Which Voting System Is Fair?

Most first choices

The first helper counts only the top line of each slip: Pokhara 12, Chitwan 10, Lumbini 8. Pokhara wins. This is plurality voting, also called first past the post: the candidate with the most first choices wins, even with fewer than half of them. It is how 165 of the 275 members of Nepal’s House of Representatives are elected, with a ballot that asks for only one choice.

But notice: 18 of the 30 students put Pokhara last. The winner is the last choice of a majority.

A runoff

The second helper says that a winner should have more than half. Nobody has 16 first choices, so she drops the place with the fewest, Lumbini, and moves each of its 8 slips to that slip’s next choice, which is Chitwan. Now Chitwan has 10 + 8 = 18 and Pokhara 12. Chitwan wins, with a majority. This is a runoff. Done in one count from ranked slips, as here, it is called instant-runoff voting or preferential voting; with more candidates the last one is dropped again and again until someone has a majority.

Points

The third helper says that second choices matter too. He gives 2 points for a first choice, 1 for a second, 0 for a third:

	1st choices × 2	2nd choices × 1	Points
Pokhara	12 × 2 = 24	0	24
Chitwan	10 × 2 = 20	8	28
Lumbini	8 × 2 = 16	12 + 10 = 22	38

Lumbini wins, by a long way. This is the Borda count, named after the French scientist who proposed it (his story is below). With more candidates, a ballot gives its last choice 0, the one above 1, the one above that 2, and so on.

Three counts, three winners, and nobody cheated. The votes didn’t decide the trip; the votes and the way of counting did.

The fourth way in the widget asks a different question. Forget counting whole slips: take the places two at a time, as if the class were choosing between only those two, and let each slip vote for whichever of the pair it ranks higher.

• Lumbini against Pokhara: the 10 Chitwan fans and the 8 Lumbini fans rank Lumbini higher. Lumbini 18, Pokhara 12.
• Lumbini against Chitwan: the 12 Pokhara fans and the 8 Lumbini fans rank Lumbini higher. Lumbini 20, Chitwan 10.
• Chitwan against Pokhara: Chitwan 18, Pokhara 12.

A candidate that beats every other one head to head is called the Condorcet winner, after the French mathematician who argued for it in 1785. Here it is Lumbini. And a candidate that loses to every other one, like Pokhara, is the Condorcet loser. So plurality voting chose the one place that a majority of the class would turn down in favor of either of the others.

Lumbini was squeezed out of the runoff because it had the fewest first choices, although it was everybody’s first or second choice. This is called the center squeeze: a candidate in the middle, liked well enough by both sides, is dropped first because neither side puts it first.

Points and pairs are related

Look at Lumbini’s 38 points again: 18 + 20 = 38, its votes in its two contests. Chitwan’s 28 is 18 + 10, and Pokhara’s 24 is 12 + 12. That is no accident. Take any one slip, say Pokhara › Lumbini › Chitwan. Pokhara is above both others on it, so it wins this slip’s vote in both of its contests: 2 votes, and it got 2 points. Lumbini is above one other: 1 contest vote and 1 point. Chitwan is above nobody: 0 and 0. On every slip, a place’s points are the number of places it is ranked above. So with three candidates, a place’s Borda points are its total votes in its two head-to-head contests.

This explains something about the Borda count: it can never elect a Condorcet loser. A Condorcet loser gets fewer than half the votes in each contest, fewer than 15 + 15 = 30 points in all, which is less than the average of 30 (90 points shared by three). The winner of the points count has at least the average, so it can’t be the Condorcet loser. Plurality voting and the runoff have no such protection.

Head to head sounds like the fairest way of all: a winner who beats everyone one to one is hard to argue with. But there isn’t always one. Suppose the next class’s 30 slips are these:

Students	1st	2nd	3rd
10	Pokhara	Chitwan	Lumbini
10	Chitwan	Lumbini	Pokhara
10	Lumbini	Pokhara	Chitwan

Pokhara beats Chitwan 20 to 10 (the first and third groups). Chitwan beats Lumbini 20 to 10. And Lumbini beats Pokhara 20 to 10. Each student has a perfectly sensible order, yet the class as a whole prefers Pokhara to Chitwan, Chitwan to Lumbini, and Lumbini to Pokhara, like rock, paper and scissors. This is Condorcet’s paradox, and a circle of preferences like it is called a cycle. Whatever is chosen, a majority would rather have something else.

