# 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

- Yes, you would be able to ask. While neutrinos move almost at speed of light, the plasma of the explosion is significally slower, 10% of speed of light tops. You will have more that hour to ask. (Note that technically, sun can't go nova, because nova is white dwarf with external source of hydrogen. It can (and will), however, go supernova, which I assume is what Randall means.) -- Hkmaly (talk) 09:19, 12 November 2012 (UTC)

I think the explanation is wrong or otherwise lacking in its explanation: The P-value is not the entire problem with the frequentist's viewpoint (or alternatively, the problem with the p-value hasn't been explained). The Frequentist has looked strictly at a two case scenario: Either the machine rolls 6-6 and is lying, or it doesn't rolls 6-6 and it is telling the truth. Therefore, there is a 35/36 probability (97.22%) that the machine is telling the truth and therefore the sun has exploded. The Bayesian is factoring in outside facts and information to improve the accuracy of the probability model. He says "Either the machine rolls 6-6 (a 1/36 probability, or 2.77%) or the sun has exploded (an aparently far less likely scenario). Given the comparison, the Bayesian believes it is MORE probable that the machine rolled 6-6 than the sun exploded, given the relative probabilities. If the latter is a 1 in a million chance (0.000001%), it is 2,777,777 times more likely that the machine rolled 6-6 than the sun exploded. To borrow a demonstration/explanation technique from the Monty Hall problem, if the machine told you a coin flip was heads, that would be 50% chance of occuring while a 2.7% chance of the machine lying, the probabilities would clearly suggest that the machine was more likely to be telling the truth. Whereas if the machine said that 100 coin flips had all come up heads (7.88x10^-31%). Is it more likely that 100 coin flips all came up heads or is it more likely the machine is lying? What about 1000 coin flips? or 1,000,000? I think the question is, whether one could assign a probability to the sun exploding. Also, I think they could have avoided the whole thing by asking the machine a second time and see what it answered. TheHYPO (talk) 19:09, 12 November 2012 (UTC)