Olya, Marko and Iryna share a flat and are choosing what to order for dinner: pizza, sushi or dumplings. They decide fairly, by majority vote, in pairs: two dishes face off, and the winner goes to a final against the third. You chair the meeting and decide just one thing: which two dishes go first. How many of the three dishes can you make the winner just by choosing the order of the votes?
All three. A majority prefers pizza to sushi, sushi to dumplings, and dumplings to pizza. It is a loop, like rock, paper, scissors: every dish loses to some other one. Whichever pair goes first, the winner of round one is beaten by the dish that waited. So the chair picks the winner by deciding which dish to hold back for the final.
The Marquis de Condorcet described this problem in 1785, while looking for a way assemblies could reach fair decisions. Below it is in its smallest form: three people and three options.
Each of them has ranked the three dishes from most to least wanted. The lists are on the right, and they don't change during the evening.
They vote in pairs. Two dishes face off, and each person votes for the one higher on their list. The dish with at least two of the three votes wins. It then meets the third dish in a final, and the winner of the final is what they order.
You don't vote and you don't persuade anyone. You only announce which two dishes go first and which one waits for the final.
How many of the three dishes can you make the winner by choosing only the order of the votes?
Most wanted at the top, least wanted at the bottom.
"They vote honestly, by majority. The order can't change anything"
Majority is the simplest and fairest rule there is. But does it always know what it wants?
Pick an answer. It counts as soon as you click.
You can change your mind, but the first choice goes into the stats.
A breakdown of your answer will appear here.
Hard to decide? Open the hints one at a time: the easy one first, then stronger ones.
First forget the order and hold all three head-to-head votes: pizza against sushi, sushi against dumplings, dumplings against pizza. Who wins each one?
Is there a dish that beats both others head to head? If so, no order can hurt it. And if there isn't?
Pick a dish and try to make it win. Which dish should go out in round one so that yours is stronger in the final? When should your dish face the vote?
The next rounds show how the order of votes picks the winner: a vote you can run yourself, your own groups of friends, thousands of random elections and the explanation. Answer round 1 first, or go on?
You chose to always show all rounds.
Choose which dish waits for the final and watch who votes for what. Try to make each of the three the winner. Further down, put together your own group of 3, 5 or 7 people and check whether it has a loop.
The number of people is odd, so a pair can never tie. Anyone's list can be changed.
Our group was picked on purpose. How often does such a loop happen by itself? The computer gives each voter a random list, all six orders equally likely, and counts the share of elections with no dish that beats every other one head to head. That is exactly when the order of votes can pick the winner.
Once you answer, the simulation code appears here so you can run it yourself.
The same simulation, short and with no libraries. Run the JavaScript in your browser console (F12) or in Node; the Python needs only the standard library.
// Each voter ranks m options at random and votes honestly. How often does no option beat every other one head to head?
function noWinner(m, n, trials) {
let cycles = 0;
for (let t = 0; t < trials; t++) {
const wins = Array.from({ length: m }, () => Array(m).fill(0)); // wins[i][j]: how many people put i above j
for (let v = 0; v < n; v++) {
const order = [...Array(m).keys()];
for (let i = m - 1; i > 0; i--) { const j = Math.floor(Math.random() * (i + 1)); [order[i], order[j]] = [order[j], order[i]]; }
for (let a = 0; a < m; a++) for (let b = a + 1; b < m; b++) wins[order[a]][order[b]]++;
}
// a Condorcet winner has a majority against every rival
const hasWinner = wins.some((row, i) => row.every((w, j) => j === i || 2 * w > n));
if (!hasWinner) cycles++;
}
return cycles / trials;
}
const exact3 = { 3: 1 / 18, 5: 5 / 72, 7: 875 / 11664, 101: 0.086901 };
for (const n of [3, 5, 7, 101]) {
console.log('options: 3, voters:', n, '→ no winner in', (100 * noWinner(3, n, 20000)).toFixed(2) + '%', 'exact', (100 * exact3[n]).toFixed(2) + '%');
}
console.log('Limit for very many voters (Guilbaud):', (100 * (1 / 4 - 3 / (2 * Math.PI) * Math.asin(1 / 3))).toFixed(2) + '%');
for (const m of [4, 5, 10]) {
console.log('options:', m + ',', 'voters: 101 → no winner in', (100 * noWinner(m, 101, 4000)).toFixed(1) + '%');
