# Benja_Fallenstein comments on How Many LHC Failures Is Too Many? - Less Wrong

12 20 September 2008 09:38PM

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Comment author: 21 September 2008 09:11:10PM 2 points [-]

Okay, it scares me when I realize that I've been getting probability theory wrong, even though I seemed to be on perfectly firm ground. But I'm finding that it's even more scary that even our hosts and most commenters here seem to be getting it backwards -- at least Robin; given that the last question in the post seems so obviously wrong for the reasons pointed out already, I'm starting to wonder whether the post is meant as a test of reasoning about probabilities, leading up to a post about how Nature Does Not Grade You On A Curve (grumble :)). Thanks to simon for pointing out the flaw -- I didn't see it myself.

Since simon's explanation is apparently failing to convince most other people here, let me try my own:

As Robinson points out, there are two underlying events. (A): The laws of physics either mean that a working LHC would destroy the world, or that it wouldn't; let p_destroyer denote our subjective prior probability that it would destroy the world. (B): Either something random happens that prevents the LHC from working, or it doesn't. There is an objective Born probability here that a randomly chosen Everett branch of future Earth at date X will have had a string of failures that kept the LHC from working. We should really consider a subjective probability distribution over these objective probabilities, but let us just consider the resulting subjective probability that a randomly chosen Everett branch will not have had a string of failures preventing LHC from working -- call it p_works.

Now, at date X, in a randomly chosen Everett branch, there are four possibilities:

1. The LHC would destroy Earth, and it fails to operate; p = p_destroyer * (1 - p_works). 2. The LHC would destroy Earth, and it works. p = p_destroyer * p_works. 3. The LHC would not destroy Earth, and it fails to operate. p = (1 - p_destroyer) * (1 - p_works). 4. The LHC would not destroy Earth, and it works. p = (1 - p_destroyer) * p_works.

Now, we cannot directly observe whether he LHC *would* destroy Earth if turned on; what we actually can "observe" in a randomly chosen Everett branch at date X is which of the following three events is true:

i. The LHC is turned on and working fine. (Aka "case 4") ii. The LHC is not turned on, because there has been a string of random failures. (Aka "case 1 OR case 3") iii. Earth is gone. (Aka "case 2")

Of course, in case iii aka 2, we are not actually around to observe -- thus the scare quotes around "observe."

simon's argument is that if we observe case ii aka "1 OR 3" aka "a string of random failures has prevented the LHC from working up to date X", then our posterior probability of "The LHC would destroy Earth if turned on" is equal to our prior probability of that proposition (i.e., to p_destroyer):

p(case 1 OR case 3) = p(case 1) + p(case 3) = p_destroyer * (1 - p_works) + (1 - p_destroyer) * (1 - p_works) = 1 - p_works

p(case 1 | the LHC would destroy Earth) = p(the LHC would destroy Earth AND it fails to operate | the LHC would destroy Earth) = 1 - p_works p(case 3 | the LHC would destroy Earth) = p(the LHC would NOT destroy Earth AND it fails to operate | the LHC WOULD destroy Earth) = 0 p(case 1 OR case 3 | the LHC would destroy Earth) = p(case 1 | the LHC would destroy Earth) + p(case 3 | the LHC would destroy Earth) = (1 - p_works + 0) = 1 - p_works

p(the LHC would destroy Earth | case 1 OR case 3) = p(case 1 OR case 3 | the LHC would destroy Earth) * p(the LHC would destroy Earth) / p(case 1 OR case 3) = (1 - p_works) * p_destroyer / (1 - p_works) = p_destroyer

- Benja