martin f. krafft: fix FAQ 4b probability (#493577)

[Martin F. Krafft]

Module: mdadm
Branch: build
Commit: b8fab3955776b1d2d948a24351e726fc11ac5d13
URL:    http://git.debian.org/?p=pkg-mdadm/mdadm.git;a=commit;h=b8fab3955776b1d2d948a24351e726fc11ac5d13

Author: martin f. krafft <madduck at madduck.net>
Date:   Wed Oct 15 10:13:23 2008 +0200

fix FAQ 4b probability (#493577)

---

 debian/FAQ |   11 +++++++----
 1 files changed, 7 insertions(+), 4 deletions(-)

diff --git a/debian/FAQ b/debian/FAQ
index e728215..e209823 100644
--- a/debian/FAQ
+++ b/debian/FAQ
@@ -137,8 +137,8 @@ The latest version of this FAQ is available here:
   I am assuming that you are talking about a setup with two copies of each
   block, so --layout=near2/far2/offset2:
 
-  In half of the cases, yes [0], and it does not matter which layout you use.
-  When you assemble 4 disks into a RAID10, you essentially stripe a RAID0
+  In two thirds of the cases, yes[0], and it does not matter which layout you
+  use. When you assemble 4 disks into a RAID10, you essentially stripe a RAID0
   across two RAID1, so the four disks A,B,C,D become two pairs: A,B and C,D.
   If A fails, the RAID6 can only survive if the second failing disk is either
   C or D; If B fails, your array is dead.
@@ -151,8 +151,11 @@ The latest version of this FAQ is available here:
 
   See also question 18 further down.
 
-  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/
+  0. it's actually (n-2)/(n-1), where n is the number of disks. I am not
+     a mathematician, see http://aput.net/~jheiss/raid10/, which gives the
+     chance of *failure* as 1/(n-1), so the chance of success is 1-1/(n-1), or
+     (n-2)/(n-1), or 2/3 in the four disk example.
+     (Thanks to Per Olofsson for clarifying this in #493577).
 
 5. How to convert RAID5 to RAID10?
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