of the sixteen transferable units depends on later rankings supplied on the A-first ballots.
The lab provides those rankings explicitly. Every B-first ballot ranks C next and A last; every C-first ballot ranks B next and A last. An A-first ballot ranks either B then C or C then B. All rankings are complete and strict, so there are no missing next choices in this example. These are additional data, not conclusions inferred from the first-preference totals. They are ordinal reports of candidate order, not approval cutoffs or numerical scores, and they do not tell us how strongly a voter favors one candidate.
Before advancing playback, identify which quantities can change with the split control. The number of voters, first preferences, seats, quota, and initial surplus all remain fixed. Only the number of A-first ballots placing B ahead of C changes. Predict that A remains elected, while the second seat may change. This is a controlled comparison within a ranked multi-seat procedure. It is not evidence that the same voters would submit these rankings under another ballot format or choose the same candidates after a real institutional reform.
The complete rankings behind the reference transfer
Records
Ranking, best first
Role in this trace
30
A > B > C
Surplus portion goes first to B
20
A > C > B
Surplus portion goes first to C
26
B > C > A
Initially unit value with B
24
C > B > A
Initially unit value with C
Reference split, 100 records total. These complete rankings are stipulated. Moving the existing split control changes only the two A-first row counts, whose sum stays fifty.
Transfer the surplus, not all the elected candidate’s value
Our bounded example starts with A 50, B 26, C 24. A wins one seat and retains thirty-four units. Its surplus is sixteen. All fifty A ballots initially have unit value, so this example transfers 16/50 = 0.32 of each to the next continuing candidate. The retained portion is 0.68 per ballot.
If thirty A ballots next prefer B and twenty next prefer C, B receives 9.6 units and C 6.4. Their new totals are 35.6 and 30.4, while A retains thirty-four. B wins the second seat. We stop when both seats are filled and do not redistribute B’s final surplus.
Split each initial ballot's value without duplicating it
The initial A ballots all have value one, so the bounded transfer spreads the surplus equally across all fifty. Sixteen divided by fifty is 0.32. Each A ballot sends 0.32 to its next continuing candidate and retains 0.68 for A. Multiplying fifty by 0.68 gives 34 retained units; multiplying fifty by 0.32 gives sixteen transferable units. Every original unit is accounted for exactly once. The count maintains numerical accounts without physically cutting ranking records into pieces.
In the reference split, thirty A-first ballots next prefer B and twenty next prefer C. The thirty records carry thirty initial units, but only their surplus fraction travels onward. B receives thirty multiplied by 0.32, or 9.6 units. C receives twenty multiplied by 0.32, or 6.4. Add those amounts to the original first-preference values rather than replacing the originals. B rises from 26 to 35.6 and C from 24 to 30.4. A's account now holds the retained 34 rather than its initial fifty.
A useful accounting check is 34 plus 35.6 plus 30.4, which equals one hundred. Another is 9.6 plus 6.4, which equals the sixteen-unit surplus. If you transferred thirty full units to B and twenty full units to C while retaining 34 for A, the total would become 134. If you removed all fifty from A without retaining anything, A's elected seat would disappear from the accounting. Fractional transfer avoids both errors by recording the retained and transferable portions of the same original ballot value.
One initial A ballot has two value accounts
Widths are proportional to value: 600 horizontal units represent one initial ballot unit. All fifty A records split this way. In the reference split, 30 × 0.32 goes to B and 20 × 0.32 to C. Retained and transferred portions sum to one per record, or one hundred over all ballots including the original B and C records.
Physical record counts differ from current value totals. After the reference transfer, B's support includes 26 B-first records at unit value and thirty A-first records contributing 0.32 each. That is 56 records contributing 35.6 current units to B. The thirty A-first records also have retained portions associated with A's seat. Summing candidate record counts would therefore double-count records appearing in these different accounts. Sum retained and current value instead. A candidate's current value is not a count of distinct people who ranked that candidate first or endorsed that candidate without qualification.
