RAID 6 with eight 8 TB drives: double parity and the odds across eight bays

RAID 5 across these eight bays would give 56 TB, one more drive's worth. RAID 6 gives that drive up in exchange for surviving a second failure, and at eight bays that's the trade you're making.

Failure odds rise with every drive you add. At an assumed 2% yearly failure rate per drive, the chance that at least one of eight drives fails within a year is 14.92%. Four drives at that rate give 7.76%. That's an assumption for planning, not a measured rate for any drive.

A single failure leaves seven survivors, and the rebuild reads all of them, 56 TB. On a quiet array, one 8 TB drive takes about 14 hours at best. At a spec rate of 1 in 10^14 bits, the chance of at least one unrecoverable read error during that read is at most about 98.87%, and at 1 in 10^15 bits it's at most about 36.11%. A spec rate is a ceiling, so treat both as worst cases, not forecasts.

ZFS reaches two drives' worth of parity with RAID-Z2. The OpenZFS docs say the raidz2 type "specifies a double-parity raidz group." On paper, RAID-Z2 gives the same 48 TB, since it keeps two drives' worth of parity too.

A RAID 10 on the same eight drives gives less space, 32 TB, and it survives a random second failure 85.7% of the time. It guarantees only 1 failure.