The Marquis de Condorcet, a French mathematician and philosopher, described the voting cycle in an essay published in 1785. His colleague at the French Academy of Sciences, the naval officer and physicist Jean-Charles de Borda, had presented his point count to the Academy in 1770, and the Academy used it to elect its own members from 1784 until 1800, when Napoleon Bonaparte had it replaced. Charles Dodgson, who wrote Alice's Adventures in Wonderland as Lewis Carroll, taught mathematics at Christ Church, Oxford, and between 1873 and 1876 wrote three pamphlets on voting methods for the votes of his college's governing body, rediscovering the cycle apparently without knowing Condorcet's work. Kenneth Arrow's impossibility theorem was his PhD thesis, published as a book in 1951, and it earned him the Nobel Prize in economics in 1972.
How to pick a winner when nobody agrees
Your class is choosing a pet. Everybody ranks every option, so all the information is right there on the ballots. The trouble starts when you count. Three fair-sounding rules can pick three different winners from the same pile, majorities can chase each other in a circle, and a famous theorem says no rule can dodge every problem at once.
Your class of 30 has to choose a class pet: a hamster, a turtle, or a fish. To keep it fair, nobody just shouts out a favorite. Every student writes down a complete ranking: first choice, second, third. The pile of 30 ranked ballots is called a preference profile, or just a profile.
Now somebody has to count. That sounds like the easy part. Here are four counting rules, and every one of them sounds fair:
- Plurality (say: ploo-RAL-i-tee): the pet with the most first-place votes wins. Most elections you have heard of work this way.
- Instant runoff: if no pet has more than half of the first-place votes, cross out the pet with the fewest, move those ballots to their next choice, and count again. Repeat until some pet has a majority.
- Borda count (say: BOR-dah): with three pets, a first place is worth 2 points, a second place 1, a third place 0. Most points wins.
- Condorcet winner (say: kon-dor-SAY): compare the pets two at a time. If one pet beats every other pet in a head-to-head majority, it wins.
The board below is loaded with a real profile of 30 ballots. Look at the four results before you touch anything. Plurality says Hamster. Instant runoff says Fish. The Borda count says Turtle, and so does Condorcet. Nobody changed their mind and nobody cheated. One pile of ballots, three different winners.
The ballot builder
Every student ranks every pet. Change anything and all four counts update.Why? Each rule reads a different part of the ballot. Plurality reads only the top line, so Hamster wins with a loyal group of 11 even though 16 students put it dead last. The Borda count reads the whole ballot, so Turtle wins: 21 students put it second and nobody puts it last. Instant runoff reads the top line and throws out the pet with the fewest firsts. That is Turtle, with 9, so it is eliminated in round one even though it would beat Hamster 19 to 11 and Fish 20 to 10 head to head. The rule you pick decides which of these facts counts.
To earn the first star, build a three-way split of your own: change at least one number or ranking and keep all three rules disagreeing, or press Reset and build one from nothing. The status line under the groups tells you what the rules are saying.
The rock-paper-scissors of elections
Chapter 1 had Rock, Paper, Scissors: rock beats scissors, scissors beats paper, paper beats rock, and no throw is best. Elections can do the same thing. Split the class into three groups of 10:
- 10 students: Hamster, then Turtle, then Fish
- 10 students: Turtle, then Fish, then Hamster
- 10 students: Fish, then Hamster, then Turtle
Now hold the head-to-head contests. Hamster against Turtle: the first and third groups put Hamster higher, so Hamster wins 20 to 10. Turtle against Fish: the first and second groups put Turtle higher, so Turtle wins 20 to 10. Fish against Hamster: the second and third groups put Fish higher, so Fish wins 20 to 10.
A clear majority prefers Hamster to Turtle, a clear majority prefers Turtle to Fish, and a clear majority prefers Fish to Hamster. Whichever pet you choose, two thirds of the class would rather switch. This is Condorcet's paradox (say: PAIR-a-dox), the rock-paper-scissors of elections.
