Two officers — where RRV parts company with the quota methods¶
Three voters, three candidates, two seats. Dana takes seat 1 under every method. Then Allocated Score and SSS seat Finn, while RRV seats Eli. The three methods charge Dana's two big backers ¼, ⅓ and 5⁄9 of their weight respectively — a ladder from harshest to gentlest — and the seat flips on the last rung.
→ Family: when the STAR-PR methods disagree · the other election: Three neighbors · the methods: Allocated Score · RRV
Level: 301 · deep dive
The ballots¶
| Dana | Eli | Finn | ||
|---|---|---|---|---|
| Voter 1 | 4 | 3 | 1 | Dana first, Eli a solid second |
| Voter 2 | 4 | 2 | 2 | Dana first, Eli and Finn equal behind |
| Voter 3 | 1 | 3 | 4 | Finn first, barely backs Dana |
| Total | 9 | 8 | 7 |
Two seats, so the Hare quota is 3 ÷ 2 = 1.5 voters.
Seat 1 is unanimous¶
Dana leads at 9, and every method elects her. Voters 1 and 2 put her there with 4 stars each; voter 3 gave her a single star.
Seat 2: how hard are Dana's backers charged?¶
Voters 1 and 2 put Dana in with 4 stars each and now owe something for it. All three methods agree they owe something; they disagree about how much, and the answer is a clean ladder.
Allocated Score — voters 1 and 2 drop to ¼. The quota is 1.5 voters and the 4-star tier holds two ballots, so it overfills; fractional surplus keeps them at a quarter. Voter 3 is never reached by the tiers at all and stays at full weight:
| weight | Eli | Finn | |
|---|---|---|---|
| Voter 1 | ¼ | 0.75 | 0.25 |
| Voter 2 | ¼ | 0.50 | 0.50 |
| Voter 3 | 1 | 3.00 | 4.00 |
| Round 2 | 4¼ | 4¾ ← Finn |
SSS — voters 1 and 2 drop to ⅓. SSS charges against a score quota rather than a ballot quota: total score 9, Hare score quota 7½, so every supporter's stars are reduced in proportion to what they gave. Voter 3 pays too, landing at ⅚ — she gave Dana a star, so she is charged for a star:
| weight | Eli | Finn | |
|---|---|---|---|
| Voter 1 | ⅓ | 1.00 | 0.33 |
| Voter 2 | ⅓ | 0.67 | 0.67 |
| Voter 3 | ⅚ | 2.50 | 3.33 |
| Round 2 | 4⅙ | 4⅓ ← Finn |
Gentler than Allocated Score, and Eli closes from 0.50 behind to 0.17 behind — but not enough.
RRV — voters 1 and 2 only drop to 5⁄9. No quota is computed at all; each ballot is simply divided by 1 + (score given to winners / max score). Voters 1 and 2 gave 4 of 5, so 1 / 1.8 = 5⁄9. Voter 3 gave 1 of 5, so 1 / 1.2 = ⅚ — the same as under SSS:
| weight | Eli | Finn | |
|---|---|---|---|
| Voter 1 | 5⁄9 | 1.67 | 0.56 |
| Voter 2 | 5⁄9 | 1.11 | 1.11 |
| Voter 3 | ⅚ | 2.50 | 3.33 |
| Round 2 | 5.28 ← Eli | 5.00 |
Voters 1 and 2 keep enough to carry their shared second choice, and Eli takes seat 2 instead.
What the ladder means¶
Line the three charges up against the same two voters:
| Method | Voters 1 & 2 charged down to | Eli − Finn | Seat 2 |
|---|---|---|---|
| Allocated Score | ¼ = 0.250 | −0.50 | Finn |
| SSS | ⅓ = 0.333 | −0.17 | Finn |
| RRV | 5⁄9 = 0.556 | +0.28 | Eli |
Nothing else differs — same ballots, same seat 1, same arithmetic afterwards. As the charge gets gentler, the two partly-satisfied voters keep more say and their second choice climbs, until at RRV it overtakes. A quota method decides those voters have had their seat and spends them down hard; a divisor method decides they are merely partly satisfied and turns them down by a proportion. The first hands seat 2 to the one voter still hungry; the second lets the partly-fed keep bidding.