B exceeds quota by 1.6 and wins the second seat. The model stops immediately because both seats are filled. It leaves the final accounts at 34, 35.6, and 30.4; it does not transfer B's final excess to C. Consequently, saying that every winner retains exactly 34 at the end would misdescribe this trace. A retains exactly a quota because its surplus had to be distributed while a vacancy remained. B's final account can exceed quota because no vacancy remains for another transfer to help fill.
Use the reference preset and advance one step to the surplus distribution. Pause and reconcile the two independent sums before advancing to the election announcement. The left panel keeps the original first preferences visible; the right panel changes to value accounts, including A's retained portion. The shared scale compares quantities; labels distinguish stages. A lower A bar does not mean A lost voters or ceased to be elected. It means sixteen units were separated from the original fifty under the stated transfer convention.
Fifty-six contributing records do not mean 56 current units
B account after reference surplus
Records
Value per record
Current contribution
B > C > A
26
1
26
A > B > C, transferred portion
30
0.32
9.6
Total currently contributing to B
56
Mixed values
35.6
Retained portion of those 30 A records
Same 30, not extra voters
0.68
20.4 retained for A
An A-first record can have retained value with A and current value with B. Counting records across accounts would duplicate people; summing value preserves the original hundred units.
Work through the mechanism
Integer Droop quota
V is the initial valid ballot count and S is the number of seats. Floor means discard the fractional remainder, not round to the nearest integer.
100 ballots, 2 seats
Initial equal-value surplus transfer
A retains value 34 and transfers value 16 across its 50 initially unit-value ballots. Each transfers 0.32 and retains 0.68. Later or statutory STV transfers can use other conventions.
Retain the quota and move the surplus
The other 20 A-first ballots next prefer C. B-first ballots rank B>C>A; C-first ballots rank C>B>A. A ballots rank the other continuing candidate third. Changing the split resets the trace. All initial ballots have unit value.
First preferences: A exceeds quota 34 and wins the first seat.
Step 0 / 2
Complete surplus and elimination trace
Stage
Value with A / B / C
Explanation
1
50 / 26 / 24
First preferences: A exceeds quota 34 and wins the first seat.
2
34 / 35.6 / 30.4
A retains value 34. Transfer its surplus 16 at 0.32 per ballot. B has 35.60; C has 30.40.
3
34 / 35.6 / 30.4
B wins the second seat. Stop: any final surplus is not redistributed in this example.
Quota 34; initial surplus 16; transfer fraction 0.32. After the surplus transfer, A’s bar is retained value, not continuing votes. Retained and continuing value sum to 100 throughout. Stop after two seats are filled; any final surplus remains with the last winner in this bounded example.
Destination
Transferred A ballots
Value per ballot
Added vote value
B
30
0.32
9.6
C
20
0.32
6.4
Total
50
0.32
16
Reference split: thirty A ballots next prefer B and twenty next prefer C. Ballot papers and vote value are distinct. A retains 34 while 16 transfers.
Follow the branch where nobody reaches quota
The elimination-needed preset sends twenty A ballots toward B and thirty toward C. At 0.32 each, their contributions are 6.4 and 9.6 units. B now has 32.4 and C has 33.6, while A retains 34. Neither continuing candidate has reached quota. The total equals one hundred, but conservation alone cannot determine the next decision. The procedure needs an elimination rule to reduce the remaining competition and direct each affected contribution to its next continuing candidate.
B is lower than C, so B leaves the contest. Its 32.4 units consist of 26 unit-value B-first ballots and twenty fractional contributions from A-first ballots. Each B-first ballot lists C before A. Each transferred A-first ballot in B's account ranks A then B then C, with A already elected and no longer continuing. Both groups therefore send their current transferable value to C. The fractional contributions keep value 0.32; elimination does not restore them to value one or transfer their retained portions away from A.
C receives all 32.4 and rises from 33.6 to 66. A still retains 34, making the total one hundred. C takes the second seat and the count stops. Its final 66 is not a claim that 66 distinct voters put C first, nor that C has majority first-choice support. It is accumulated current value after transfers and an elimination. The model leaves that large final value with C because distributing a final surplus would serve no remaining vacancy. Do not append another transfer merely to make every elected account equal the quota.
