On Wed, Dec 06, 2006 at 01:11:06PM -0600, Mike McCarty wrote:
> Andrew Sackville-West wrote:
> >On Wed, Dec 06, 2006 at 02:52:29PM +0100, Johannes Wiedersich wrote:
> >
> >>Question: how likely is it that both disks develop bad blocks, while
> >>none of them is damaged? I'm no expert on this, but I guess a better
> >>strategy might be to rotate backups on two disks, and use (and check:
> >>fsck and smartctl) them reguarly.
> >
> >
> >
> >if the chance of a disk failure is (say) 1% in the time alloted, then
> >the chance of having a failure with disks is 2%. THe change of any one
>
> I don't follow this reasoning. Are you presuming independence of the
> failures and identical probabilities? If so, then this is the way to
> compute it:
>
> 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? ;-)
> 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 :)
>
> >particular disk failing is still 1%, it the odds of A failure in the
> >system as a whole that goes up. So with more disks you're more likely
> >to have failures of some kind, but the per disk failure stays the same
> >and the odds of losing ALL of them goes the other way. The odds of
> >losing BOTH disks is .1%. the question becomes, which one has
> >failed... ^^
^^
-------------------------^^
> 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.
A
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