September 2, 2012
The electoral map looks better for Obama than you think
Vtgenie at DailyKos has an interesting analysis of Nate Silver‘s latest state-by-state polling numbers. (Nate Silver is, of course, the Eric Clapton of poll analysts.)
Here’s VTgenie’s explanation:
…starting with states that each candidate is guarranteed to win, work your way down the probabilities adding electoral votes as you go. So for example, Obama currently is given a 100% chance of winning nine states (go Vermont!) and Maine, district 1, followed by 11 states between 90 and 100%, ending at Pennsylvania with 91%. Assuming Obama will win all of these, we proceed through the next four states, all between 70 and 80%, ending with Ohio at 71% for a total of 275 electoral votes.
Notice that to get to 270, using only Obama’s highest probability states, we never had to use a state with lower than 70%. On the other hand, doing the same process for Romney, using only his highest probability states, he has to use states for which he has a less than 50% chance of winning– in some cases, much less. Put another way, the easiest path to 270 for Romney– in terms of current probabilities– runs through three states between 30 and 40%, ending with Ohio, at only a 29% chance of winning. That’s gotta hurt.
But wait! There’s more…
While Romney has no path to victory using states where his chance of winning is greater than 50%, Obama has not just one but many such paths.
I can’t make it any clearer than that.
VTgenie notes that of course this depends on the accuracy of Nate Silver’s numbers, and things may change, etc. But it seemed trenchant to me. Of course, my math skills do not allow me to have an opinion. Is VTgenie’s analysis right?