Disgraceful fraud of the Iowa caucuses, cont.

In a previous entry I said, based on nothing but an educated guess, that without the vote of Independents in the Iowa caucuses, Ron Paul “probably would have had less than 10 percent of the total,” instead of the 21 percent that he did get. Now I read Jim Geraighty at NRO who writes:

With no seriously contested Democratic caucus to compete for the votes of independents, the caucus turned into yet another sales pitch for closed primaries. According to the entrance polls, 38 percent of caucus-goers had never voted in a GOP caucus before; of those, by far the largest share, 37 percent, voted for Ron Paul. Among the registered so-called independents who took part in the caucus, 48 percent voted for Ron Paul, way ahead of anyone else. Next highest was Romney with 16 percent.

We don’t have all the figures yet, but if, as reported at the Wall Street Journal yesterday, 25 percent of the caucus participants were Independents, and if Paul got 48 percent of their votes, then about 12 percent of the total votes in the primary were those of Independents who voted for Paul. Which means that without the absurdity and outrage of non-Republicans being allowed to vote in the Republican caucuses, Paul would have won less then ten percent of the total instead of 21 percent, exactly as per my guess.

Posted by Lawrence Auster at January 04, 2012 03:08 PM | Send

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