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/ 184.108.40.206 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)
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
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)
- Our sun will not go supernova, as it has insufficient mass. It will slowly become hotter, rendering Earth uninhabitable in a few billion years. In about 5 billion years it will puff up into a red giant, swallowing the inner planets. After that, it will gradually blow off its lighter gasses, eventually leaving behind the core, a white dwarf. 220.127.116.11 01:58, 15 November 2012 (UTC)
- Please don't edit others' comments on talk pages; it's considered quite rude. On a talk page, discourse is meant to be conducted, by editors for the betterment of the article. For constructive discourse to occur, a person's words must be left in tact. The act of censorship hurts the common goal of betterment. Per Wikipedia, the authoritative source on how a wiki works best: "you should not edit or delete the comments of other editors without their permission." lcarsos_a (talk) 17:38, 13 November 2012 (UTC) Note: much of this conversation has been removed at the request of the authors.
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)
Another source of explanation: http://stats.stackexchange.com/questions/43339/whats-wrong-with-xkcds-frequentists-vs-bayesians-comic --JakubNarebski (talk) 20:12, 12 November 2012 (UTC)
The P-value really has nothing to do with it. If I think that there is a 35/36 chance that the sun has exploded, then I should we willing to take any bet that the sun has exploded with better than 1:35 odds. For example, if someone bets me that the sun has exploded in which they will pay me $2 if the sun has exploded and I will pay them $35 if it hasn't, then based on my belief that the sun has exploded with 35/36 probability, then my expected value for this bet is 2*35/36 - 35 * 1/36 = 35/36 dollars and I will take this bet. Clearly I would also take a bet with 1:1 odds - my estimated expected value in the proposed bet in the comic would be 50*35/36 - 50 * 1/36 = $49 (approximately), and I would for sure take this bet. The Bayesian on the other hand has a much lower belief that the sun has exploded because he takes into account the prior probability of the sun exploding, so he would take the reverse side of the bet. The difference is that the Bayesian uses prior probabilities in computing his belief in an event, whereas frequentists do not believe that you can put prior probabilities on events in the real world. Also note that this comic has nothing to do with whether people would die if the sun went nova - the comic is titled "Frequentists vs Bayesians" and is about the difference between these two approaches. -- 18.104.22.168 (talk) (please sign your comments with ~~~~)
The Labyrinth reference reminds me of an old Doctor Who episode (Pyramid of Mars), where the Doctor is also faced with a truthful and untruthful set of guards. Summarized here: http://tardis.wikia.com/wiki/Pyramids_of_Mars_(TV_story) Fermax (talk) 04:49, 14 November 2012 (UTC)
People have gone over this already, but just to be a bit more explicit: Let NOVA be the event that there was a nova, and let YES be the event that the detector responds "Yes" to the question "Did the sun go nova?" What we want is P(NOVA|YES)=P(YES|NOVA)*P(NOVA)/P(YES) Suppose P(NOVA)=p is the prior probability of a nova. Then P(YES|NOVA)=35/36, P(NOVA)=p, and P(YES)=p*35/36+(1-p)*1/36=1/36+34/36 So then P(NOVA|YES)=35p/(1+34p). If p is small, then P(NOVA|YES) is also small. In particular, the Bayesian statistician wins his bet at 1:1 odds if p<1/36, which is probably the case. If the Bayesian statistician wants 95% confidence that he'll win his bet, then he needs p<1/666. =P