Texas Rep. Ron Paul secured his first popular vote victory of 2012 this weekend in the Virgin Islands. But although he won the most votes in the territory’s Republican presidential caucuses, he didn’t walk away with the most delegates.
According to results released by the Republican Party of the Virgin Islands, Paul received 112 votes, or 29 percent, to Mitt Romney’s 101 votes, a 26 percent share. Rick Santorum received 23 votes and Newt Gingrich attracted 18 votes.
Paul senior adviser Doug Wead declared victory in a statement posted to his website, but expressed irritation that most media outlets reported the Saturday announcement by Herb Schoenbohm, the territory’s GOP chair and a Romney supporter, that Romney had won the contest, before learning that Paul had indeed clinched the most votes.
Romney emerged with four delegates from the Virgin Islands caucuses and Paul came out with just one. Delegates were elected individually, accounting for the disparity between the popular vote and the delegate count. One uncommitted delegate decided to support Romney after the vote.
In his victory declaration, Wead wrote that it was ironic that Paul’s popular vote victory was nullified by the efforts of a deft delegate-focused rival — especially as the Paul campaign claims to have perfected the strategy of snatching delegate-count victories in states where it lost the popular vote.
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