Numbers can never do this. If x > y and y > z, then x > z, always. That property is called transitivity (say: tran-zi-TIV-i-tee), and it is so automatic that you never think about it. Majorities are not numbers. "The majority prefers" sounds like one opinion held by one very large person, but a majority is a different set of people every time you ask a different question, and three different majorities can contradict each other without a single student being confused.
Load the cycle into the board and look at the head-to-head table: every pet wins one contest and loses one. Then break it. Move as few students as you can from one group to another until some pet beats both of the others. How few is enough? Find a reason before you check.
How few students?
Six. Every contest is 20 to 10. Moving one student changes a head-to-head score by at most 1 on each side, so a 10-vote gap shrinks by at most 2 per student moved. Five students can only reach 15 to 15, and a tie is not a win. Six can flip one contest: move six from the Fish group to the Hamster group and Hamster beats Fish 16 to 14 while still beating Turtle 20 to 10, so Hamster is the Condorcet winner. The board counts how many students you moved.
Voting is a game
So far the students have been honest. Now suppose they know the counting rule ahead of time and want the result they like best. The rule is plurality, and the profile is this:
- 12 students: Hamster, then Turtle, then Fish
- 10 students: Turtle, then Hamster, then Fish
- 8 students: Fish, then Turtle, then Hamster
Honest count: Hamster 12, Turtle 10, Fish 8. Hamster wins, and the 8 Fish fans get their last choice. But watch what happens if the Fish group votes Turtle first, even though every one of them likes Fish better. Turtle 18, Hamster 12. Turtle wins, which the Fish group prefers to Hamster. By lying about their favorite, they got a result they like more.
The Fish group's trick
Rule: plurality. Most first-place votes wins.Grown-ups call this strategic voting. It is why, under plurality, a candidate people genuinely like can end up with almost no votes: supporters abandon it so their ballot will "count" between the two front-runners. The Fish group did nothing wrong. They played the game the rule handed them, and a voting rule is a game: every voter picks a ballot, the rule turns the ballots into an outcome, and every voter cares about the outcome.
Could a cleverer rule stop this? No. Allan Gibbard in 1973 and Mark Satterthwaite in 1975 proved, separately, that with three or more candidates, every voting rule that is not a dictatorship (a rule that just copies one particular voter's ballot) can be gamed: there is always some profile where some voter does better by ranking dishonestly. The trick may be rare and hard to spot, but it exists.
Your turn. The rule below is the Borda count and the honest votes are already in: Turtle wins with 40 points. Change one group's ranking so that group ends up with a pet it likes more than Turtle.
Find the trick
Rule: Borda count. 2 points for a first place, 1 for a second, 0 for a third.The Borda count is a weighted sum. With n pets, a first place is worth n − 1 points, a second place n − 2, and so on down to 0 for last. With three pets:
In the first profile, Turtle has 9 firsts and 21 seconds, so Turtle scores 2 × 9 + 1 × 21 = 39. Check it against the board: Hamster 25, Fish 26, and 25 + 39 + 26 = 90 = 30 × 3, because every ballot hands out exactly 2 + 1 + 0 = 3 points. That is a useful check.
The pairwise count. To compare Hamster and Fish, walk through the groups and give each group's whole size to whichever of the two it ranks higher. Groups 1 and 4 put Hamster above Fish: 11 + 3 = 14. Groups 2 and 3 put Fish above Hamster: 10 + 6 = 16. Fish beats Hamster 16 to 14. With n pets there are n(n − 1)/2 contests: 3 for 3 pets, 6 for 4.
How many ballots are possible? With 3 pets there are 3 choices for first place, then 2 for second, then 1 left: 3 × 2 × 1 = 6 rankings. With 4 pets: 4 × 3 × 2 × 1 = 24. Mathematicians write n × (n − 1) × ... × 1 as n! and say "n factorial". 5! = 120.
Arrow's theorem. In 1951 Kenneth Arrow asked for a rule that turns everyone's rankings into one ranking for the whole class, and wrote down three things any reasonable rule should do:
- Unanimity. If every single student ranks Hamster above Turtle, the class ranking must put Hamster above Turtle.
