# Obama's Fuzzy Math

Campaigns toss around lots of numbers, many of which get published without any double-checking. And with delegates we’re usually not sticklers, since everyone has their own count . But one number being floated today struck us as just plain wrong.

Obama campaign manager David Plouffe said on a conference call today that if March 4 is a tie, Clinton would need to win 75 percent of the remaining pledged delegates. This seemed high to me, given all the monkeying around we’ve been doing with
**
Slate
**
’s
delegate calculator
.

The campaign claims Obama currently leads by 162 pledged delegates. (That’s about right .) They point out that after March 4 there are another 611 pledged delegates up for grabs. So far, so good.

But if Clinton wins 75 percent of those 611 remaining delegates (giving her 458 delegates), that means Obama will win 25 percent (giving him 152 delegates). If you add that to Obama’s current lead of 162, he will have 314 delegates total. In other words, Clinton:
**
458
**
. Obama:
**
314
**
. She would have a lead of 144 pledged delegates. That’s a lot more than "catching up" (epecially when coupled with her current
lead among superdelegates
).

A little algebra* shows that if Clinton and Obama split the March 4 delegates, she would actually have to win about
**
63 percent
**
of the remaining delegates to tie it up.

*
Got a problem with our math?
Let us know
.
*

*OK, here goes: If
*
x
*
= percentage of delegates Obama wins after March 4, then the percentage Hillary wins is (1-
*
x
*
).
*
y
*
is the number of delegates each candidate will have won between now and the end of the primaries, including Obama’s current 162 delegate lead. So,

611(1-
*
x
*
) =
*
y
*

611
*
x
*
+ 162 =
*
y
*

*
x
*
, and you’ll find that
*
x
*
= 0.37, which means (1-
*
x
*
) = 0.63.