What should a counting method do with these slips? Notice something about them: if you rename the places, Pokhara as Chitwan, Chitwan as Lumbini and Lumbini as Pokhara, the three piles just swap places and the slips are the same as before. Now suppose a method treats every voter alike (only how many slips of each kind matter, not whose they are) and every place alike (renaming the places renames the winner in the same way). If it chose Pokhara here, then renaming would make it choose Chitwan from the very same slips. It can’t choose both. So:

No method that treats every voter alike and every place alike can always choose one winner. With these slips it can only call a tie (and then perhaps toss a coin), or break one of its own rules: let one person’s slip decide, or always favor one place.

How often does a cycle happen? If every voter picked one of the six possible orders at random (each equally likely, each voter on their own), the chance of a cycle would be 12 in 216 (5.6%) with 3 voters, and about 8.8% with a very large number. Real voters aren’t random. A review of 37 studies of 265 real elections found a cycle in 25 of them, and in one collection of 189 ranked elections held in the United States from 2004 to 2022, only one had a cycle. So cycles are rare in big public elections, but they are a real possibility, and in a committee of a few people they happen.

Here is the same counter with every kind of slip, six in all, and the four winners shown together. Change how many students hand in each kind.

Perhaps we should first agree on what a fair count must do, then choose a method that does it. Four rules that sound reasonable:

1. Majority: if more than half the voters put one candidate first, that candidate wins.
2. Condorcet: if one candidate beats every other head to head, that candidate wins.
3. No spoilers: whether A or B wins shouldn’t depend on whether C, who doesn’t win, is on the ballot. (Mathematicians call this independence of irrelevant alternatives.)
4. More support never hurts: if some voters move the winner up their slips and change nothing else, the winner still wins. (Monotonicity.)

Spoilers

Plurality voting breaks rule 3. Suppose Chitwan had been ruled out before the vote (say the hotels there were full). The Chitwan fans would vote for their next choice, Lumbini, and Lumbini would beat Pokhara 18 to 12. With Chitwan on the slip, Pokhara wins. Chitwan, which loses either way, decides which of the other two wins: it is a spoiler.

The Borda count breaks rules 1, 2 and 3. Take a small club of five: 3 vote Pokhara › Chitwan › Lumbini and 2 vote Chitwan › Lumbini › Pokhara. Pokhara has 3 of the 5 first choices, a majority. But the points are Pokhara 3 × 2 = 6, Chitwan 3 + 2 × 2 = 7, Lumbini 2: Chitwan wins on points. Now take Lumbini, which nobody puts first, off the slips: Pokhara wins 3 to 2. Lumbini, last by far, was a spoiler.

More support that loses

Instant-runoff voting obeys rule 1, but the class trip showed it breaking rule 2, and it breaks rule 4 too, in the strangest example of this topic. Twenty-one friends rank the same three places:

Friends	1st	2nd	3rd
6	Pokhara	Chitwan	Lumbini
8	Chitwan	Lumbini	Pokhara
7	Lumbini	Pokhara	Chitwan

Round 1: Pokhara 6, Chitwan 8, Lumbini 7. Pokhara is out, and its 6 slips go to Chitwan: Chitwan 14, Lumbini 7. Chitwan wins.

Now two of the Lumbini fans, hearing how good Chitwan’s jungle walks are, move Chitwan from last to first: Chitwan › Lumbini › Pokhara. They have only helped Chitwan. Count again. Round 1: Pokhara 6, Chitwan 10, Lumbini 5. This time Lumbini is out, and its 5 remaining slips go to their next choice, Pokhara: Pokhara 11, Chitwan 10. Pokhara wins.

