vdev/pool math with combined raidzX vdevs...

[Martin Simmons]

>>>>> On Fri, 6 Jul 2012 03:08:46 -0400, Zaphod Beeblebrox said:
> 
> Is there some penalty for not googling some basic stats course?  OK.
> This is from memory (hint: you probably should google).
> 
> p(f) ... the probably of failure of one drive over some unit time (say
> one year).  A two drive RAID-0 array has probability p(2dr0) = 2 *
> p(f) + p(f).  That is (for the logic guys): the array fails if either
> drive fails.

This looks wrong to me because it can be > 1 if p(f) > 1/3 :-)

I think it should be

p(2dr0) = (2 * p(f) * (1 - p(f))) + (p(f) * p(f))

I.e. the probability of either drive failing and the other one not failing
plus the probability of both drives failing simultaneously.

Or another way of thinking about it:

p(2dr0) = 1 - ((1 - p(f)) * (1 - p(f)))

I.e. the inverse of the probability of neither drive failing

__Martin