# Difference between revisions of "1132: Frequentists vs. Bayesians"

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In the comic, the likelihood that the detector is lying is much higher than the likelihood of the Sun exploding. Therefore, one should conclude that a single "yes" result is probably a false positive. The bit about "p < 0.05" comes from a naive interpretation of {{w|P-value}}, a scientific research standard where a result is presumed to be valid if there is less than a 5% chance that it is merely random coincidence. There is a 1/36 chance of rolling two sixes on 2d6. | In the comic, the likelihood that the detector is lying is much higher than the likelihood of the Sun exploding. Therefore, one should conclude that a single "yes" result is probably a false positive. The bit about "p < 0.05" comes from a naive interpretation of {{w|P-value}}, a scientific research standard where a result is presumed to be valid if there is less than a 5% chance that it is merely random coincidence. There is a 1/36 chance of rolling two sixes on 2d6. | ||

− | The title and the last two frames suggest that "frequentist" interpretation of statistics is somehow wrong, which is not actually the case. The Bayesian interpretation and the frequentist interpretation are not mutually exclusive and neither is wrong. It's the misuse of P values that causes the character in the panel labeled "Frequentist Statistician" to conclude that the sun has exploded, not the use of frequentist statistics. (In particular, P-values are usually used only for numerical values that are known to fall along a specific distribution – in this case, it is used to determine the significance of a discrete event, which is wrong.) | + | The title and the last two frames suggest that "frequentist" interpretation of statistics is somehow wrong, which is not actually the case. The Bayesian interpretation and the frequentist interpretation are not mutually exclusive and neither is wrong. It's the misuse of P values that causes the character in the panel labeled "Frequentist Statistician" to conclude that the sun has exploded, not the use of frequentist statistics. (In particular, P-values are usually used only for numerical values that are known to fall along a specific distribution – in this case, it is used to determine the significance of a discrete event, which is wrong.) In fact, the labels on the bottom two panels were applied as an after-though, according to Munroe's post [http://web.archive.org/web/20130117080920/http://andrewgelman.com/2012/11/16808/#comment-109366 here]; he states his intention was "to illustrate a case where naïve application of that significance test can give a result that’s obviously nonsense." |

[[File:Nate Silver Tweet.png|[email protected]: If you think it's a toss-up, let's bet. If Obama wins, you donate $1,000 to the American Red Cross. If Romney wins, I do. Deal?|right]] | [[File:Nate Silver Tweet.png|[email protected]: If you think it's a toss-up, let's bet. If Obama wins, you donate $1,000 to the American Red Cross. If Romney wins, I do. Deal?|right]] |

## Revision as of 23:55, 23 August 2013

## Explanation

This is another comic about the accuracy of presidential election predictions that used Bayesian statistical models, such as Nate Silver's *538* and Professor Sam Wang's *PEC*. Thomas Bayes studied conditional probability - the likelihood that one event is true when given information about some other related event. From Wikipedia: "Bayesian interpretation expresses how a subjective degree of belief should rationally change to account for evidence".

In the comic, the likelihood that the detector is lying is much higher than the likelihood of the Sun exploding. Therefore, one should conclude that a single "yes" result is probably a false positive. The bit about "p < 0.05" comes from a naive interpretation of P-value, a scientific research standard where a result is presumed to be valid if there is less than a 5% chance that it is merely random coincidence. There is a 1/36 chance of rolling two sixes on 2d6.

The title and the last two frames suggest that "frequentist" interpretation of statistics is somehow wrong, which is not actually the case. The Bayesian interpretation and the frequentist interpretation are not mutually exclusive and neither is wrong. It's the misuse of P values that causes the character in the panel labeled "Frequentist Statistician" to conclude that the sun has exploded, not the use of frequentist statistics. (In particular, P-values are usually used only for numerical values that are known to fall along a specific distribution – in this case, it is used to determine the significance of a discrete event, which is wrong.) In fact, the labels on the bottom two panels were applied as an after-though, according to Munroe's post here; he states his intention was "to illustrate a case where naïve application of that significance test can give a result that’s obviously nonsense."

