Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Path: utzoo!mnetor!uunet!seismo!husc6!rutgers!ucla-cs!zen!ucbvax!degas.Berkeley.EDU!wallace From: wallace@degas.Berkeley.EDU (David E. Wallace) Newsgroups: sci.philosophy.tech Subject: Re: The nature of knowledge (probabilities) Message-ID: <19843@ucbvax.BERKELEY.EDU> Date: Thu, 23-Jul-87 21:14:21 EDT Article-I.D.: ucbvax.19843 Posted: Thu Jul 23 21:14:21 1987 Date-Received: Sat, 25-Jul-87 10:52:18 EDT References: <3587e521.44e6@apollo.uucp> <680@gargoyle.UChicago.EDU> Sender: usenet@ucbvax.BERKELEY.EDU Reply-To: wallace@degas.Berkeley.EDU.UUCP (David E. Wallace) Distribution: world Organization: University of California, Berkeley Lines: 36 Keywords: knowledge belief truth certainty In article <2099@mulga.oz.R> lee@mulga.UUCP (Lee Naish) writes: >Suppose each page of the book was simply a list of 100 numbers >which (should) add up to 1000. Suppose also that the book source >was on-line and with the appropriate tools all the numbers added >by the computer and the result was 349999. The probability of there >being an error is extremely high (say 0.999). What do you believe is >the probability of an error on any given page? If you say 1/350 then >the probability of an error in the book should be, according to >simple probability theory, 1-(349/350)^350 = 0.63. If you say 10/350 ^^^^^^^^^^^^^^^ This is the problem: see below. >(or whatever is needed to get the 0.999 figure) then the expected >number of errors greatly increases (which I think is unreasonable). > >How can this paradox be resolved without admitting inconsistent >beliefs? Simple: the formula you cite only applies if the probabilities are independent. The global knowledge you possess of the overall sum means that the probabilities of errors on the separate pages are not independent, so the formula doesn't apply. To take a somewhat cleaner example, if I have 350 identical sealed boxes on the table and tell you that there is a red ball in one (and only one) of the boxes, the probability that there is a red ball in any given box is clearly 1/350, before any of the boxes have been inspected. But the probability that there is a red ball in *some* box (assuming you can trust the conditions of the problem) is 1, not 0.63. If you now open one of the boxes, the probability that there is a red ball in the *second* box you inspect will either rise to 1/349 (if you find the first box empty) or drop to zero (if you find the ball in the first box), because the probabilities are not independent. For them to be independent, the probability of finding a ball in the second box would have to remain the same regardless of what you found in the first one. Dave Wallace UUCP: ...!ucbvax!wallace ARPA: wallace@degas.Berkeley.EDU