The delegates don’t add up for Newt

This is a great reference for all your delegate allocation questions.

After some brief consultation, I noticed that CA, NY, MD, DE, CT, MT, UT, PR, NJ,  and VA are all either strictly winner-take-all or winner-take-all if the winner gets a majority.

That’s 500 delegates in the bag for Mitt, right there.

He has about 100 delegates so far, and is projected to have about 250 by Super Tuesday. So just based on those numbers, Mitt already basically has 250 + 500 = 750 delegates in the bank for all intents and purposes. 1144 are needed. That means Mitt has a mere 400 delegates to find elsewhere to clinch the nomination. This is not going to be a problem.

Now, let’s look how it stacks up for Newt:

Not a single Newt-friendly state is winner-take-all except AL and MS. IN is a bit confusing as to how delegates are allocated, but my money’s on Mitt there regardless.

Newt has been talking about Texas deciding the race. He’s been saying he plans to take the race until Texas and see where things shake out for him at that point. But…

Let’s just say Gingrich, Paul and Santorum stay in until Texas for this example.

TX has 155 delegates, which are totally proportional.

Say the result looks like this:

Gingrich – 50%
Romney – 30%
Paul – 15%
Santorum – 5%

In the best-case scenario for Newt, he gets about 78 delegates, and Mitt gets about 50. That’s a 28-delegate pickup.

That’s nothing. Montana has only 26 delegates, but it’s winner-take-all. Imagine Texas being only as decisive as Montana?

So, in short, Mitt is the nominee. Or at least Newt isn’t. Santorum looks like the final not-Romney candidate, but the math is almost as bad for him anyway.

Tags: , ,


One Response to “The delegates don’t add up for Newt”

  1. Frozone
    February 5, 2012 at 2:49 pm #

    Nothing adds up with Newt…

Leave a Reply