What makes the Swedish Election Act fail?

Jan LankeDept of Statistics, Lund University

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The Swedish parliamentary election process:an overview

Three time periods:

P1. 1911--1948; d’HondtP2. 1952--1968; modified Sainte-LaguëP3. 1970--2010; modified Sainte-Laguë, adjustment seats

I shall concentrate on period 3; however, a few comments on the first two periods will be given later on.

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The 1970 Election ActStep 1 of 6

S1. 310 seats allocated to constituencies method: Hamilton (= greatest remainders);

criterion: number of registered voters

Crucial point: number as of what date? Answer: as of Nov 1, the year before the election That causes quite a bit of trouble, which I shall avoid by ignoring it

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The 1970 Election ActStep 3 of 6

S3. Seats to parties within constituencies method: modified Sainte-Laguë

modification: divisors 1.4, 3, 5, ... instead of 1, 3, 5

Why modified?

I shall return to that.

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The 1970 Election ActStep 4 of 6

S4. Adjustment seats to partiesmethod: modified Sainte-Laguë

Why modified?

Completely pointless, since a party that is permitted to take part here has at least 4% of the votes nationwide, and even if [very unlikely!] it has got no seat among the 310, it will get more than one among the 349 and so is not influenced by the modification.But on the other hand: the modification causes no harm, either.

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The 1970 Election ActStep 5 of 6

S5. Adjustment seats to constituencies within partiesmethod: (before 1991) modified Sainte-Laguë

Why modified?That gives small constituencies a disadvantage!

My interpretation: complete black-out on the part of those who wrote the Election Act.(Other persons have a less charitable interpretation.)

current method: modified Sainte-Laguë; however, if the modification is about to come into action, it is not to be implemented [!].

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The 1970 Election ActStep 6 of 6

S6. Seats to persons within parties and constituenciesmethod: essentially d’Hondt

procedure: fairly complicateddescription of procedure in ValL: inordinately complicated (ValL = Election Act)

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The 1970 Election ActWhat can go wrong?

S3. 310 seats to parties in 29 constituencies S3.5. For each party, find the total number of seats S4. 349 seats to parties with Sweden as one constituency Give each party a number of adjustment seats equal to the difference between S4 and S3.5

Complication: What if that difference is negative?

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The 1970 Election Act What if it goes wrong?

What if a party is to be given a negative number of adjustment seats?

The party keeps all the seats that it has got among the 310 [what else could be done?], and some other party gets fewer adjustment seats than it should have.

That is what happened in 1988 (one seat misplaced), and in 2010 (four seats misplaced). In none of these cases the majority was changed by this, but in 2010 it was close.

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The future, 2014 onwardsWhat can be done?

Reasonable question: What can be made to decrease the risk of such outcomes? The reason for the mishap clearly is that the 310 seats are distributed among the parties in a way that does not properly reflect the distribution of the votes.

And why is that so?

My explanation: the modification factor 1.4 is the culprit.

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Why is Sainte-Laguë modified?On the history of 1.4

1911-1948: d’Hondt, cartels permitted

during the later part of that period(h)+(fp)+(bf) [non-socialist] in cartel(s) and (k) [socialist] not in cartel

(during the former part: other constellations)

in 1951: (s) and (bf) form a coalition government

with the 1952 election approaching, a cartel between the opposition and one party in office was considered politically impossible.

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On the history of 1.4, cont’d

Something had to be done, quickly.

Note: in Sweden ValL is an ordinary act, not part of the constitution; it can be changed by a simple decision in the parliament.

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On the history of 1.4, cont’d

The task was to find a procedure that1. does not encourage cartels2. gives as far as possible the same result as d’Hondt with

(h)+(fp)+(bf) in cartel

Another way of formulating 2 is

2’. gives (k) the same disadvantage as d’Hondt with (k) outside cartel

Note: among the five parties, (k) was the smallest

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On the history of 1.4, cont’d

Many suggestions were discussed.

Final choice: Modified Sainte-Laguë, first divisor 1.4

In passing: the proponent of that method was Sten Wahlund, (bf) politician and also professor of Statistics.

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On the history of 1.4, cont’d

This Election Act, which was taken as a provisional act, was in force in 1952. The election resulted in a second chamber with

(h)+(fp)+(bf) = 115, (s) = 110, (k) = 5i.e. neither socialists nor non-socialists had a majority.However, the first chamber had a socialist majority,and (s)+(bf) remained in office.

Ironically: if d’Hondt with cartels had been used, the outcome would, ceteris paribus, have been a non-socialist majority.

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On the history of 1.4, cont’dTo sum up:

the factor 1.4 was in 1952 introduced to give small parties some disadvantage

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A new Election Act in 1970

The provisional act of 1952 in fact remained in force until 1970 when a one-chamber parliament was elected by means of a totally new Election Act, one particular aspect of which I shall now comment on.

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Why did things go wrong in 1988 and 2010?

The distribution of the 310 seats did not match the distribution of the 349 seats.

The 349 seats were in principle distributed by means of ordinary, i.e. unmodified, SL.

Thus the distribution, within constituencies, was made in a slightly indequate way: the distorsion from the distribution of votes was too large to be corrected by the 39 adjustment seats.

