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German Electoral Politics$

Geoffrey K. Roberts

Print publication date: 2006

Print ISBN-13: 9780719069901

Published to Manchester Scholarship Online: July 2012

DOI: 10.7228/manchester/9780719069901.001.0001

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(p.137) Appendix 6 d'Hondt or Hare–Niemeyer? That is the question

(p.137) Appendix 6 d'Hondt or Hare–Niemeyer? That is the question

Source:
German Electoral Politics
Publisher:
Manchester University Press

For Bundestag and Land parliament elections, there is a requirement that seats be allocated proportionately among qualifying parties (i.e. parties which have secured at least 5 per cent of list votes or, in Bundestag elections, won three constituencies). However, vote shares will always mean that parties will have entitlements to fractional parts of seats. How can a fair method be devised to decide which parties obtain the last two seats when they have, say, vote shares of 44.2, 42.6, 7.7 and 5.5 per cent? In a parliament of 100 seats the first party will win 44 seats, the second party 42, the third party 7 and the fourth party 5. This leaves two seats still to be allocated.

The d’Hondt method, used in Bundestag elections until a change in the Electoral Law in 1985, lists the total votes for parties in order, then divides the total vote by successive integers (1, 2, 3 …). For this result the outcome would be:

Party

A

B

C

D

Vote

44.2

42.6

7.7

5.5

Divided by 2

22.1

21.3

3.55

2.55

Divided by 3

14.7

14.2

2.57

1.83, etc.

Taking the highest quotients, Party A wins the first seat, Party B the second, Party A the third (22.1), Party B the fourth (21.3), etc.

Which party wins the 99th and 100th seats ? The vote-share of Party A divided by 45 (i.e. one more than the 44 seats already allocated) gives a quotient of 0.9822. That of Party B divided by 43 gives a quotient of 0.9906. Party C has a quotient of 0.9625 (i.e. 7.7 divided by 8) and Party D a quotient of 0.9167 (5.5 divided by 6). So Party B wins the 99th seat, Party A the 100th seat.

(p.138) Using Hare–Niemeyer, the outcome is different. This method uses an equation:

Appendix 6 d'Hondt or Hare–Niemeyer? That is the question

So, using the same percentages given in the d’Hondt example, the outcomes would be as before: Party A 44 seats, Party B 42, Party C 7 and Party D 5, but the remaining two seats would be awarded to the parties with the highest fractional remainders: so Party C (0.7 remaining after seven seats awarded) and Party B (0.6 remaining) would obtain the 99th and 100th seats. Comparing the results:

A

B

C

D

d’Hondt

45

43

7

5

Hare-Niemeyer

44

43

8

5