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Hillary Clinton, Fairy PrincessCan we please stop pretending she has a plausible chance to win the nomination?

By Timothy Noah
Posted Friday, May 2, 2008, at 7:21 PM ET

(Continued from page 1)

That means whoever ends the primary season with a plurality of delegates is all but certain to win the nomination, unless the plurality is so paper-thin as to be meaningless. According to Slate's delegate calculator, Clinton needs to win, on average, 70 percent of the vote in every remaining contest in order to surpass Obama on pledged delegates. Remember when I said there was no way Obama would capture 71 percent? There's no way Clinton's going to capture 70 percent, either.

OK, let's see how Hillary can get close enough to call it a tie. If she gets within about 30, that's pretty close, right? To do that, she needs to win, on average, 65 percent of the vote in every remaining contest. That's still in the realm of extreme improbability. How about 60 percent? That's a difference of 74 delegates, which is starting to sound like too many to justify throwing up your hands and declaring, "Close enough for government work." And, anyway, that's still too improbable to take very seriously. Do I hear 55 percent?* Which is to say: What if she wins every remaining contest, on average, by the 10-point spread she achieved in Pennsylvania? (It was really 9 points, but everybody thinks it was 10, so let's say 10.) OK, that's possible. Difficult to achieve, but possible. But that puts Obama 115 delegates ahead of Clinton. That is definitely too large a plurality to shrug off as a virtual tie.

But what if the superdelegates decide the will of the people resides in the popular vote? I doubt they will, because the popular vote seriously undercounts Obama's support in the caucus states. Even if they ignore that shortcoming, though, Hendrik Hertzberg has demonstrated that Obama right now has a plurality of 611,520 votes. That's not likely to change, because all the big-population states have already voted. Even if you toss in the delegates from Florida's unsanctioned primary, Obama maintains a plurality of 316,748. Add in Michigan and Clinton acquires a plurality of 121,783. But it's insane to count Michigan, because Obama wasn't even on the ballot there. (It is merely unfair to count Florida, because Obama was on the ballot there; in Florida, the problem is that neither candidate campaigned there.)

Hertzberg posited that a mental compromise might be reached in counting Michigan's popular vote by giving Obama all the "uncommitted votes." Clinton has on occasion tried to argue that the Michigan primary wasn't a Soviet-style election because most of the uncommitteds should be considered Obama supporters. OK, then, Hertzberg reasoned; let's include Florida and Michigan in the tally but count the Michigan uncommitteds for Obama. That leaves Obama with a margin of 188,439. If Clinton were to win every remaining contest by 10 points on average, Hertzberg calculated that she still would lose the popular vote by 161,520 votes.

That is, assuming Florida and Michigan went uncounted. Toss in Florida, and Clinton gains a popular-vote plurality of 133,252. But this scenario depends on three improbable contingencies: The superdelegates decide it's fair to equate the popular vote in primaries and caucuses with the popular will (which it isn't); Clinton wins by 10 percent everywhere from now on (possible but unlikely); and the superdelegates decide it's fair to consider the popular vote in Florida (doubtful).

That leaves Option 3, which is for Clinton to convince the already-pledged primary delegates and/or superdelegates that they must change their minds. This has happened in the distant past; Charles Peters cites in the latest issue of the Washington Monthly the Republican convention nominating the last-minute entrant Wendell Willkie in 1940. But that was in a different era, when much less than a third of all convention delegates were chosen by primary; everyone else was, in effect, a superdelegate. Ted Kennedy tried and failed to turn Carter's primary delegates (there were no superdelegates) at the 1980 convention. Mondale turned a few of Hart's primary delegates in 1984, but he already had a delegate plurality, which made his job a lot easier; he just needed to turn that plurality into a majority. At the 2008 convention, Clinton's position would be comparable to Kennedy's in 1980, not to Mondale's in 1984.

What would it take for Clinton to start a stampede? A massive, catastrophic drop in the polls for Obama. But the only way for that to happen is for Clinton to tear into Obama so viciously, Lee Atwater-style, that she destroys her own reputation, causing her to lose the general election and very likely her Senate seat, too. Not going to happen. Clinton is determined, but she isn't insane.

That exhausts the possibilities. Not one of them is plausible. So, please, let's stop pretending there's much suspense about who the nominee will be. As an arithmecrat, I will not consider anyone the winner until a candidate achieves 2,025 delegates. But neither am I obliged to believe Hillary Clinton has a decent shot. She doesn't.

Momentucracy vs. Arithmecracy Archive:
April 23, 2008: "Hillary Clinton, Ex-Arithmecrat"
March 6, 2008: "Agony of the Arithmecrats"
Feb. 6, 2008: "Triumph of the Arithmecrats"
Feb. 1, 2008: On the Media interview about momentucracy and arithmecracy, New York Public Radio
Jan. 30, 2008: "Momentucrats vs. Arithmecrats, Part 2"
Jan. 28, 2008: "Momentucrats vs. Arithmecrats"
Jan. 21, 2008: "Is Obama Winning?"
Dec. 11, 2007: "Whose Nominee Is It, Anyway?"

Correction, May 3, 2008: In the original column, a 60 percent to 40 percent split was erroneously described as a 10-point difference. (Return to the corrected sentence.)

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Timothy Noah is a senior writer at Slate.
Illustration by Rob Donnelly.
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