Bug#493577: Wrong probability of survival in FAQ 4b

[Per Olofsson]

Package: mdadm
Version: 2.6.7-3
Severity: minor

Hi,

/usr/share/doc/mdadm/FAQ.gz question 4b says:

"4b. Can a 4-disk RAID10 survive two disk failures?
[...]
In half of the cases, yes [0],
[...]
0. it's actually 1/(n-1), where n is the number of disks. I am not
   a mathematician, see http://aput.net/~jheiss/raid10/"

I do not think any of these numbers are correct. If you have a
four-disk RAID10 array, the probability of it surviving two disk
failures should be 2/3.

The reason is simple. Let's use the same example as in the FAQ: you
have two pairs of disks, A,B and C,D. Disk A fails. There are now
three disks left that can fail (three possibilities). In the case that
either C or D fails, the array survives (two possibilities). That
gives a survival chance of exactly 2/3.

I think the number from the web page quoted is the chance of the array
*failing*. Indeed, it says:

"The chance of system failure in a RAID 1+0 system with two drives per
mirror is 1/(n - 1). "

-- 
Pelle