How to Calculate the D’Hondt Method: Writing the Quotient Table, a Worked Seat Allocation, How Many More Votes Win a Seat, Sainte-Laguë and Adams Compared, and Japan’s Best-Loser Ratio

The short answer: divide votes by 1, 2, 3… and give seats to the largest quotients

The D’Hondt method is a way of allocating seats in proportional representation devised by the Belgian jurist Victor D’Hondt. In Japan it decides how many seats each party (list) wins in the proportional representation part of both the House of Representatives and the House of Councillors, and many other countries use it too. All it takes is division and sorting.

This article uses the following example: 6 parties sharing 10 seats. The votes add up to exactly 100,000, so you can read the vote shares directly.

Party A takes 40% of the seats with 34% of the votes, while parties E and F get nothing despite holding 10% of the votes between them. Let’s go through the quotient table to see why, and where the line was drawn.

How to write a D’Hondt table (the quotient table)

The quotient table lists parties across the top and divisors (1, 2, 3…) down the side, and each cell holds votes ÷ divisor. As for how far down to go, one row beyond the number of seats the largest party is likely to win is enough. For 10 seats in this example, going up to 5 is plenty.

1. 1

   Fill in the cells by dividing the votes by 1, 2, 3…

   The ÷1 row is just the votes. When a division does not come out even, one decimal place is enough for comparing the order.

2. 2

   Number the quotients ①, ②, ③… starting with the largest in the whole table

   The order is 34,000 (A÷1) → 25,000 (B÷1) → 18,000 (C÷1) → 17,000 (A÷2) → 13,000 (D÷1) → 12,500 (B÷2) → 11,333.3 (A÷3) → 9,000 (C÷2) → 8,500 (A÷4) → 8,333.3 (B÷3).

3. 3

   Everything up to the seat count (here ⑩) wins; count them per party

   Party A has ①④⑦⑨ for 4 seats, Party B has ②⑥⑩ for 3, Party C has ③⑧ for 2 and Party D has ⑤ for 1, making 10 seats in total.

4. 4

   Check the next largest quotient after ⑩ (the runner-up)

   The 11th is Party A’s 34,000 ÷ 5 = 6,800. If there were 11 seats, this one would go to Party A.

The 10th and last quotient, 8,333.3, works like the “price of one seat in votes”, and each party wins as many seats as it can pay that price for. Party E’s largest quotient is 6,000 and Party F’s is 4,000; neither reaches 8,333.3, so both get no seats.

A one-division check: divide by the “votes per seat” and round down

Pick any number d between the last winning quotient (8,333.3) and the runner-up quotient (6,800). Dividing each party’s votes by d and dropping the decimals always gives the D’Hondt result. This is another way of looking at the method, and it lets you check a result without writing out the table.

Any d above 6,800 and up to 8,333.3 works, so choosing a round number like 8,000 lets you do it in your head. Put the other way round, the D’Hondt method searches for a d at which the rounded-down values of votes ÷ d add up to exactly the number of seats. Because it rounds down, no party’s leftover votes short of a full seat ever turn into a seat. That is also why it favours large parties, as a later section explains.

• If the total comes out below the number of seats, d is too large. Divide again by a slightly smaller number
• If the total comes out above the number of seats, d is too small. Divide again by a slightly larger number
• If no d gives exactly the number of seats, the last seat is a tie that has to be decided by lot

How many more votes would have won one more seat

If the other parties’ votes stay the same, a party gains a seat when its next quotient (votes ÷ next divisor) beats the last winning quotient of the other parties. For Party E the next divisor is 1, so 8,334 votes, just over 8,333.3, would have done it: 2,334 more votes would have taken Party B’s third seat and given Party E one seat.

Party B’s own case is a little different. Its fourth quotient (votes ÷ 4) does not have to beat its own third-seat quotient of 8,333.3, but the lowest winning quotient of any other party, which is Party A’s fourth-seat quotient of 8,500. So Party B needs 34,001 votes, or 9,001 more. Because the quotient to beat differs from party to party, the trick when working it out by hand is to look at the smallest winning quotient among the other parties.

If two quotients tie for the last seat, it is decided by lot

When two parties have the same quotient for the last seat, the D’Hondt calculation alone cannot decide it. Under Japan’s Public Offices Election Act, the returning officer draws lots to decide which party gets the seat in such a case.

For example, if 3 parties with 100, 100 and 50 votes share 3 seats, the quotients are 100 (first party ÷1) and 100 (second party ÷1), followed by three quotients of 50: first party ÷2, second party ÷2 and third party ÷1. The third seat could go to any of the three, so it is decided by lot. With hundreds of thousands of votes this almost never happens in a real election, but it is surprisingly common when the numbers are small, such as when allocating committee seats in a council.

