Santorum Needs Gingrich to Stay in the Race

In his Washington Post column, Eugene Robinson explains why he thinks the "subtleties of the delegate math" mean that Newt Gingrich's remaining in the race for the GOP nomination will ultimately benefit Santorum.  

The “throw Newt from the train” people think that the math is on their side, but it isn’t.

It’s true that from the primaries and caucuses held so far, we know that the Romney vote is much smaller than the anti-Romney vote. In Ohio, for example, Romney managed a slim victory with 38 percent to Santorum’s 37 percent. But Gingrich, meanwhile, drew nearly 15 percent. Add those voters to Santorum’s, and Romney would have suffered a shattering defeat.

Santorum and Gingrich are both campaigning on the premise that Romney is not a genuine conservative. Both candidates draw support from self-described “very conservative” Republicans. Since Gingrich — who supposedly had a “Southern strategy” for winning the nomination — couldn’t even beat Santorum in Alabama and Mississippi, it’s clear who would have the better chance against Romney, mano a mano. Ergo, Newt, hasta la vista.

But this logic ignores the subtleties of the delegate math. Sorry to inflict a flurry of numbers, but here goes: To win the nomination, a candidate needs the support of 1,144 convention delegates. According to projections from the Associated Press, at this point Romney has 481 delegates, Santorum has 252, Gingrich has 128, and Ron Paul has 48.

By AP’s count, 1,356 delegates remain up for grabs in the remaining primaries and caucuses. That’s right: We haven’t even reached the halfway point of this seemingly endless slog to the convention in Tampa. 

Read Eugene Robinson's entire column at the Washington Post.