The Remainer’s Friend and Nemesis
As we come up to the European Parliamentary elections, there has never been a more important time to cast a vote for a seat at the European Parliament. Britain’s European election turnout has traditionally been extremely low. Always lower than the EU average. Probably a proxy for the disconnectedness of the UK public with activities on the continent prior to 2016.
This leads to a dominance of engaged parties and those parties have unfortunately, been on the right and far-right of the political spectrum, including Nazi sympathisers, fraudsters and racists.
The upcoming European elections are probably the most important in the UK’s history. Potentially the last time Britain will participate in them. Yet, very little is known about the way the D’Hondt system calculates votes and apportions seats to parties.
D’Hondt (or Jefferson) Method
Named after 19th century Mathematician Victor D’Hondt, or Thomas Jefferson, who used the method in the House of Representatives in 1791, the method uses a quotient system to apportion seats to a Parliament. In Europe, the method is employed along party lines, not candidates. Meaning the electorate vote for parties with those parties selecting candidates.
The D’Hondt method is regarded as having an advantage for larger parties. Smaller parties may lose out significantly due to rounding and mathematically is not quite as clean a proportional representation process as it could be.
The formula cited in wiki is incomplete. I’ll come on to that later
How does it work?
After the electorate has cast their vote and they are totalled up, the seats are allocated in cycles starting from a 0-th seat (total votes) to the number of seats in the region. With each seat reducing the denominator.
Taking for example, the North West of England. Labour received 594,063 votes in the 2014 election. There are 8 seats, so the quotients are calculated as
0. quot = 594,063 / (0 + 1) = 594,063
- quot = 594,063 / 2 = 297,032
- quot = 594,063 / 3 = 198,021
- quot = 594,063 / 4 = 148,516
- quot = 594,063 / 5 = 118,813
- quot = 594,063 / 6 = 99,011
- quot = 594,063 / 7 = 84,866
- quot = 594,063 / 8 = 74,258
- quot = 594,063 / 9= 66,007
However, this is only one party. In the event they were the only ones who ran, this would be the end result. Of course, if only one party ran, this would not be democratic. So more than one party must run and the quotients calculated for each, before the parties are allocated their seats.
As an example, the image shows the top 8 quotients across all “seat-cycles” for seats zero to 8 allocated to the Parliament as representatives of the North West of Britain. UKIP received 3 seats, as they occupied 3 spaces in the top 8. This step is also why the previous equation was incomplete, since it needs to sort the equation’s results by quotient, for all parties and take the top 8 from them.
The highlights in the above image shows what happened with UKIP, Labour, Conservatives, Greens and LibDems. In that election, remaining LibDem and Green votes were wasted. Since neither of them achieved enough to gain a seat (an inconvenient truth with proportional representation is that votes are still wasted — there are a fixed number of seats). The picture looks very different in the event Green and LibDem votes had combined, as I explained on twitter (6 seat breakdown shown for brevity). Whether that is LibDems standing down and siding with Green or vice-versa.
For completeness, as an unashamed bug-bear, the elections and campaigners will give away whether voters and [remain] campaigners are Country or Party first, by the way they advocate voting. This election will be interesting for that alone.