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> it means that 6 times out of 10 Obama would win and 4 times out of 10 Romney would win. If Obama won all 10 times, by his own account he would be wrong.

That's a very frequentist perspective. I think the Bayesian interpretation is a little more sensible. Given a prior estimate, updated with the information we have, it is logical to assume that Obama has a better chance of winning.

That is, Nate Silver's 60% doesn't mean that if the election in a given state were repeated, we'd see different results four times out of 10, but rather that his information in making the prediction was incomplete.




Nate Silver himself said (in a radio interview that I cannot locate despite much Googling :< ) that if he said something would happen 60% of the time and it was 10 out of 10 that he his figure was wrong and should have said 100%.


But you can't run the same election 10 times. Without hearing the interview, I'm going to guess he's talking about multiple states, all with 60% chance of winning, which puts us in a different place entirely.

I guess I'm not disagreeing with the main point of your post. Just your use of the phrase "something would happen 60% of the time", as being oddly frequentist, when a Bayesian perspective is more appropriate here.


> But you can't run the same election 10 times.

That makes assessment difficult, but is irrelevant for interpretation. If something happens 60% of the time, it happens with the same probability as drawing a red ball from an urn with 6 red balls and 4 green balls, whether it happens once or several times.

Trying to determine the quality of a model using some observed data is inherently a frequentist exercise -- a dyed-in-the-wool Bayesian would take the election results, use them to update his or her posterior distributions, and carry on happily. (I know that no one would actually do this; no one who analyzes data is really a "pure" Bayesian or a "pure" frequentist).




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