Elimination preserves the value of each parcel
Twenty-to-B branch
A retained
B current
C current
After initial surplus
34
26 + 6.4 = 32.4
24 + 9.6 = 33.6
Remove B: original B records
34
26 units leave
33.6 + 26 = 59.6
Remove B: A-first fractional portions
34
6.4 units leave
59.6 + 6.4 = 66
After the complete elimination
34
0
66; elected, then stop
The two parcel rows explain one elimination operation, not additional playback steps. B-first records rank C next; A > B > C records skip elected A and removed B. No retained A value moves.
Contrast this with the ten-to-B preset. B gets 3.2 and C gets 12.8 from A's surplus, yielding 29.2 and 36.8. C reaches quota directly, so there is no elimination step. The same elected pair, A and C, emerges through different intermediate decisions. A final winner list cannot show whether someone reached quota immediately, received a surplus, or survived an exclusion. The complete count table records that branch.
Find the boundary using the available split slider, which moves in steps of five ballots. At twenty to B, C leads after transfer but remains below quota, so elimination is necessary. At twenty-five to B, B has 34 and C 32, so B reaches quota exactly. Calculate the change before moving the control: shifting five A ballots from C toward B moves 1.6 units from C to B, changing their difference by 3.2. It does not add new ballot value. A's retained 34 remains unchanged throughout this comparison.
Changing the split resets playback because the old sequence no longer describes the selected ballots. Use the complete trace to check the new intermediate totals rather than assuming that a step number has the same meaning across scenarios. A quota-reaching preset needs no exclusion, while the elimination-needed preset has an additional decision. There is no exact B/C tie at an integer split in this bounded example. The equality point would require 21.875 A ballots next preferring B, so the available ballot counts never call for a tie-breaking decision.
Available splits lead to different branches
A records next preferring B / C
B after surplus
C after surplus
Second-seat decision
10 / 40
29.2
36.8
C reaches quota directly
20 / 30
32.4
33.6
Eliminate B; C finishes at 66
25 / 25
34
32
B reaches quota exactly
30 / 20
35.6
30.4
B exceeds quota
Always 100 ballots, quota 34, surplus 16, transfer fraction 0.32. Moving five A records toward B moves 1.6 units from C to B. The resulting B-minus-C difference changes by 3.2 units.
Do not generalize one transfer convention to every STV rule
If neither remaining candidate reaches quota in this example, eliminate the lower one and transfer its remaining value to the other. All rankings are complete, so no value exhausts. The lab changes only how the fifty A-first ballots divide their second choices; the initial first-choice totals remain fixed.
Real STV implementations differ in transfer methods, which parcels are included, precision, rounding, tie handling, and the order of exclusions. Later transfers can involve already-fractional ballots. This model demonstrates one initial equal-value surplus, not a general statutory STV implementation or a prediction of actual voters’ later choices.
Audit what this transferable count does and does not establish
The equal transfer fraction assumes A's fifty ballots initially have unit value. In a longer count, a candidate might receive records at several current values through earlier transfers. A new surplus would then require a specified convention about included parcels, weights, rounding, and transfer order. Reusing sixteen divided by fifty without checking those inputs would not extend this model correctly. The lesson implements one initial surplus followed, when needed, by one elimination. It does not implement arbitrary sequences of surpluses or every procedure called STV.
Complete rankings remove exhaustion from this particular trace. If incomplete rankings were allowed, an affected contribution could have no next continuing candidate. That would require a separate exhausted-value account rather than deletion or reassignment by guesswork. With fractional transfers, exhausted value and the number of exhausted records would also need separate labels. This is a paper extension, not a control in the current lab. It neither replaces the complete-ranking assumptions nor gives ballot-completion instructions for any jurisdiction.
For a concrete conservation thought experiment, suppose the twenty A-first records contributing 6.4 to B in the elimination-needed branch had no usable preference after B. If that optional-ranking variant were permitted, B's 26 unit-value original ballots would still move to C, while 6.4 would exhaust. C would finish with 59.6, A would retain 34, and exhausted value would be 6.4. These sum to one hundred. Twenty exhausted fractional contributions are not twenty exhausted full units. The original complete-ranking trace instead sends their 6.4 onward and finishes C at 66.
