European elections

How our MEPs will be elected: the dos and the D’Hondts of PR

This week’s European elections will give the UK’s palate a rare taste of that most acquired of sensations: PR.

The whole of the country gets to sample the joys (or woes, depending on your opinion) of proportional representation only once every five years. If you live in Scotland, Wales, Northern Ireland or London you’ll be more familiar with this system of voting. But it’s only for the EU elections that the entire UK takes the plunge.

The version of PR used to allocate seats is called the D’Hondt method. Mr D’Hondt, as his name suggests, hailed from across the Channel – Belgium, to be precise. He conceived of it in 1878, and it’s been going strong ever since. It’s used in countries as widespread as East Timor, Peru and Finland. D’Hondt has also been used in the UK for the London Assembly elections and the allocation of ministers in the Northern Ireland Assembly.

Its virtues, aside from it being – as its name suggests – proportional, include the simplicity of voting. You just put a cross in the box of your preferred party, and that’s it. No ranking by number, no multiple forms, just one mark on the ballot paper.

Its application is a little more involved, but nowhere near as complicated as it is often portrayed.

At its heart is a calculation based on dividing the number of votes a party has received by the number of seats it has been allocated, plus one.

Here’s an example of how the D’Hondt system works.

Let’s say the number of votes received by the top five parties in the North West England region in Thursday’s election ends up as follows:

Labour 420,000
Ukip 360,000
Conservative 320,000
Green 170,000
Lib Dems 120,000

(For a comparison, here are the actual results for this region in the last elections in 2009.)

The NW England region has eight seats to allocate.

As with all the other regions, the first seat goes automatically to the party with the greatest number of votes.

1st seat: Labour

Then the maths kicks in. D’Hondt’s formula is applied, meaning Labour’s total gets divided by one (the number of seats already allocated) plus one. In other words, divided by two. The same goes for the other parties, but as they’ve yet to get any seats, their totals are divided merely by one – meaning their totals stay the same.

This gives us figures of:
Lab 210,000; Ukip 360,000; Con 320,000; Green 170,000, Lib Dems 120,000

I’ve highlighted Ukip in bold, because it is now the party with the highest total. And as such, it gets a seat:

2nd seat: Ukip

To decide all the remaining seats, the same formula is applied each time. But given Ukip now has a seat, its original total gets divided by two, as does the total for Labour. This produces the following figures:

Lab 210,000; Ukip 180,000; Con 320,000; Green 170,000; Lib Dems 120,000

The Tories are now showing the highest total, so they get a seat.

3rd seat: Conservative

On we go. The Tory vote now has also to be divided by two – the number of seats they’ve already won plus one. This gives us:

Lab 210,000; Ukip 180,000; Con 160,000; Green 170,000; Lib Dems 120,000

And look: Labour is back on top. So it gets another seat, its second for this region:

4th seat: Labour

But now the calculation changes. While Ukip and the Tories still get their original vote divided by two, Labour’s is now divided by three: the number of seats it has already won (two), plus one. So we now have:

Lab 140,000; Ukip 180,000; Con 160,000; Green 170,000; Lib Dems 120,000

Ukip is back in first place, so it gets another seat – its second for the north-west:

5th seat: Ukip

Both Labour and Ukip’s original vote is now divided by three, the Tories’ by two, and the others by one. This leaves us with:

Lab 140,000; Ukip 120,000; Con 160,000; Green 170,000; Lib Dems 120,000

Finally, something for the Greens.

6th seat: Green

We divide again: by three for Labour and Ukip as before, but now by two for the Tories and the Greens. We divide by one for the Lib Dems:

Lab 140,000; Ukip 120,000; Con 160,000; Green 85,000; Lib Dems 120,000

The Tories are back at the top, so they get the seventh and penultimate seat for this region:

7th seat: Conservative

That leaves us with one last calculation. As Labour, Ukip and the Tories all now have two seats, their original vote is divided by three. The Greens’ vote is once again divided by two, while the Lib Dems is divided by one. And we’re left with…

Lab 140,000; Ukip 120,000; Con 106,666; Green 85,000, Lib Dems 120,000

8th seat: Labour

Labour scoops the last seat.

This would give us a notional line-up of MEPs for NW England that went:
Labour 3 (+1)
Ukip 2 (+1)
Conservative 2 (-1)
Green 1 (+1)
Lib Dems 0 (-1)

Follow all that?

Even if you didn’t, I hope you got a sense that something proportional and, yes, fair was going on.

Somewhat ironically, it’s thanks only to a system like D’Hondt that a party officially against proportional representation – the Conservatives – gets a share of seats roughly equating to its overall share of the vote, despite the chance of it coming third in the overall poll: an outcome that, in a general election fought under the first-past-the-post system, would leave it unjustly under-represented.

Note: the MEPs for the Northern Ireland region aren’t being elected using the D’Hondt system of PR. The single transferable vote is being used instead, meaning electors will rank the candidates in order of preference on the ballot paper. When it comes to allocating the seats, the lowest-scoring candidates are eliminated first, with their vote being redistributed among the remaining candidates until – in this case – three winners emerge.

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Discussion

2 responses to ‘How our MEPs will be elected: the dos and the D’Hondts of PR

  1. Love the blog.

    Another way of looking at your example is that 140,000 votes is required to secure each seat (with remainders ignored, as they are not enough for one more seat). So:

    Labour 420,000 / 140,000 => 3
    Ukip 360,000 / 140,000 => 2
    Conservative 320,000 / 140,000 => 2
    Green 170,000 / 140,000 => 1
    Lib Dems 120,000 / 140,000 => nil (0.857 is not enough)

    With several multiple-member seats, this course favours the larger parties, as there is an implicit minimum thresold to secure a seat in each constituency.

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