}
# Each voter ranks m options at random and votes honestly. How often does no option beat every other one head to head?
import math, random
def no_winner(m, n, trials):
cycles = 0
for _ in range(trials):
wins = [[0] * m for _ in range(m)] # wins[i][j]: how many people put i above j
for _ in range(n):
order = random.sample(range(m), m)
for a in range(m):
for b in range(a + 1, m):
wins[order[a]][order[b]] += 1
# a Condorcet winner has a majority against every rival
has_winner = any(all(j == i or 2 * wins[i][j] > n for j in range(m)) for i in range(m))
if not has_winner:
cycles += 1
return cycles / trials
exact3 = {3: 1 / 18, 5: 5 / 72, 7: 875 / 11664, 101: 0.086901}
for n in [3, 5, 7, 101]:
print("options: 3, voters:", n, "→ no winner in", f"{100 * no_winner(3, n, 20000):.2f}%", "exact", f"{100 * exact3[n]:.2f}%")
print("Limit for very many voters (Guilbaud):", f"{100 * (1 / 4 - 3 / (2 * math.pi) * math.asin(1 / 3)):.2f}%")
for m in [4, 5, 10]:
print("options:", f"{m},", "voters: 101 → no winner in", f"{100 * no_winner(m, 101, 4000):.1f}%")
Each of the three friends knows what they want: their list is consistent. But "the majority" is a different pair of people each time, and together their decisions close into a loop.
“In plain words” is hidden on all puzzles.
Each arrow points from winner to loser. Olya and Iryna put pizza above sushi, Olya and Marko put sushi above dumplings, Marko and Iryna put dumplings above pizza. Each of the three majorities is fair, but they are three different pairs of people, and together they make a loop.
This card explains the core but not everything: why it is the chair who picks the winner is the next card.
Say dumplings wait and pizza and sushi go first. Pizza beats sushi. But in the loop, pizza is beaten by exactly the dish that waited: dumplings. It works with any pair: the winner of round one is always the dish the third one beats.
| Waits for the final | Round one | Winner |
|---|
So the chair doesn't guess or persuade. They simply choose whom to hold back until the end.
If Olya likes pizza more than sushi and sushi more than dumplings, she likes pizza more than dumplings too. That is how each person works. It seems "the will of the majority" should behave the same way. But a majority isn't a person: in each pair it is made up of different people. Condorcet noticed that even when every voter is consistent, the collective result can be inconsistent.
It can also go differently. If one dish beats every other one head to head, it is called the Condorcet winner. Then the order doesn't matter: it wins every vote, whether it goes first or last. The chair gains power only where there is no such winner.
“Rigorous” is hidden on all puzzles.
A list of three dishes can be written in 3 · 2 · 1 = 6 ways. Three people give 6 · 6 · 6 = 216 different groups. A loop happens only when each dish is first for someone, second for someone and third for someone, and the lists go "round the circle" as our friends' do.
The loop can run two ways: pizza → sushi → dumplings → pizza, or the reverse. Each has three lists, which can be handed to three people in 3 · 2 · 1 = 6 ways. That makes 2 · 6 = 12 groups with a loop, so 12/216 = 1/18 ≈ 5.6%. The other 204 have a Condorcet winner.
| people | groups | with a loop | share |
|---|---|---|---|
| 3 | 216 | 12 | 5.56% |
| 5 | 7776 | 540 | 6.94% |
| 7 | 279 936 | 21 000 | 7.50% |
| 9 | 10 077 696 | 785 820 | 7.80% |
| 101 | 6101 | – | 8.69% |
| endless | – | – | 8.77% |
Every row is counted exactly: all groups are enumerated, with each of the six lists equally likely. The share grows slowly and levels off at 8.77%. So even in large elections with three options, about one in eleven has no Condorcet winner.
The loop: x beats y, y beats z, z beats x. Round one between x and y goes to x, and x loses to z. The winner of the agenda is always the one who waited.
Three voters, three options: two directions of the loop, six ways to hand out three lists.
The limit for endlessly many voters (Guilbaud, 1952). Each of the three pairs is nearly a normal variable, and two pairs sharing a dish are correlated with coefficient −1/3.
“Deeper” is hidden on all puzzles.
Maybe the pairwise scheme is to blame, and another rule would do? Kenneth Arrow proved in 1951 that it wouldn't. Let a rule turn voters' lists into a joint list and meet three demands: if everyone puts x above y, so does the group; the order of x and y in the joint list depends only on how people rank x and y; and there is no dictator whose list always becomes the joint one. With three or more options no such rule exists.
Condorcet's paradox is the smallest example of what breaks: pairwise majority meets all three demands, but doesn't always produce a list, because it goes round in a circle.
Duncan Black found an important exception in 1948. Suppose the options can be placed on a single axis, say a tax from low to high, and each voter has a favourite point on it, liking options less the further they are from it. Then there is no loop, and the Condorcet winner is the median voter's favourite. The ice cream sellers on the beach rests on the same idea.
Our dishes don't fit on one axis: each of them is last for someone. Whichever dish you put in the middle, someone ranks it last, so they like both ends more than the middle.