- Independence. Whether Hamster is above Turtle in the class ranking depends only on how the students compare Hamster with Turtle, not on where Fish sits on anyone's ballot.
- No dictator. There is no student whose ballot simply becomes the class ranking no matter what everyone else says.
Arrow proved that with three or more candidates, no rule satisfies all three. This is Arrow's impossibility theorem. Which one does the Borda count break? The second. Suppose 18 students rank Hamster, Turtle, Fish and 12 rank Turtle, Fish, Hamster. Borda: Hamster 36, Turtle 42, Fish 12, so Turtle is on top. Now the fish tank breaks and Fish is off the ballot. Nobody's opinion about Hamster versus Turtle has changed, but the count is now Hamster 18, Turtle 12, and Hamster is on top. A pet that could not win changed who did. Plurality breaks the same condition; that is exactly what the Fish group's trick exploited. Head-to-head majority passes all three, but as the cycle shows, it sometimes produces no ranking at all, and Arrow's theorem asks for a ranking every time.
There is no such thing as "what the class wants" until you say how you are counting. The counting rule is part of the decision, and every rule is a game the voters can play. Choose the rule first, out loud, before anyone votes.
1. Twelve students vote. 5 rank Hamster, Turtle, Fish. 4 rank Turtle, Fish, Hamster. 3 rank Fish, Turtle, Hamster. Compute the Borda count. Who wins under plurality?
Answer
Hamster: 2 × 5 = 10. Turtle: 2 × 4 + 1 × (5 + 3) = 16. Fish: 2 × 3 + 1 × 4 = 10. Turtle wins the Borda count with 16. Check: 10 + 16 + 10 = 36 = 12 × 3. Plurality picks Hamster with 5 first-place votes.
2. Eighteen students vote. 7 rank Hamster, Fish, Turtle. 6 rank Turtle, Fish, Hamster. 5 rank Fish, Turtle, Hamster. Is there a Condorcet winner?
Answer
Hamster against Turtle: 7 to 11, Turtle wins. Hamster against Fish: 7 to 11, Fish wins. Turtle against Fish: 6 to 12, Fish wins. Fish beats both, so Fish is the Condorcet winner, even though Fish has the fewest first-place votes and would be eliminated first under instant runoff.
3. With four pets, how many different rankings are possible? How many of them put Hamster first?
Answer
4! = 4 × 3 × 2 × 1 = 24 rankings. If Hamster is fixed in first place, the other three pets can be arranged in 3! = 6 ways, so 6 of the 24 put Hamster first. With five pets it would be 5! = 120.
Add the fourth pet in the ballot builder and build a profile where the plurality winner is the Condorcet loser: the pet that loses every head-to-head contest. Prove it with the pairwise table. (The Condorcet card names a Condorcet loser whenever there is one.) Then say in one sentence what such a profile has to look like.
Answer
One profile that works: 10 students rank Gecko, Hamster, Turtle, Fish. 7 rank Hamster, Turtle, Fish, Gecko. 7 rank Turtle, Fish, Hamster, Gecko. 6 rank Fish, Hamster, Turtle, Gecko. Gecko wins plurality with 10 first-place votes, the most. But in every head-to-head contest Gecko gets exactly its own 10 fans against the other 20, so it loses 10 to 20 to Hamster, to Turtle, and to Fish. Gecko is the Condorcet loser. (Hamster is the Condorcet winner: it beats Turtle 23 to 7 and Fish 17 to 13.)
The shape: one pet is the favorite of the biggest single group and the last choice of everyone else. If that group is smaller than half the class, its pet loses every contest, and if the rest of the class is split among the other pets, it still wins plurality.
Stars in this chapter
Earn them by doing the clever thing, not by clicking around.
- Build a profile where plurality, instant runoff, and Borda pick three different winners
- Build a Condorcet cycle from scratch
- Find the strategic vote in the Borda election