Must-read from Nate Silver: Romney has only 2% chance of winning election if he loses Florida

By on September 12, 2012

The New York Times‘ Nate Silver applied his presidential race forecasting models to the Sunshine State and the importance of Florida has never been more clear:

“Florida now ranks a clear second on our list of tipping point states, those most likely to provide the decisive votes in the Electoral College, behind only Ohio.”

Moreover, because of some “choppy” polling recently for President Obama in three other battleground states – Colorado, Iowa, and Virginia – Florida now represents the Obama campaign’s clearest path to victory.

“If Mr. Obama wins Florida, he could lose each of Virginia, Colorado and Iowa, along with Ohio, Wisconsin, and North Carolina, and either New Hampshire or Nevada, and still win the Electoral College.”

But what Silver stresses above all else is that for Mitt Romney, Florida is the whole ball of wax:

“Imagine that the election is very close: the popular vote is within one percentage point either way. This condition occurred roughly 3,000 times out of the 25,000 simulations that I ran in the forecast model on Monday.”

“For each of the top 12 states on our tipping point list, I looked up the probability of Mr. Romney winning the election conditional upon losing the state in these 3,000 simulations. If Mr. Romney has great difficulty winning the Electoral College without the state in a close election, we can fairly describe it as a must-win.”

“These simulations estimate that Mr. Romney has only a 2 percent chance of winning the election if he loses Florida — even assuming that the election is very close over all. Losing its 29 electoral votes just presents too daunting a challenge for him, given his inability so far to penetrate into states like Pennsylvania that could plausibly substitute for it.”

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