• Ties happen more often when the “votes” are small whole numbers, such as the number of members in each council group
• When using it for committee seats, agree in advance how ties will be settled (by discussion, by lot and so on)
• The calculator flags any tie and lists the party with more votes first

Why the D’Hondt method favours large parties

As the checking section showed, D’Hondt is the same as dividing votes by the “votes per seat” d and rounding down. On average each party loses about half a seat to rounding down, and the more parties there are, the more these losses add up. Those seats are handed out again across the board, which makes d smaller, and when d shrinks, parties with more votes gain more seats from the division. So seats drift towards the large parties.

The effect is stronger when there are fewer seats. With the same 6 parties and only the number of seats changed, the results are as follows.

Party E, with 6% of the votes, only wins its first D’Hondt seat once there are 15 seats. A first seat needs more votes than the last winning quotient, and the fewer the seats, the higher that number. Proportional representation for Japan’s House of Representatives is split into 11 regional blocks, some with as few as 6 to 28 seats, so small parties face a de facto threshold, a vote share they need just to win any seat.

Comparing Sainte-Laguë, Adams, Hare and other methods

Proportional seat allocation methods fall into two families: divisor methods, which divide votes by a fixed series of numbers, and largest remainder methods, which divide votes by a quota (the votes needed for one seat) and hand out the remaining seats in order of the largest leftovers. D’Hondt is a divisor method, and changing the series of divisors gives a different method.

The Gallagher index squares each party’s “vote share − seat share”, adds them up, halves the sum and takes the square root. The closer it is to 0, the more closely seats follow the votes. In this example Sainte-Laguë and Hare are the most proportional at 6.56. Because Sainte-Laguë’s divisor for a second seat is 3, large parties find it harder to win their second and third seats, and those seats go to smaller parties instead.

• Hare counts 10,000 votes as one seat, first hands out the whole parts (3, 2, 1, 1, 0, 0, which is 7 seats), and gives the remaining 3 seats to the largest leftovers: Party C (8,000), Party E (6,000) and Party B (5,000)
• Droop’s quota is a smaller 9,091, so the whole parts are again 3, 2, 1, 1, 0, 0, and the remaining 3 seats go to the largest leftovers: Party C (8,909), Party B (6,818) and Party A (6,727). The result matches D’Hondt
• Adams favours small parties (or less populous prefectures), and Japan’s House of Representatives uses it to decide how many single-member districts each prefecture gets and how many proportional seats each block gets
• Modified Sainte-Laguë raises the first divisor to 1.4 so that very small parties find it harder to win a first seat. It has been used in Nordic countries such as Norway

Elsewhere, countries such as Spain, Portugal and the Netherlands use D’Hondt, while New Zealand and the German Bundestag use Sainte-Laguë. Every method aims to share seats in proportion to votes; the difference lies in how the fractions are handled, which changes where the last few seats go.

How Japanese elections use the D’Hondt method

• In the House of Representatives, voters write a party name for the proportional vote, and each block allocates party seats by D’Hondt
• In the House of Councillors, voters write either a party name or a candidate’s name. Both are added together as the party’s votes, and seats are allocated by D’Hondt
• If a party has fewer candidates on its list than the seats it wins, the extra seats go to other parties. This has actually happened in the House of Representatives
• Local council elections in Japan are not proportional, but councils often use D’Hondt to share committee seats and posts among groups according to their size

The best-loser ratio and the order of list revival

In Japan’s House of Representatives, a candidate can stand in both a single-member district and on the proportional list (dual candidacy). Once D’Hondt fixes a party’s seats, winners come from the top of the list, but when dual candidates share the same list position, they win in order of their best-loser ratio (sekihairitsu). A candidate who loses the district but wins through the list is said to be revived through proportional representation.

Suppose a party wins 3 seats in a block and its list looks like this. Position 1 is a list-only candidate, and position 2 holds 5 dual candidates.

• A candidate who wins their district is removed from the proportional list, and those below move up
• A candidate whose district votes fall short of one tenth of the valid votes cannot win a list seat, whatever their position or ratio (Z falls short of 20,000, one tenth of 200,000)
• If two best-loser ratios are equal, the order is decided by lot
• The House of Councillors’ proportional vote ranks candidates by personal votes, so the best-loser ratio is not used

Since the best-loser ratio measures “what share of the winner’s votes the candidate got”, it is highest for those who lost close races. Note that in a district where the winner won with few votes, a candidate can have a high ratio even with a small number of votes.

