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To: Utmost Certainty

“Wrong. Santorum would have to get 65% of the remaining 1384 delegates to reach 1,144. That is not going to happen.”

So is that MORE or LESS than the number of delegates Gingrich will have?

I’m so confused by the Historian’s Math that the Gingrich supporters use to tabulate the victory that they are so certain their candidate will achieve with delegates at the convention.


64 posted on 03/14/2012 4:25:23 AM PDT by RFEngineer
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To: RFEngineer; Utmost Certainty; Cincinatus' Wife

You are looking at it in the wrong way.

Let’s look at utmost certainty’s claim that Santorum would need to win 65 per cent of the remaining delegates.

Just ask yourself if that is possible, whether Newt is there or he isn’t there, for Santorum to beat his opponents or opponent by that amount, going forward.

I haven’t assessed those numbers, don’t know if they’re dead on, but it makes common sense to me. Romney has a BIG delegate lead, and don’t ever forget that. Romney’s problem is his big lead doesn’t mean he’s a front runnner in the true sense of that word.

So many, many do not like, do not trust, and do not want Mitt, that he is very, very weak as a front runner.

But mathematically, the claim is that Santorum cannot get to the nomination in Romney’s place.

Neither can Newt.

But what kind of an awkward position does that put Romney in, if he can’t get there either, and if two thirds or more of the primary voters DO NOT WANT HIM.

Meanwhile, Newt is driving the debate and taking on all comers, that are responsible for putting the country in the horrible position it’s in.

And Santorum is the religion canidate. Make no mistake. Other than that, he cannot hold a candle to Newt Gingrich.

I see you don’t appreciate Newt.

That’s too bad.

I thank the good Lord he’s there.


66 posted on 03/14/2012 4:43:12 AM PDT by txrangerette ("HOLD TO THE TRUTH...SPEAK WITHOUT FEAR" - Glenn Beck)
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