October 27, 2012


Mitt Romney's Very Long Odds of Winning the Election (GABRIEL SNYDER, OCT 26, 2012, Slate)

[F]or the last three presidential election cycles, I've found charting out the electoral vote odds a useful way to cut through the clutter of overheated punditry and keep perspective on the state of play in an election. 

While most of the campaign coverage is dominated by familiar cliches of the race being too close to call, nail-biting, and down-to-the-wire, there is arguably a lot more known about the Nov. 6 vote than there is unknown. According to the Constitution, the presidential election is actually 51 separate races. Each of those races is a winner-takes-all contest for a certain number of votes in the Electoral College. (Note: My Atlantic colleague Chris Heller points out this is not entirely correct since Nebraska and Maine award some of their electoral votes proportionally. For the purpose of this math, I assumed those states are winner-take-all.) The presidential candidate who gets 270 or more electoral votes will be the next President. We know the outcome of the vast majority of those 51 contests: New York, California, and Hawaii, and so on, will award their electoral votes to Obama, while Wyoming, Oklahoma, Utah, etc., will award their electoral votes to Romney. In these 41 "known" races, Obama has a huge lead over Romney: 237 electoral votes to 191. [...]

In the nine remaining toss-up states -- Nevada, Colorado, Iowa, Wisconsin, Ohio, New Hampshire, Virginia, North Carolina, and Florida -- there are 110 electoral votes up for grabs, but because Obama needs only 33 of those votes to win re-election, he wins in the vast majority of the possible scenarios. As a mathematical exercise, Romney has just 76 paths to victory out of the 512 possible combinations.

Posted by at October 27, 2012 10:56 AM

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