Fun With Popular Vote Numbers

Last night, Clinton announced that she’s “winning the popular vote.” It’s a claim she’s been making since Pennsylvania , but now that Obama has won a majority of pledged delegates, it’s really her last plausible argument for the nomination.

As this blog has noted before, the popular vote is a flawed metric because the number of people who participate in caucuses is much smaller than the number of people who vote in primaries. But let’s set that objection aside. What’s the logic undergirding Clinton’s claim?  

1) If you count the vote in all primaries and caucuses sanctioned by the Democratic National Committee, Obama leads by about 552,000 votes, according to an estimate by Real Clear Politics.

2) If you add in Florida, whose primary was not DNC-sanctioned, and where the candidates agreed not to campaign, Obama’s lead drops to 257,000.

3) If you further add in Michigan, whose primary was not DNC-sanctioned, where candidates agreed not to campaign, and where Obama’s name did not appear on the ballot , then Clinton leads by 71,000.

But:  

4) In Michigan, “uncommitted” received 40 percent of the vote, which seems kind of high. If we count Michigan’s 238,000 uncommitteds for Obama, then he leads by 167,000. Presumably the Clinton campaign isn’t making this final calculation. In the past, though, her camp has explained away the embarrassingly large proportion of Michigan uncommitteds—remember, Clinton was the only major candidate on the ballot— by pointing out that Obama’s Michigan supporters urged primary voters to pull the lever for “uncommitted.” Indeed, on May 10 Clinton spokesman Howard Wolfson said Clinton would be willing to give Obama all the Michigan uncommitteds , provided Obama dropped his opposition to seating Michigan and Florida.

Hendrik Hertzberg of the New Yorker , a strong proponent of the popular-vote metric, has argued that if you’re going to count Michigan, you have to take this last step. (Incidentally, Hertzberg later discovered that his own calculations understated Obama’s support.)

5 ) An additional variable is whether you count all the caucuses. Four caucus states—Iowa, Nevada, Maine, and Washington— never reported their popular votes. So the calculations above are based on estimates. If you don’t count these estimates, Clinton’s 71,000 lead rises to 181,000 votes .

6 ) But isn’t it inconsistent to argue for enfranchising Florida and Michigan while simultaneously disenfranchising Iowa, Nevada, Maine, and Washington? Yes, that’s inconsistent. So let’s disenfranchise Florida and Michigan in addition to the four caucus states. That gives the popular vote lead back to Obama by 442,000.

7 ) OK, now let’s put Florida and Michigan back in but give Obama the Michigan uncommitteds, per Hertzberg’s recommendation and Wolfson’s May 10 comment. That also gives the popular vote lead back to Obama, this time by about 57,000.