How the Democrats do their political arithmetic

By Arun Kumar, IANS

Washington : Democrats in Iowa went back to school after brushing up their maths and armed with calculators to pick their candidate in the first nomination contest of the 2008 US presidential race.

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For unlike Republicans, who simply hold straw polls, the Democrats must get their calculations right to ensure that they pick only a ‘viable’ candidate from among the eight party hopefuls, a candidate with at least 15 percent backing.

So as they gathered in middle schools, community halls or gymnasiums in 1,781 precincts across the western state on a chilly Thursday night, the Democrats huddled among themselves to discuss the merits of all candidates before moving into the corner of a Hillary or an Obama or an Edwards or a long shot Kucinich.

Presiding officers then took out their calculators and scribble sheets to see who gets how many delegates to vote in the state’s 99 county conventions and put the writing on the wall.

For example, if 29 registered Democrats show up in a precinct on caucus night, a preference group must get the support of 15 percent of 29 or 4.35. As fractions can’t vote, one needs at least five (whether the fraction is above or below 0.5) to stay in the game.

If no candidate stays above the 15 percent watermark, then it’s time for round two after a 15-minute break for realignment. Caucus goers move afresh into corners of their choice leaving aside the unviable.

If a precinct was assigned 11 delegates and Clinton got say eight of the 29 attendees, how many delegates would she get?

The calculators come out again to solve the new equation: eight supporters multiplied by 11 delegates equals 88; 88 divided by 29 equals 3.03. That amounts to three delegates. Now the numbers are rounded out again as one learnt to do back in school.

Sounds complicated? No wonder the Democratic party machine held training sessions for the field staff all over Iowa last month to help them with their homework.