Trailhead : Popular Vote

Slate V: Hillary's Last Ad

Christopher Beam — Tue, 03 Jun 2008 03:40:00 GMT

Hillary Clinton released a new TV spot on Sunday touting the 17 million people who have voted for her in the Democratic primaries. That's "more than any primary candidate in history," the narrator tells us.

Well, sort of. Over at Slate V, we've got a remix of the ad pointing out the big old asterisk next to Clinton's claim that she's winning the popular vote. Watch it here.

The Popular Vote Chronicles: Shifting on "Uncommitted"

Christopher Beam — Thu, 22 May 2008 17:19:00 GMT

On a conference call just now, Clinton adviser Harold Ickes articulated the campaign’s position on Michigan’s "uncommitted" delegates: Obama shouldn’t get them. Over at Politico, Avi Zenilman points out how this would hinder Obama’s attempts to win the pledged-delegate count.

But it also affects the popular-vote tally. Namely, it justifies Clinton’s declaration that she’s "winning the popular vote," since she counts her own votes in Michigan but not "uncommitted."

At a breakfast with reporters earlier this month, Clinton strategist Howard Wolfson reportedly suggested that they’d be willing to give Obama the "uncommitted" delegates. Yesterday we wondered why, if they were willing to give him the delegates, they were unwilling to give him the popular votes.

But today, Ickes took a harder stance: "It is presumptuous to assume that each and every one of those delegates is an Obama supporter," he said. He described a different scenario: Rather than going to Obama, the "uncommitted" delegates would attend the August convention as just that—"uncommitted." The campaign would then make their case to the delegates at the convention.

It’s still a stretch to say you’re winning the popular vote while counting a state where your opponent wasn’t on the ballot. But now at least the logic is internally consistent—"uncommitted" delegates shouldn’t go to Obama, nor should "uncommitted" votes.

How To Legitimize the Popular Vote

Christopher Beam — Wed, 21 May 2008 18:53:00 GMT

Now that we’ve laid out Hillary Clinton’s logic for how she’s winning the popular vote, it’s worth examining whether and how she can turn this from a tenuous argument into a compelling case.

Right now, superdelegates aren’t buying it, most likely because no one thinks Michigan should be counted toward the popular vote, especially if you’re not counting the "uncommitted" votes for Obama. So if she’s going to persuade them, she needs a lead that doesn’t rely on counting Michigan’s dubious vote in order to put her ahead.

The upcoming votes in Montana and South Dakota won’t help her much. South Dakota currently has 194,000 active registered Democrats. Even if turnout is as high as 50 percent and she wins with a 60-40 split, she’ll only net around 20,000 votes. Turnout in Montana is expected to be in the mid-100,000 range, which likewise won’t net Clinton more than a few thousand votes. And those are optimistic scenarios.

Puerto Rico looks like her best shot. The island has 2.3 million registered voters, all of whom are eligible to vote in the Democratic primary. (Puerto Rico’s parties don’t align with U.S. parties.) I’ve seen turnout estimates of 80 percent. That seems high, but say it’s true and 1.84 million people vote. If Clinton wins a 60-40 split, that would net her about 370,000 votes.

So it’s possible that in the next three contests, Clinton could net as many as 400,000 votes. Obama currently leads by 257,000 votes, counting Florida and all caucus states but not Michigan. Which means Clinton could still come out ahead by more than 100,000 votes. That sort of margin isn’t Florida-proof—it relies on counting Florida’s popular vote in order to hold. But counting Florida could become standard practice if the DNC's Rules and Bylaws Committee decides to seat the state's delegation on May 31. If all that happens, Clinton could make a reasonable case to superdelegates that more people voted for her than for Obama.

Update 6:16 p.m.: Clinton's task is even harder if you factor in the enigmatic Texas caucuses. Check out the revised analysis here.

Fun With Popular Vote Numbers

Christopher Beam — Wed, 21 May 2008 18:20:00 GMT

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.