Using it to share out committee seats and posts

D’Hondt is useful beyond elections, whenever seats are shared out according to the size of groups. In a 40-member council whose groups have 18, 12, 6 and 4 members, committee seats are shared out as follows.

With 5 or 7 committee seats, D’Hondt rarely gives a seat to the groups of 6 and 4, while Sainte-Laguë and Hare do. At 10 seats, a quarter of the council, all three methods agree. When the choice of method changes the result, agreeing in advance which one to use avoids arguments. The same idea works for sharing out officers in a parents’ association or a union, or club representatives by branch or year group.

Splitting a number into whole parts “in proportion” comes up outside elections too. For handling the fractions when splitting costs by ratio, see how to decide who pays more when splitting a bill, and for the margin of error in opinion polls, see how to calculate sample size.

Common mistakes when calculating D’Hondt

• Multiplying the vote share by the seats and rounding: 34% × 10 = 3.4 → 3 seats. Under D’Hondt Party A gets 4 seats, which differs from simple rounding
• Using 1, 3, 5… as divisors: that is Sainte-Laguë. D’Hondt divisors go up by 1: 1, 2, 3, 4…
• Skipping the ÷1 row: the votes themselves (÷1) are quotients too, and the first seat usually comes from this row
• Comparing quotients within one party: you compare all the quotients in the table, not the order within a party
• Not checking the quotient after the last seat (the runner-up): if it equals the last seat’s quotient, the seat goes to a lottery
• Rounding quotients before comparing them: dropping the fraction, as in 8,333.3 vs 8,333, can swap the order. Compare them with decimals

FAQ

Can you explain the D’Hondt method simply?

Divide each party’s votes by 1, 2, 3, 4…, sort all parties’ quotients from largest to smallest, and award one seat per quotient from the top until the seats run out. The number of winning quotients a party has is its number of seats. For 6 parties with 34,000, 25,000, 18,000, 13,000, 6,000 and 4,000 votes and 10 seats, the result is 4, 3, 2, 1, 0 and 0.

How many divisors do I need in a D’Hondt table?

Going one beyond the number of seats the largest party is likely to win is enough. If the top party looks set for 4 of 10 seats, writing up to ÷5 lets you check the runner-up as well. If unsure, writing as many divisors as there are seats is always safe.

Is there a quick way to check a D’Hondt result?

Divide each party’s votes by a number between the last winning quotient and the runner-up quotient, then drop the decimals. If the total equals the number of seats, the result is right. In the example, dividing by 8,000 gives 4.25, 3.125, 2.25, 1.625, 0.75 and 0.5, which round down to 4, 3, 2, 1, 0 and 0, or 10 seats.

Why does the D’Hondt method favour large parties?

D’Hondt is equivalent to dividing votes by the votes per seat and rounding down. When the seats lost to rounding down are redistributed, parties with more votes gain more of them. In the example, Party A with 34% of the votes gets 40% of the seats, whereas Sainte-Laguë, which divides by 1, 3, 5…, gives it 3 seats (30%).

What is the difference between Sainte-Laguë and D’Hondt?

The divisors. D’Hondt divides by 1, 2, 3, 4…, while Sainte-Laguë divides by 1, 3, 5, 7…. Under Sainte-Laguë, large parties find it harder to win a second seat and beyond, so more seats go to small parties and the gap between vote and seat shares is smaller. In the example, D’Hondt gives 4, 3, 2, 1, 0, 0 and Sainte-Laguë gives 3, 3, 2, 1, 1, 0.

How do I work out how many more votes a party needed for a seat?

Find the votes at which the party’s next quotient (votes ÷ next divisor) beats the last winning quotient of the other parties. In the example, Party E (6,000 votes, 0 seats) faced a last quotient of 8,333.3, so 8,334 votes, or 2,334 more, would have won it a seat.

What happens if quotients tie under D’Hondt?

If several parties tie on the quotient for the last seat, it is decided by lot; the party with more votes does not automatically win. For 3 parties with 100, 100 and 50 votes and 3 seats, all three have a quotient of 50 for the third seat, so it goes to a lottery.

How is the best-loser ratio calculated, and when can a candidate be revived through the list?

The best-loser ratio is the candidate’s votes in their district ÷ the district winner’s votes × 100. Dual candidates sharing a list position win in order of this ratio. However, a candidate who wins their district is removed from the list, and one whose votes fall short of one tenth of the valid votes cannot win a list seat.