The math gets a little less obvious with something like RAID-5 since there isn't a

*designated* parity drive. The parity data gets distributed among all the drives. But for RAID-5 you'll always need one

*additional* drive to do an array.

Minimum number is three drives. Your data space is the capacity of

all the drives in the array minus the capacity of

one drive in the array. (Note: All the drives in a pure RAID-5 array must have

*identical* capacity.)

Assuming 1TB drives, a three drive array would have a 2TB data space (i.e. 3TB-1TB). So 33% of the total capacity is 'lost' to parity. But a five drive array would have a 4TB data space (i.e 5TB-1TB) with only 20% total capacity lost to parity. And that percentage decreases with each drive added.

So as you can see, RAID-5 is

*least* economical with only three drives - but it becomes

*increasingly* economic with each drive added after that - up to the capacity of the RAID controller.

The tradeoff is that a standard RAID-5 array can only tolerate

one drive failing at a time without experiencing data loss. As

**skwire** pointed out earlier, the general rule is you can have as many drives

*simultaneously* fail as you have parity drives. And in the case of RAID-5, you basically only have one "drive" for parity even though it isn't a specific drive.

In practice, it's a little more complicated than that. But not much.