*To*: debian-user@lists.debian.org*Subject*: Re: backup archive format saved to disk*From*: Mike McCarty <Mike.McCarty@sbcglobal.net>*Date*: Thu, 07 Dec 2006 08:41:23 -0600*Message-id*: <[🔎] 45782813.6030703@sbcglobal.net>*In-reply-to*: <[🔎] 20061206194450.GK24701@localhost.localdomain>*References*: <[🔎] 20061204202016.GB3777@pluto> <45754189.2080705@physik.blm.tu-muenchen.de> <[🔎] 20061205220135.GA1410@pluto> <[🔎] 4576CB1D.300@physik.blm.tu-muenchen.de> <[🔎] 20061206164344.GE30683@localhost.localdomain> <[🔎] 457715CA.1040003@sbcglobal.net> <[🔎] 20061206194450.GK24701@localhost.localdomain>

Andrew Sackville-West wrote:

On Wed, Dec 06, 2006 at 01:11:06PM -0600, Mike McCarty wrote:Andrew Sackville-West wrote:

[Mike wrote]

Let p be the probability of failure of each disc, independently of the other. There are four mutually independent events which comprise the space. Both discs may fail [Pr = p^2]. The first disc may fail, while the second does not [Pr = p(1-p)]. The second disc may fail, while the first does not [Pr = (1-p)p]. Both discs may survive [Pr = (1-p)(1-p)]. So, the probability that at least one disc fails is 1-(1-p)(1-p). For p = 0.01, that is 0.0199.--------------------^^^^ you aren't using an old pentium are you? ;-)

If p = 0.01, then (1-p) = 0.99, and (1-p)(1-p) = 0.9801, so that 1 - (1-p)(1-p) = 0.0199 exactly.

I'll grant you this is not markedly different from 2%, but it is also not simply 2p.huh. been a long time since math class for me. I was under the impression that the probability was 2p. chance of one failing is added to the chance of the other failing give the chance of both failing. But I'll take your word for it :)

If the probability of the first one failing is 60% and the probability of the second one failing is 60%, then I suppose you believe that the probability of one or the other or both failing is 120%. The probability in this case works out to 1-(0.4)(0.4) = 0.84. The way to think of it is that it is 1 - Pr(none fails). This extends to any number of independent events. The problem with trying to add those probabilities is that the events are not disjoint, so the probabilities do not add. You can however decompose it into events A, B, and C which are disjoint, as A = {disc one fails, but disc two does not}, B = {disc two fails, but disc A one does not} and C = {both discs fail}. This is all the ways one or more discs fail, and the events are disjoint. Pr(A) = p(1-p) Pr(B) = (1-p)p Pr(C) = p^2 So, Pr(at least one disc fails) = p - p^2 + p - p^2 + p^2 = 2p - p^2 This technique gets more complicated as the number of events increases. NB: 1 - (1-p)^2 = 1 - (1 - 2p + p^2) = 2p - p^2, as computed above. [snip]

I don't follow this reasoning. The probability of both discs failing (if they do so independently) is not 0.1%, but rather 0.01%. A partially failed disc is usually easy to detect, since they have FEC on them. A completely failed disc is even easier to detect :-)yeah, that was me shuffling decimals before coffee. I was doing p^2 (.01*.01=.0001 and turning that into .1%). I should know better than to think that early in the morning.

:-) Mike -- p="p=%c%s%c;main(){printf(p,34,p,34);}";main(){printf(p,34,p,34);} This message made from 100% recycled bits. You have found the bank of Larn. I can explain it for you, but I can't understand it for you. I speak only for myself, and I am unanimous in that!

**Follow-Ups**:**Re: backup archive format saved to disk***From:*Andrew Sackville-West <andrew@farwestbilliards.com>

**References**:**backup archive format saved to disk***From:*Douglas Tutty <dtutty@porchlight.ca>

**Re: backup archive format saved to disk***From:*Douglas Tutty <dtutty@porchlight.ca>

**Re: backup archive format saved to disk***From:*Johannes Wiedersich <johannes@physik.blm.tu-muenchen.de>

**Re: backup archive format saved to disk***From:*Andrew Sackville-West <andrew@farwestbilliards.com>

**Re: backup archive format saved to disk***From:*Mike McCarty <Mike.McCarty@sbcglobal.net>

**Re: backup archive format saved to disk***From:*Andrew Sackville-West <andrew@farwestbilliards.com>

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