Nugget from 538: 99.6% Chance the Map Will be Different than 2012 (user search)
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  Nugget from 538: 99.6% Chance the Map Will be Different than 2012 (search mode)
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Author Topic: Nugget from 538: 99.6% Chance the Map Will be Different than 2012  (Read 1553 times)
palandio
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« on: July 12, 2016, 01:17:39 PM »

This means almost nothing; even if the map is same, they can just say "we're in the 0.4%".

So by your logic, probabilities mean nothing unless they're either 100% or 0%?

Perhaps not nothing, but I find little utility in this type of statistic.

Right, that's what I've been struggling with. What is this supposed to mean? If it's just a snapshot of where the race stands, cool. But it purports to be more. It purports to be a prediction, and it purports to tell us that if we ran this election 5 times, then on average, Trump would win. Is that true?
That is in reality a deeply philosophical question about the nature of probability. The frequentist concept of probability is more or less "If we would run the election 1000 times, Trump would win 200 times". The concept that I personally prefer in many situations is the concept of probability introduced by the Bayesian school and De Finetti: Probability is subjective and encodes the lack of information that you have. Despite the lack of information on an experiment, you can still bet on its outcome. That is if Nate Silver was a bookmaker and he was forced to offer you fair odds on the outcome of the election, he would say:
"If you bet 1.00$ on Trump and he wins, I'll give you 5.00$. If you bet 1.00$ on Clinton and she wins, I'll give you 1.25$."
If he is right and Trump has a 20% chance of winning, and you bet 1.00$ on Trump, then his expected revenue is 0.8*(1.00$-0.00$)+0.2*(1.00$-5.00$)=0.00$. Similarly when you bet on Clinton his expected revenue would be 0.00$.
If he would offer different odds, he would give you the chance to make profit (on average).
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Well, when talking about the 0.4% we are approaching another phenomenon of probability. How likely is it that you throw your dart into the center of the dartboard? Not so likely after all. But at the same time the center of the dartboard is more likely to be hit than any other area of half a square-inch in its surroundings, at least if you don't systematically deviate in one direction. The 2012 outcome can be thought of as the center of the dartboard, and the 2016 outcome will likely be somewhere near the center of the dartboard, but probably not in it exactly.
At the same when you will have thrown your dart at the dartboard, every possible exact position will have been quite unlikely before you threw the dart.
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Of course, that's the challenge behind it.
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