A paper extension distinguishes exhausted records and value
After B elimination at the 20/30 split
Complete rankings in lab
Optional-ranking thought experiment
A retained value
34
34
C current value
66
59.6
Exhausted fractional contributions
0 records, 0 units
20 records, 6.4 units
Total retained + current + exhausted
100
100
The optional variant omits usable later preferences on twenty A-first records contributing 0.32 each to B. It is not an available lab setting or a statement of any jurisdiction’s ballot-validity rules. C still exceeds the unchanged quota of 34.
Do not confuse the quota's denominator with a continuing-vote majority. Our quota was fixed from the original hundred valid ballots and two vacancies. An instant-runoff majority calculation for one office can instead compare current votes with half the continuing total at each round. Those are different decisions with different denominators. In this STV example, A is elected with fifty initial units, B can be elected with 34 at a boundary split, and two members eventually represent the district. None of those facts establishes a nationwide party-proportional legislature.
The ranking split also has no automatic behavioral interpretation. A different split could reflect genuinely different preferences, a different nominated candidate set, or strategic reporting, but the count cannot identify which explanation applies. Nor does a candidate's transferred value measure policy agreement or approval intensity. The selected AEC explanation supports the distinction between quota, surplus, and exclusion, while its fuller procedures include details outside this model. The historical handbook likewise supplies institutional context, not a current legal specification for the invented candidates or empirical proof of a reform's effects.
Audit a new trace by identifying the open vacancies before each operation. Check the quota inputs, separate retained and continuing accounts, preserve current value during elimination, and stop when the stated vacancies are filled. Then ask what information a broader conclusion would require. Predicting party representation needs candidate affiliations and more districts; assessing population equality needs population counts; preferring this rule to another needs explicit objectives. The worked count establishes how these complete rankings fill two seats under one bounded procedure. Correct arithmetic leaves those questions separate.
Check the operation before extending the conclusion
Stage or claim
Required check
Boundary of this example
Initial quota
100 valid ballots, 2 seats; floor(100/3) + 1
Quota fixed at 34
A surplus
50 × 0.68 retained; 50 × 0.32 transferred
One initial equal-value surplus
B exclusion when needed
Move 26 + 6.4 current units to C
Complete rankings; no exhaustion
Final vacancy filled
Stop at two elected candidates
Do not distribute final surplus
Party or population representation
Candidate affiliations, other districts, population basis
Not established by this count
The trace establishes two elected candidates under a bounded procedure. Its current-value totals are neither majority first-choice counts nor population shares, and the trace is not a statutory counting implementation.
Exercises
Conserve the vote value
After A’s surplus transfers in the 30/20 split, add A’s retained value and the new B and C totals. What would transferring all fifty A votes get wrong?
Hint
Keep the value retained for A’s elected seat in the accounting.
Worked answer
Retained and continuing value occupy one hundred units in total, without double-counting the surplus.
Account
Value
A retained
34
B after transfer
35.6
C after transfer
30.4
Total
100
Retained and transferred portions sum to the original value.
The total is 34 + 35.6 + 30.4 = 100. Transferring fifty while also retaining thirty-four would create thirty-four extra units. The fraction transfers only the surplus, not an additional full vote from each A ballot.
Change later preferences
Only ten of A’s fifty ballots next prefer B; forty prefer C. Which candidate wins the second seat after the surplus transfer?
Hint
The surplus stays sixteen and the transfer fraction stays 0.32.
Worked answer
Candidate
Starting value
Transferred value
New value
B
26
3.2
29.2
C
24
12.8
36.8
The second-choice split changes; the electorate and surplus do not.
B receives 10 × 0.32 = 3.2 and rises to 29.2. C receives 40 × 0.32 = 12.8 and rises to 36.8, exceeding quota. C wins the second seat. Unchanged first choices do not imply an unchanged STV outcome.
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 quota, surplus transfer value, exclusions, and exhausted votes. Our example covers one initial equal-value surplus, not every statutory transfer or rounding rule.
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.