# Talk:1132: Frequentists vs. Bayesians

Something should be added about the prior probability of the sun going nova, as that is the primary substantive point. "The neutrino detector is evidence that the Sun has exploded. It's showing an observation which is 35 times more likely to appear if the Sun has exploded than if it hasn't (likelihood ratio of 35:1). The Bayesian just doesn't think that's strong enough evidence to overcome the prior odds, i.e., after multiplying the prior odds by 35 they still aren't very high." - http://lesswrong.com/r/discussion/lw/fe5/xkcd_frequentist_vs_bayesians/ 209.65.52.92 23:51, 9 November 2012 (UTC)

Note: taking that bet would be a mistake. If the Bayesian is right, you're out $50. If he's wrong, everyone is about to die and you'll never get to spend the winnings. Of course, this meta-analysis is itself a type of Bayesian thinking, so Dunning-Kruger Effect would apply. - Frankie (talk) 13:50, 9 November 2012 (UTC)

- You don't think you could spend fifty bucks in eight minutes? ;-) (PS: wikipedia is probably a better link than lmgtfy: Dunning-Kruger effect) -- IronyChef (talk) 15:35, 9 November 2012 (UTC)

Randall has referenced the Labyrinth guards before: xkcd 246:Labyrinth puzzle. Plus he has satirized p<0.05 in xkcd 882:Significant--Prooffreader (talk) 15:59, 9 November 2012 (UTC)

A bit of maths. Let event N be the sun going nova and event Y be the detector giving the answer "Yes". The detector has already given a positive answer so we want to compute P(N|Y). Applying the Bayes' theorem:

- P(N|Y) = P(Y|N) * P(N) / P(Y)
- P(Y|N) = 1
- P(N) = 0.0000....
- P(Y|N) * P(N) = 0.0000...
- P(Y) = p(Y|N)*P(N) + P(Y|-N)*P(-N)
- P(Y|-N) = 1/36
- P(-N) = 0.999999...
- P(Y) = 0 + 1/36 = 1/36
- P(N|Y) = 0 / (1/36) = 0

Quite likely it's not entirely correct. Lmpk (talk) 16:22, 9 November 2012 (UTC)

Here's what I get for the application of Bayes' Theorem:

- P(N|Y) = P(Y|N) * P(N) / P(Y)
- = P(Y|N) * P(N) / [P(Y|N) * P(N) + P(Y|~N) * P(~N)]
- = 35/36 * P(N) / [35/36 * P(N) + 1/36 * (1 - P(N))]
- = 35 * P(N) / [35 * P(N) - P(N) + 1]
- < 35 * P(N)
- = 35 * (really small number)

So, if you believe it's extremely unlikely for the sun to go nova, then you should also believe it's unlikely a Yes answer is true.

I wouldn't say the comic is about election prediction models. It's about a long-standing dispute between two different schools of statisticians, a dispute that began before Nate Silver was born. It's possible that the recent media attention for Silver and his ilk inspired this subject, but it's the kind of geeky issue Randall would typically take on in other circumstances too. MGK (talk) 19:44, 9 November 2012 (UTC)

I agree - this is not directed at the US-presidential election. I also want to add, that Bayesian btatistics assumes that parameters of distributions (e.g. mean of gaussian) are also random variables. These random variables have prior distributions - in this case p(sun explodes). The Bayesian statistitian in this comic has access to this prior distribution and so has other estimates for an error of the neutrino detector. The knowlege of the prior distribution is somewhat considered a "black art" by other statisticians.

My personal interpretation of the "bet you $50 it hasn't" reply is in the case of the sun going nova, no one would be alive to ask the neutrino detector, the probability of the sun going nova is always 0. Paps