The D'Hondt method/Jefferson method is a highest averages voting system (or apportionment method). While D'Hondt is usually described using successive division to calculate various quotients, Jefferson is usually described using a quota. It is known that the two systems produce the same results, but why?

First, an overview of how each is usually described:

Jefferson (quota) method

Imagine that there is an election with 230 voters, 4 parties (A, B, C and D) and 8 seats. The votes cast are as follows:

Party Votes
Party A 100
Party B 80
Party C 30
Party D 20

Under the Jefferson method [BaYo78], we determine a quota \( Q \) and divide the number of votes for each party by \(Q\). Discarding any remainder (i.e. rounding down), the quotient is the number of seats allocated to that party. The quota \(Q\) is chosen so that the number of seats allocated equals the correct number of seats.

For example, with quotas from 27 to 19, the table below shows the number of seats won by each party:

Quota 27 26 25 24 23 22 21 20 19
Party A 3 3 4 4 4 4 4 5 5
Party B 2 3 3 3 3 3 3 4 4
Party C 1 1 1 1 1 1 1 1 1
Party D 0 0 0 0 0 0 0 1 1
Total 6 7 8 8 8 8 8 11 11

As the table shows, any quota of more than 25 results in too few seats being allocated, whereas any quota of less than 21 results in too many seats being allocated. Any quota between 21 and 25, however, results in the correct number of seats being allocated.

Therefore, under the Jefferson method, the result is:

Party Votes Seats
Party A 100 4
Party B 80 3
Party C 30 1
Party D 20 0

D'Hondt (quotient) method

Imagine that the same election is being counted using D'Hondt. Under the D'Hondt method, we calculate a quotient for each party \( q = \frac{V}{s + 1} \), where \(V\) is the number of votes and \(s\) is the number of seats allocated to that party so far (initially 0).

We allocate the each seat successively to the party with the highest quotient at that point, updating the quotients throughout:

Seat 1 2 3 4 5 6 7 8
Party A s = 0
q = 100		 s = 1
q = 50		 s = 1
q = 50		 s = 2
q = 33.33		 s = 2
q = 33.33		 s = 3
q = 25		 s = 3
q = 25		 s = 3
q = 25
Party B s = 0
q = 80		 s = 0
q = 80		 s = 1
q = 40		 s = 1
q = 40		 s = 2
q = 26.67		 s = 2
q = 26.67		 s = 2
q = 26.67		 s = 3
q = 20
Party C s = 0
q = 30		 s = 0
q = 30		 s = 0
q = 30		 s = 0
q = 30		 s = 0
q = 30		 s = 0
q = 30		 s = 1
q = 15		 s = 1
q = 15
Party D s = 0
q = 20		 s = 0
q = 20		 s = 0
q = 20		 s = 0
q = 20		 s = 0
q = 20		 s = 0
q = 20		 s = 0
q = 20		 s = 0
q = 20
Highest \(q\) A B A B A C B A

This is often illustrated using a simple tabular approach. The table below shows the value of \(q\) for each party for each corresponding value of \(s\).

\(s\) 0 1 2 3 4
Denominator (\(s+1\)) 1 2 3 4 5
Party A 100 50 33.33 25 20
Party B 80 40 26.67 20 16
Party C 30 15 10 7.5 6
Party D 20 10 6.67 5 4

Seats are awarded for the 8 largest quotients in the table, shown in bold.

Therefore, under the D'Hondt method, the result is the same as in the Jefferson method.

Why are the results the same?

Note that in the Jefferson method, we divide votes by a quota to obtain a number of seats, whereas in D'Hondt, we divide votes by a seat-like number to obtain a quota-like number. Indeed, in D'Hondt, the smallest quotient to win a seat (25 in the example) effectively serves as the quota in the Jefferson method.

Note that the preceding table essentially demonstrates dividing each party's number of votes by 25. For example, for Party B, \(\frac{80}{1} = 80\), \(\frac{80}{2} = 40\) and \(\frac{80}{3} = 26.67\) are at least 25, but \(\frac{80}{4} = 20\) is less than 25. \(\frac{80}{25}\) must therefore be somewhere between 3 and 4, and following Jefferson's approach, rounds down to give 3 seats.

References

[BaYo78] Balinski ML, Young HP. The Jefferson method of apportionment. SIAM Review [Internet]. 1978 Apr [cited 2018 Aug 17];20(2):278-84. Available from: https://www.jstor.org/stable/2029901