There has been some speculation out there that counting the number of donors to a campaign can determine and multiplying it by a given voter-to-donor ratio, the number of votes can be determined. According to Jim Babka, Paul would need a voter:donor ratio of 22:1 to 28:1 in order to capture 1/3rd of the caucus vote and win (22:1 at 80,000 GOP caucus-goers, and 28:1 with 100,000 GOP caucus-goers). Although voter:donor ratios are hard to come by, I looked at the number of donors prior to the 2004 Democratic Iowa caucus and the number of votes that candidate received (via OpenSecrets.org).

In the ‘04 Iowa Democratic Caucus, the candidate with the worst (lowest) voter:donor ratio was Dennis Kucinich, with 49 Iowa donors prior to the vote — which yielded him 1588 votes and a voter:donor ratio of 32:1. If Ron Paul even achieved this low ratio (with his ~1,200 Iowa donors), he would win hands down.

Update: I’ve gotten a lot of e-mails asking where the 1,200 Iowa donors number came from. Well, Jim Babka extrapolated 1,200 donors from the overall Q4 donor count at RonPaulGraphs.com. RonPaulGraphs.com has 68,419 Q4 donors, but the campaign says that there has been 130,000 donors overall. As a result, RonPaulGraphs.com’s numbers represent about 52.6% of all donors — apply that to RPG’s numbers and you get about 1,200 total donors. Babka’s piece slightly underestimated the total number of Paul’s Iowa donors (it’s actually 1,270) — but 1,200 is a safe, conservative number.