The Washington Post’s Chris Cillizza takes a hard look at Mitt Romney’s winning map.
Such as it is:
A detailed analysis of Romney’s various paths to the 270 electoral votes he would need to claim the presidency suggests he has a ceiling of somewhere right around 290 electoral votes…
Romney’s relatively low electoral-vote ceiling isn’t unique to him. No Republican presidential nominee has received more than 300 electoral votes in more than two decades. (Vice President George H.W. Bush won 426 electoral votes in his 1988 victory over Massachusetts Gov. Michael Dukakis…)
In 2000, Bush won 271 electoral votes – one more than he needed to claim the presidency. In eking out that victory, Bush not only carried the South and Plains states with a near sweep but also claimed wins in swing states such as Nevada, Colorado, Missouri and the major electoral-vote prizes of Ohio and Florida…
In 2004, Bush won reelection with 286 electoral votes, losing New Hampshire from his 2000 map but adding wins in Iowa and New Mexico.
Under the 2012 map, Romney would win 292 electoral votes if he replicated the Bush 2004 victory. But New Mexico seems like a very tough place to win – not to mention the fact that he would again need to carry Ohio, Florida, Colorado and Nevada as well as North Carolina and Virginia.
Cillizza does note that the number of safe Republican states means Romney’s minimum guaranteed electoral vote count looks a little better in relation to his “ceiling”–meaning the victory scenarios involve combinations of only a few swing states, most including Colorado. One caveat for that exists if Romney can pick up any of several states that Republican presidential candidates have not carried recently, like Michigan or Pennsylvania.
But just about any way you look at it, we’re crucial to the outcome of this thing.