Wednesday, February 06, 2008

McCain to get 95% of California’s delegates? Adios Willard!



McCain to get 95% of California’s delegates?
Adios Willard!


We’re looking at a delegate breakdown closer to
160-10, which qualifies as, shall we say, an objectively bad day for Mitt Romney, even among serious analysts whose pundit fu is, of course, to be trusted. It has, at last, come to this.

Update: CNN puts Maverick at 615 delegates but with only 56 in his column thus far from California. Team
McCain says the real number is closer to 775.
Which means…

Speaking with reporters today, McCain adviser Charlie Black said,
“To date, we have 775 delegates, Romney has 284, Huckabee has 205. It takes
1,191 to clinch the nomination. There are 963 left to be chosen, so Romney or
Huckabee would have to have all of them — all of them — to get to 1,191. Now you
can’t do that because a majority of those 963 are chosen in proportional
primaries, which means you’d have to get 100% if the vote to get them all.

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