Arrow’s theorem

So every method so far breaks some rule. Is there a cleverer one that breaks none? In 1950 the American economist Kenneth Arrow proved that there isn’t. Take any method that turns everybody’s ranked slips into a ranking for the group, works for every possible set of slips, and has:

• Unanimity: if every voter ranks A above B, so does the group.
• No spoilers: the group’s order of A and B depends only on how each voter ranks A and B.

Then, if there are three or more candidates, the method is a dictatorship: there is one voter whose slip always decides the group’s ranking, whatever everyone else writes. This is Arrow’s impossibility theorem. Since nobody wants a dictator, and unanimity is the mildest rule there is, every sensible ranked method must allow spoilers somewhere. The rock, paper, scissors slips show the trouble in miniature. The full proof takes a few pages of careful reasoning; the economist John Geanakoplos has written three short versions of it, free to read.

The theorem covers only methods that use rankings. Giving each candidate a grade, or ticking every candidate you approve of (approval voting), escapes it, though an approval voter must decide where to draw the line between yes and no.

With two candidates, there is a perfect answer

All this trouble needs at least three candidates. With only two there is no spoiler and no cycle, and in 1952 the mathematician Kenneth May proved that majority voting is the only method that treats every voter alike, treats both candidates alike, and responds to support (if A was winning or tied and a voter switches from B to A, A now wins). The idea: treating voters alike means only the two counts matter, and treating candidates alike means that swapping every vote swaps the result, so equal counts must give a tie. Now suppose A could win with fewer votes, say 4 to B’s 6. Swapping every vote, B would win when A has 6 and B has 4. But responding to support says that if 2 of B’s voters switch to A, A still wins, and now A has 6 and B has 4: the same counts, two different winners. So fewer votes never win, and (by the same step from a tie) more votes always do.

• Nepal: the House of Representatives has 275 members. 165 are elected in single-member constituencies by first past the post, the plurality count, and 110 by proportional representation, with the whole country as one constituency: voters choose a party, and the seats are shared among parties in proportion to their votes. As of the election of 5 March 2026, a party needs at least 3% of the valid votes to get any of those seats, and the Election Commission shares them by a modified form of the Sainte-Laguë method (Question c shows the plain method).
• Australia elects its House of Representatives by preferential voting, the instant runoff, first used at a by-election in December 1918. The change came after a by-election in Swan the same year, where the Labor candidate won with 34.4% of the votes because two candidates of the other side split theirs (the first question below). Ireland elects its president the same way.
• The center squeeze, for real: in Alaska’s special election for the US House of Representatives on 16 August 2022, held by instant runoff, Nick Begich had the fewest first choices (27.8%) and was dropped first, and Mary Peltola beat Sarah Palin in the final round, 51.5% to 48.5%. Yet a majority of voters ranked Begich above each of his two rivals: he was the Condorcet winner, squeezed out like Lumbini.
• Points counts: Slovenia elects two of its 90 members of parliament, those for its Italian and Hungarian minorities, by the Borda count; Nauru uses a variant with fractions; the Eurovision Song Contest gives each jury’s favorites 12, 10, 8 down to 1 point; and sports awards such as baseball’s Most Valuable Player are decided by points from ranked ballots.
• Approval voting picks the two finalists for mayor and other city offices in St. Louis, in the United States, who then face each other in a plain two-way vote: first used in 2021.

• How Do You Cut a Cake Fairly for Three? (Grades 6–8) Fairness made precise for sharing instead of choosing.
• How Do You Share an Inheritance Fairly? Dividing things that can’t be cut, among people who value them differently.
• Should You Cooperate or Cheat? Game theory: when each person’s sensible choice leaves the group worse off.
• Can 1,000 People Speak for 3 Crore? How an opinion poll, which asks before the vote, can be right within a few percent.
• When Will Nepal’s Population Stop Growing? Another model of the real world: which formula fits Nepal’s censuses, and what each predicts.
• Grade 10 Math: Sets The unions and intersections behind approval voting’s ticks.