Note where the two quota methods differ from each other, since it is easy to miss: Allocated Score leaves voter 3 at full weight because the score tiers never reach her, while SSS charges her ⅚ for the single star she gave Dana. Same winner here, different books — and Three neighbors is the election where that same bookkeeping gap changes the seat.
Neither is a malfunction. This is the Balinski–Young trade showing up in three ballots: the quota methods guarantee that a quota-sized faction can force a seat and pay with the non-monotonicity behind the Alabama paradox; RRV is coherent and monotone and pays by failing the Hare Quota Criterion. Here that abstraction has a face: voter 3 is a one-person faction just over a 1.5 quota's worth of unspent weight, and only the quota methods guarantee she gets the seat.
The counts¶
Allocated Score → Dana, Finn
--- Allocated Score Voting Method (2 winners) ---
[Allocated Score Voting]
Tabulating 3 ballots to fill 2 seats.
Dana,Eli,Finn
4, 3, 1
4, 2, 2
1, 3, 4
[Allocated Score Voting: Round 1]
The highest-scoring candidate wins a seat.
Dana -- 9 -- First place
Eli -- 8
Finn -- 7
Dana wins a seat.
[Allocated Score Voting: Round 1: Ballot allocation round]
Allocating 1+1/2 ballots.
[Allocated Score Voting: Round 1: Ballot allocation round: Round 1]
Allocating 2 ballots at score 4.
This allocation overfills the quota. Returning fractional surplus.
Allocating only 75.00% of these ballots.
Keeping these ballots, but multiplying their weights by 1/4.
2 ballots reweighted from 1 to 1/4.
[Allocated Score Voting: Round 2]
The highest-scoring candidate wins a seat.
Finn -- 4+3/4 -- First place
Eli -- 4+1/4
Finn wins a seat.
[Allocated Score Voting: Winners — Allocated Score Voting Method (2 winners)]
Dana
Finn
Sequentially Spent Score → Dana, Finn
--- Sequentially Spent Score Voting Method (2 winners) ---
[Sequentially Spent Score]
Tabulating 3 ballots to fill 2 seats.
Dana,Eli,Finn
4, 3, 1
4, 2, 2
1, 3, 4
[Sequentially Spent Score: Round 1]
The highest-scoring candidate wins a seat.
Dana -- 9 -- First place
Eli -- 8
Finn -- 7
Dana wins a seat.
[Sequentially Spent Score: Round 1: Ballot allocation round]
Total score is 9, Hare score quota is 7+1/2, giving back surplus.
Reducing each ballot's stars by their vote * 1/6.
Reweighted 3 ballots:
2 ballots voted 4, stars reduced from 5 to 5/3, reweighted to 1/3.
1 ballot voted 1, stars reduced from 5 to 25/6, reweighted to 5/6.
[Sequentially Spent Score: Round 2]
The highest-scoring candidate wins a seat.
Finn -- 4+1/3 -- First place
Eli -- 4+1/6
Finn wins a seat.
[Sequentially Spent Score: Winners — Sequentially Spent Score Voting Method (2 winners)]
Dana
Finn
Reweighted Range Voting → Dana, Eli
--- Reweighted Range Voting Method (2 winners) ---
[Reweighted Range Voting]
Tabulating 3 ballots to fill 2 seats.
Dana,Eli,Finn
4, 3, 1
4, 2, 2
1, 3, 4
[Reweighted Range Voting: Round 1: Score round]
The highest-scoring candidate wins a seat.
Dana -- 9 -- First place
Eli -- 8
Finn -- 7
Dana wins a seat.
[Reweighted Range Voting: Round 1: Reweighing Ballots]
Reweighted 3 ballots:
2 ballots reweighted from 1 to 5/9.
1 ballot reweighted from 1 to 5/6.
[Reweighted Range Voting: Round 2: Score round]
The highest-scoring candidate wins a seat.
Eli -- 5+5/18 -- First place
Finn -- 5
Eli wins a seat.
[Reweighted Range Voting: Winners — Reweighted Range Voting Method (2 winners)]
Dana
Eli
Related¶
- The companion election: Three neighbors — the same trick applied to the two quota methods
- The methods: Allocated Score · Sequentially Spent Score · Reweighted Range Voting
- At a bigger size: the Lackner & Skowron shadow election splits Allocated Score and RRV on a real academic profile