Labour Loser Wins Again
In the individual member section of Labour's barmy electoral college - the votes cast were as follows:
Tom Harris - 3.444%
Johann Lamont - 12.183%
Ken Macintosh - 17.707%
Total - 33.33%
Now the reason that the total adds up to 33.33% instead of 100% - is that party members have only one-third of the votes - which sounds completely bonkers because it is completely bonkers.
But in most other political parties this would have produced the following result:
Tom Harris - 10%
Joahann Lamont - 37%
Ken Macintosh - 53%
Total - 100%
So Ken Macintosh won an overall majority in the ballot of individual Labour party members in Scotland - of whom there are less than 20,000 these days.
And no wonder because they don't even get to elect their own leader.
In the Alice in Wonderland world of the Labour party two more sections of the 'electoral college' come into play - one for parliamentarians (MSPs and MPs) and the other for trade unions.
So out of the total number of ballot papers sent out - well over 300,000 according to Labour - less than 20,000 are for individual Labour party members - 100 or so are for MSPs and MPs - and around 300,000 are for non-Labour party members in the trade unions.
Which means that 20,000 votes - has the same value as 100 votes - has the same value as 300,000 votes - or to put it plainly some votes in the Labour party are much more equal than others.
Not everyone votes of course which distorts the picture even further - but the turnout figures have still to be released for each section - and will make interesting reading at some point.
Ironically, Labour's new deputy leader in Scotland - Anas Sarwar - has been elected by ordinary party members who voted for him by a majority of 61% - despite a trade union campaign to elect one of his rivals.
So Scottish Labour has ended up in exactly the same position as the UK Labour party - they have a new leader - but one who has been rejected by ordinary party members.
