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 -- To UNSUBSCRIBE, email to [EMAIL PROTECTED] with a subject of "unsubscribe". Trouble? Contact [EMAIL PROTECTED]