Effective thresholds in MMP when there is no threshold
Abolishing the 5% threshold in MMP (as I advocate) doesn’t mean that a party getting just one vote picks up one in 120 seats. It’s fairly intuitive that there is still an “effective threshold”: a number of votes that parties must get to earn their first seat. That then begs the question: How many votes is enough?
The answer depends on the method used to translate the party vote to seats in Parliament. New Zealand (and a number of other countries) uses a method called the Sainte-Laguë method. Another common method is the d’Hondt method. In this post I’ll assume you’re familiar with at least one of them (they are very similar); if you’re not, Wikipedia explains them reasonably well.
The Sainte-Laguë method is more sympathetic to smaller parties than the d’Hondt method, so we expect the Sainte-Laguë effective threshold to be lower. The report of the 1986 Royal Commission on the Electoral System lists the thresholds in Addendum 2.1, on page 74. The threshold for an N-member House, when there are k parties other than the one-seat party, is V/(2N − k + 1) for the Sainte-Laguë method and V/(N + 1) for the d’Hondt method.
I couldn’t find proofs of these effective thresholds, so I derived those results myself. That proof is in a PDF file here.
That then helps me to find the effective threshold of a modified Sainte-Laguë method that I support, which is the same one that they use in Norway and Sweden. In this method, the first divisor is changed to 1.4 (instead of 1). The threshold is then V/(5(2N − k)/7 + 1). More generally, if the first divisor is changed to m, then the effective threshold is V/((2N − k)/m + 1).
What does that even mean?
Those formulae don’t really mean much at first glance. The best way to find meaning is to compare them to V/N. That is, in a 120-seat house, how does the “effective threshold” compare to 1/120 of all party votes, or 0.83%?
To make life easier, we’ll make an approximation: we’ll assume that N is much larger than k, i.e. there are many more seats than parties, which is generally true. We’ll also use the fact that N >> 1.
Then the Sainte-Laguë effective threshold is approximately V/2N. That means, in order to get one seat out of 120, you need roughly half of 1/120th of the party vote, or 1/240th of the party vote, or about 0.42%.
For the modified Sainte-Laguë method, it’s roughly mV/2N. Basically that means you take 1/120th of the vote and multiply it by m/2. For example, if m = 1.4, then you need about 70% of 1/120th of the vote, which is about 0.58%.
The d’Hondt threshold, roughly V/N, is just 1/120th of the party vote (or marginally less), or about 0.83%.
It seems fair to me that a party falling just short of 1/120 of the party vote should get one seat in Parliament. But awarding them a seat for achieving just half of that seems a bit unfair—and disproportional—to me. The effective threshold should be enough to be “close-ish” to 1/120. I would put “close-ish” at about 70% of 1/120 of the party vote, which is why the modified Sainte-Laguë method used by Norway and Sweden seems sensible to me.
Proof
