Wednesday, October 11, 2006

Liberal math: 3 Quebecers = 8 Albertans

I just read this post over on CalgaryGrit. It would appear that the delegate selection process for the federal Liberal Party is a bit of a fiasco. There are allegations of voting irregularities, and forged signatures on delegate forms.

In other words, it's pretty much par for the course in the natural governing party.

In addition to the scandals, it would appear that even Liberals are beginning to notice the regional favoritism in their party. The rules of the party are such that approximately 3 Quebec members have the same representation as roughly 8 Alberta members in terms of delegate selection. Commenter Jason Bo Green weighs in:

Wow, interesting numbers.

I'm inclined to agree with you, man - it does look like everyone bombed out in Quebec.

And - pardon my English, but just what the fuck is going on with Alberta getting so royally screwed on say/votes yet again?

Honestly, I'm amazed (and grateful) Alberta sticks it through with us ...

Jason, for your information, Albertans are not sticking with you, or the Liberals. According to the numbers CalgaryGrit posted, your party has a grand total of 3,299 voting members in the Province of Alberta. There are individual ridings in this province with more Conservative members than that. I have also heard that the number of members expected to vote in the provincial Progressive Conservative leadership race could go as high as 200,000.

The federal Liberal Party is completely dominated by eastern Canadians, as your leadership race shows. Albertans simply aren't interested in it (other than a couple of opportunists like CalgaryGrit and daveberta).

Ask yourself this: who is the last leader of the Liberal Party of Canada to come from Alberta, British Columbia, Saskatchewan, or Manitoba?

Got you stumped? Let's make it easier: who is the last western Canadian candidate to get a single vote at a federal Liberal Party leadership convention?

Think hard now. There must be someone.