Ron Paul Wins at Least Six Delegates in Wyoming Caucus
Garners 21 percent of the total vote, defying expectations and affirming delegate-win strategy
LAKE JACKSON, Texas – 2012 Republican Presidential candidate Ron Paul gained six delegates as a result of winning 21 percent of the vote in the Wyoming caucus.
The end result of the contest is that if the next stage of the Wyoming nominating process goes as planned, Dr. Paul will take at least six of the 26 delegates tied to the caucus process with him to the Republican National Convention.
Caucusing in the state occurred between February 9th and 29th. The vote total of 439 votes cast for Ron Paul out of 2,108 total votes cast was especially notable in that the Paul campaign had no paid staff on the ground in Wyoming. In addition, the organization purchased no advertising and held no events, yet the candidate and campaign have stunned establishment expectations by winning five of 23 counties.
Also notable is that Ron Paul received more than two and a half times the vote total of would-be Republican frontrunner Newt Gingrich, once considered one of the “two” in a “two man race.”
“Ron Paul winning six delegates in Wyoming using few resources is an extraordinary outcome and it affirms that our delegate-attainment strategy can help Dr. Paul secure the Republican nomination,” said Ron Paul 2012 National Campaign Chairman Jesse Benton.
“The race to see who will face President Obama in a head-to-head matchup, a scenario that best favors Paul, is a 50-state one and ours is one of only two campaigns with the stamina, organization, and resources to get to Tampa on the way to the White House,” said Mr. Benton, referring to the site of the Republican National Convention and to Mitt Romney.
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