As Josh Wills pointed out, isnatesilverawitch.com has a dissapointing methodology: it doesn't use any statistics!
To answer this question more quantitatively, we made a couple assumptions:
- Nate's final predictions represent the true probability of Obama winning the state
- For example, Obama had a 50.3% chance of winning Florida
- Election outcomes in one state don't depend on another state
- So Obama winning Ohio doesn't effect the Nevada outcome
- If Nate Silver is a witch, then he would get every state correct
- In Bayesian terms, \(P(\text{AllStatesRight} | \text{NateIsAWitch}) = 1\)
This probability was calculated by executing the following javascript code on fivethirtyeight.com:
nytg.stateData.forecast[0].states .map(function(x) {return {"stateAbbreviation": x[0], "pObamaWinsState": x[5]/100};}) .filter(function(x) {return isNaN(parseInt(x.stateAbbreviation[1]));}) .map(function(x) {return [x.pObamaWinsState, 1-x.pObamaWinsState].max();}) .reduce(function(x,y) {return x*y;})
This tells us Nate Silver had a 12.53% chance of correctly calling each state.
From there, we have a straightforward application of Bayes theorem: $$P(H|D)=\frac{P(D|H)P(H)}{P(D)}=\frac{P(D|H)P(H)}{P(D|H)P(H)+P(D|H')P(H')}$$ with $$\begin{align*}&H \text{, the hypothesis that Nate Silver is a witch}~\\ &D \text{, the data that Nate accurately predicted all states}~\\ &P(D|H) = 1 \text{ (if Nate was a witch he would've nailed each state)}~\\ &P(D|H') = 0.1253 \text{ (probability of Nate predicting all states given our model)}~\\ &P(H) \text{, your prior belief that Nate is a witch}~\\ &P(H') = 1-P(H)\end{align*}$$ Thus, we come to a different conclusion - Nate Silver is probably not a witch for any reasonable priors.