As early as the 13th century the philosopher Ramon Llull described elections for monasteries by comparing candidates in pairs. In 1785 the Marquis de Condorcet, in his "Essay on the Application of Analysis to the Probability of Majority Decisions", proposed choosing whoever beats every rival head to head, and himself showed that there may be no such candidate. Charles Dodgson, better known as Lewis Carroll, analysed such loops in an 1876 pamphlet on voting for an Oxford college and proposed electing no one if they did not go away. His pamphlets were forgotten for almost a century until Duncan Black reprinted them. And Kenneth Arrow turned the observation into a theorem and in 1972 shared the Nobel Prize in Economics with John Hicks.
Each new option makes a loop more likely. Other counting rules don't save the day. And the order of votes in a parliament has already buried a bill that had a majority.
With more options the chair puts them in a queue: the first two face off, the winner meets the next one, and so on to the end. Tap an option in the diagram and the page will find a queue that makes it win, or tell you there is none.
The options that can be made to win are exactly those in the "top cycle": the smallest group whose every member beats everyone outside it. If there is a Condorcet winner, the group is just that one option. In a simulation with 1001 voters and random lists, there is no Condorcet winner in about 17.5% of elections with 4 options, 25% with 5, 31% with 6 and 49% with 10.
When the options form a continuum in several dimensions, say a tax and a budget at once, and there is no Condorcet winner, Richard McKelvey proved more in 1976: with a chain of honest majority votes the chair can lead an assembly from any option to any other.
We are used to "above" working in a chain: if x is above y and y above z, then x is above z. For height, weight or one person's tastes, it is. But "a majority prefers" is made of different people in different pairs, and the chain breaks. In the same way "beats" breaks in rock, paper, scissors, in intransitive dice and in Penney's game, where every coin triple has a stronger rival.
So a group has no "true" winner that a rule merely needs to find. The three fair rules on the left give three different trips from the same ballots, and head-to-head duels give none: Odesa beats the Carpathians, Lviv beats Odesa, and the Carpathians beat Lviv. Choosing the rule is already choosing the result.
In the US House of Representatives amendments are voted before the bill itself: first the bill with the amendment against the bill without it, then the winner against "pass nothing". The order is set by the rules, not a chair, but it works just as in our puzzle.
In 1956 the House was debating a bill on federal money for building schools. Congressman Adam Clayton Powell proposed an amendment: no money for states that fail to comply with the Supreme Court's school desegregation rulings. The amendment passed 225 to 192, and then the whole bill failed 224 to 194. You can check this in the roll-call records of the 84th Congress.
The political scientist William Riker explained it in 1958 with a loop of three blocs, as in the diagram on the left: each bloc is less than half the House, while any two together have a majority. His explanation is still debated: some Republicans may have backed the amendment on purpose to sink the bill, and Gerry Mackie argues in "Democracy Defended" (2003) that there was no true loop. But the trap itself is real: an amendment that makes a bill unacceptable to part of its supporters is called a "killer amendment".
Wherever a decision is built from pairwise comparisons and there may be no Condorcet winner.
| Where | What is compared in pairs | What the order decides |
|---|---|---|
| Committees and parliaments | amendments, the bill and "nothing" | the order of amendments and when the bill itself is put |
| Knockout tournaments | teams in matches | the bracket: if teams beat each other in a loop, whoever draws the bracket can shape the champion |
| Ranked-ballot elections | candidates on the ballots | in the 2022 Oakland District 4 school director race there was no Condorcet winner, and the result depended on the counting rule |
| Seat apportionment | states or parties | another paradox of a fair rule: the Alabama paradox, where a bigger house takes a seat away from a state |
Other puzzles where sensible individual choices add up to a surprising joint result: Braess's paradox, where a new road slows everyone down, and guess 2/3 of the average, where everyone has to predict the others.
Nicolas de Condorcet, a mathematician and secretary of the Paris Academy of Sciences, published his "Essay" on the probability of majority decisions four years before the French Revolution. Both the paradox and the "Condorcet winner" are named after it.
Charles Dodgson, the author of "Alice in Wonderland", printed a pamphlet, "A Method of Taking Votes on More than Two Issues", for the meetings of his college, Christ Church, Oxford. If the majorities went round in a loop, he advised electing no one.
Kenneth Arrow set out his impossibility theorem in the book "Social Choice and Individual Values", which grew out of his doctoral thesis. In 1972 he became the youngest winner of the Nobel Prize in Economics up to that time.
Georges-Théodule Guilbaud worked out in 1952 how often a very large election with three options and random tastes has no Condorcet winner: 1/4 − (3/2π)·arcsin(1/3), a little under one election in eleven.
On 5 July 1956 the US House first adopted the Powell amendment to the school construction bill (225 to 192), then rejected the bill itself: only 194 voted for it, 224 against.
In the school director race in Oakland's District 4, California, there was no Condorcet winner, as the mathematician David McCune showed. That election was also first counted with an error, and the winner was finally settled in court in 2023.
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