The final panel is a tongue-in-cheek reference to the absurdity of the premise. If the sun did explode, he won't need to pay out the bet because the Earth and everyone on it would be destroyed. But if it didn't explode, then he'll win $50. It also likely refers to a well-publicized bet that Nate Silver tried to make with Joe Scarborough regarding the outcome of the election:

The title text refers to a classic series of logic puzzles known as Knights and Knaves, where there are two guards in front of two exit doors, one of which is real and the other leads to death. One guard is a liar and the other tells the truth. The visitor doesn't know which is which, and is allowed to ask one question to one guard. The solution is to ask either guard what the other one would say is the real exit, then choose the opposite. Two such guards were featured in the 1986 Jim Henson movie *Labyrinth*, which is referenced in the text.

## Transcript

- Did the sun just explode? (It's night, so we're not sure)

- [Two statisticians stand alongside an adorable little computer that is suspiciously similar to K-9 that speaks in Westminster typeface]
- Frequentist Statistician: This neutrino detector measures whether the sun has gone nova.
- Bayesian Statistician: Then, it rolls two dice. If they both come up as six, it lies to us. Otherwise, it tells the truth.
- Frequentist Statistician: Let's try. [to the detector] Detector! Has the sun gone nova?
- Detector:
*roll*YES.

- Frequentist Statistician:
- Frequentist Statistician: The probability of this result happening by chance is 1/36=0.027. Since p<0.05, I conclude that the sun has exploded.

- Bayesian Statistician:
- Bayesian Statistician: Bet you $50 it hasn't.

**add a comment!**⋅

**add a topic (use sparingly)!**⋅

**refresh comments!**

# Discussion

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)

- 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. 50.0.38.245 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.

- 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

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. -- 171.64.68.120 (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)

This is actually an example of the Base rate fallacy. --71.199.125.210 04:04, 19 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

It's cute to attempt to connect this to the U.S. presidential election, but it's far likelier that it's a reference to Enrico Fermi taking bets at the Trinity test site as to whether or not the first atomic bomb would cause a chain reaction that would ignite the entire atmosphere and destroy the planet. I'll bet you $50 it is. 71.229.88.206 21:29, 7 March 2013 (UTC)

I don't like the explanation at all. Some of the discussion posts give a good view on this. I'd like to share my thought about the last panel, though. The page reads as if the punch line is about the fact that you cannot spend the money if the sun was going to explode; but why does the bayesian propose this bet and not the frequentist - no reason for this. I think there is a better explanation for this panel: there are several proofs that bayesian probabilities result in "rational" behaviour: They state that if you act according to bayes' rule you cannot be cheated in betting. 108.162.254.179 17:11, 6 March 2014 (UTC)

The last panel may refer to Nate Sliver's view expressed in his book The Signal and the Noise that if one believes one's prediction to be true one should be confident to bet on it. --Troy0 (talk) 18:46, 6 July 2014 (UTC)

Please excuse my ignorance, but how is two sixes rolled on fair dice 31/32? (In the explanation: "the detector is telling the truth (31 in 32)") --Pudder (talk) 17:06, 9 December 2014 (UTC)

- Just a missreading, not stupid. The detector is telling the truth when you dont role 2 sixes. roling 2 sixes is 1/6 * 1/6 or 1/36. So not roling is 35 in 36, wait oops 36 not 32, thanks. 108.162.216.209 17:39, 9 December 2014 (UTC)

I have always thought that the suggested Bet is also a reference to the Dutch Book argument for judging and accounting for probabilities underlying Bayesian interpretations of probability theory. 141.101.98.100 22:11, 12 August 2015 (UTC)

The likelyhood of a solar explosion may be wrong. Since the detector I'd only used at night, the event is twice as likely to occur than listed. That said, there's a 50% chance of the event never being detected, so I'm not sure. Any one more knowledgeable than I care to comment? Mikemk (talk) 06:47, 5 April 2016 (UTC)

Huh. I thought that the last panel was pragmatism: "If the sun goes nova, $50 doesn't matter; I'll be dead. If the sun hasn't, I get $50!"