Clinton math

In the immediate aftermath of Puerto Rico's primary, Hillary Clinton's campaign made some counterintuitive arguments about turnout and the popular vote.

Published June 1, 2008 11:44PM (EDT)

In San Juan, as we were waiting for Hillary Clinton to come address a small crowd of supporters after her victory in Puerto Rico's primary, her campaign aides attempted to teach reporters some new math.

First, though network projections say only about 400,000 of the island's 2.4 million registered voters cast ballots in the primary, the Clintonistas say turnout wasn't low. And they said it exactly that directly. "The turnout is not low," spokesman Mo Elleithee told me. That's despite the fact that various Clinton supporters had said, over the last couple of months, that she might win the state by 400,000 or 500,000 votes, or that turnout would at least be around a half a million. Until it became clear that the low -- er, sorry, not low -- turnout would make it unlikely, the Clinton camp had been expecting to use Puerto Rico's primary to take the lead in the national popular vote.

This first piece of new math rests on two arguments. One, that turnout of 25 percent or so isn't that low in comparison to the local primaries in March. That's the baseline the Clinton camp prefers to the 1980 Democratic presidential primary here, probably because in 1980, more than 800,000 Puerto Ricans turned out for the Jimmy Carter-Ted Kennedy showdown. And two, that turnout would have been much higher if Barack Obama had spent as much time here as Clinton did. That's probably true, but if Obama had campaigned much here, Clinton wouldn't have beaten him by the 2-1 margin she racked up in an election where her supporters were the only people who bothered to show up.

But that's all beside the point, because the Clinton campaign has also declared that the decision made Saturday night to seat Michigan's delegation to the Democratic National Convention means that state's primary is now officially counted in the national popular vote, even though Clinton was the only major candidate whose name was on the ballot. (Not that there is any official count of the national popular vote.) In the minds of Clinton's spinners, that gives Clinton an extra 328,309 votes -- so she leads no matter how many votes she nets in Puerto Rico, and so who cares about turnout in the commonwealth? I asked Elleithee if the DNC's decision also means Obama should be credited with some of the votes cast for "Uncommitted" in the Michigan primary. You may not be surprised to learn that Elleithee thought that was too complicated, that the "easiest" -- and also most accurate -- way of counting was simply to tally up all the votes cast for Clinton in all the elections, no matter who else was on the ballot, then add up all the votes cast for Obama and compare the two totals.

Presumably, aides will explain all this new math in more detail on the campaign's seven-hour flight tonight from San Juan to Sioux Falls, S.D., but unfortunately -- or fortunately, depending on the way you look at it -- I'm heading back to Washington, D.C., instead, so I won't get to hear more of it.

By Mike Madden

Mike Madden is Salon's Washington correspondent. A complete listing of his articles is here. Follow him on Twitter here.

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