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Romney's strong Iowa odds

WASHINGTON — As we close in on the Iowa caucuses, it’s becoming clearer that everything is going right for Mitt Romney. He may not win Iowa, but right now it appears he’ll emerge looking strong, and if the polling holds up, he’ll be the overwhelming favorite to win the nomination.

The reason things are looking so good for Romney is that Ron Paul and Newt Gingrich are remaining competitive — at the expense of the one candidate who could give Romney a real run for the nomination: Rick Perry. As the only other candidate in the race with conventional credentials and mainstream conservative positions on public policy, if Perry were to become competitive with Romney, it wouldn’t just be inconvenient for him; Perry could win. Fortunately for Romney right now, Perry still appears to be mostly going nowhere, but all he would have to do is rally to third to return to almost full viability.

By contrast, if Gingrich remains viable, he would be less of a threat to Romney than Perry would. He scares Republican politicians and leaders of GOP-aligned groups because he would be a wild card in the White House. Ron Paul doing well in Iowa isn’t a threat to Romney either: Many GOP-aligned groups oppose most of his positions on public policy. Indeed, in a two-person race against Paul, Romney could campaign on being tough against terrorists and drugs while preserving popular portions of the federal government, all of which would help position him for the general election.

In this light, the polling in Iowa shows almost a perfect finish for Romney. According to Nate Silver’s latest numbers, Romney and Paul are close for first, with Gingrich hanging on for third and Michele Bachmann besting Perry for fourth. Romney couldn’t ask for a better scenario. Now Silver’s projections are based only on current polls, so the late effects of Perry’s spending could easily still shake things up. But as long as Perry remains dead in the water, Romney — whatever happens in Iowa — will continue cruising toward the nomination.