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What can be done?

Two possibilities:1. increase the number 392. change the factor 1.4

or a combination of these two actions.

Starting with the 2010 election I have studied the question:for values 1.00(0.05)1.50 of the modification factor,which is the smallest number of adjustment seats that would have worked?

More precisely: ... smallest a such that all nA ≥ a would have ... [remember Alabama!]

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Result for 2010

f W nA 1.00 52 1.05 33 1.10 28 1.15 22 1.20 29 1.25 29 1.30 31 1.35 38 1.40 58 1.45 58 1.50 63

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Results for 1970-2010

I have performed the same computations for all the 13 elections that we have had with the 1970 Election Act.

As a way of summarizing the results I have, for each of the studied values of the modification factor, checked which was the largest of the required numbers of adjustment seats in these 13 elections.

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Results for 1970-2010

f W nA year 1.00 52 2010 1.05 46 1994 1.10 41 2006 1.15 33 1998 1.20 33 1998 1.25 51 1988 1.30 51 1988 1.35 51 1988 1.40 58 2010 1.45 58 2010 1.50 63 2010

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Results for 1970-2010

020

4060

1970 1980 1990 2000 2010year

1.00 1.201.40

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Why do we have 1.4 instead of 1.0?

The purpose of choosing 1.4 was not, as in 1952, to be unkind to small parties;rather, it was to see to it that a moderate number, say 39, of adjustment seats would suffice.

But, as we have seen, that failed.

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Why do we have 1.4 instead of 1.0?

The choice 1.4 is a bit hard to understand, since extensive simulations by Fröberg & Sundström rather gave the impression that a lower value would be preferable.

Is the reason simply that 1.4 was a number well known to politicians?

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What if we had had 1.2 instead of 1.4?

In 1988 and 2010 the distribution of seats among the parties would have differed from what 1.4 gave

butin the other 11 elections no such differences would have occurred.

However,the distribution of seats among the constituencies would in some cases have changed,

but not very much.

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Number of seats moved between consistuencies if 1.4 were replaced by 1.2

1970 21973 01976 01979 11982 11985 01988 4

1991 11994 81998 22002 32006 42010 10

Don’t look at the figures for 1988 and 2010!

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A comparison between two parameter combinations

• nA = 39 fW = 1.4• nA = 39 fW = 1.2

We know how many seats move when we change 1.4 to 1.2.

But which of the two combinations gives the ”best” result?

That of course depends on our interpretation of ”best”.

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A suggested criterion function

variables: v = votes, s = seatsindices: c = constituency, p = party

vcp , scp for c=1(1)C, p=1(1)Pvc. , v.p , v.. , sc. , s.p , s.. marginals (s.. = 349)

A natural idea is to form qcp = vcp s.. / v.. and then compare{ scp } with { qcp }, and perhaps also { sc. } with { qc . }, e.g. by forming

Q_CP = c,p (scp - qcp)2 and Q_C = c (sc. - qc.)2

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Results when comparing 1.4 with 1.2:Q_PC when fW=1.4 (left), 1.2 (right)

1970 12.69 12.49 1973 12.03 12.03 1976 11.71 11.71 1979 12.56 12.42 1982 15.01 14.92 1985 12.71 12.71 1988 16.87 16.46 1991 17.87 17.82 1994 32.18 23.45 1998 24.94 22.77 2002 22.38 20.56 2006 28.03 23.49 2010 40.50 31.21

In all 10 cases where there is a winner, it is 1.2

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Results when comparing 1.4 with 1.2:Q_C when f W=1.4 (left), 1.2 (right)

1970 18.28 15.47 1973 15.23 15.23 1976 13.40 13.40 1979 10.61 10.57 1982 12.91 15.83 1985 8.86 8.86 1988 14.65 22.15 1991 12.68 9.43 1994 25.33 19.78 1998 21.74 17.80 2002 14.81 8.99 2006 30.94 22.66 2010 42.40 23.99

Three ties, two for 1.4, and eight for 1.2.

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Another criterion function

scp, qcp, sc., qc. as before

Instead of sums of squares, consider the number of cells where quota is violated, i.e. where

s < q or s > q Thus consider

V_CP = c,p (|scp – qcp|>1), V_C = c (|sc. – qc.|>1)

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Results when comparing 1.4 with 1.2:V_PC when fW=1.4 (left), 1.2 (right)

1970 7 5 1973 4 4 1976 3 3 1979 3 3 1982 3 5 1985 2 2 1988 4 6 1991 2 1 1994 10 7 1998 8 5 2002 5 3 2006 8 7 2010 10 9

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Results when comparing 1.4 with 1.2:V_C when fW=1.4 (left), 1.2 (right)

1970 0 0 1973 0 0 1976 0 0 1979 0 0 1982 1 1 1985 0 0 1988 0 0 1991 0 0 1994 2 0 1998 1 0 2002 2 1 2006 1 0 2010 2 1

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Another idea

Use 349 adjustment seats, i.e. skip what I called Step 1 in the distribution process.

However, that idea is not likely to raise much enthusiasm in this audience, and even less among politicians, so I abstain from giving the results of the computations I have performed.