>The sub result for 4 simultaneous drive failures in 156 hours = 1.89187284e-13. That means the probability of it NOT happening in 156 hours is (1 – 1.89187284e

>The sub result for 4 simultaneous drive failures in 156 hours = 1.89187284e-13. That means the probability of it NOT happening in 156 hours is (1 – 1.89187284e-13) which equals 0.999999999999810812715 (12 nines). Minor nitpick, this ignores the possibility of more than 4 failures, although this error only affects the fourth digit after the nines. Much more egregious is the following: >there are 56 “156 hour intervals” in a given year This is too simplistic, there are in fact infinitely many 156-hour intervals in a year, some of them just happen to overlap. This overlap can't simply be ignored because even if none of their 56 disjoint intervals contain 4 events this does not rule out the possibility of there being 4 events in some 156 hour interval they didn't take into account. In fact failing to take into account even one of the infinitely many intervals creates a blind spot (consider what happens if the drives happen to fail precisely at the start and end of a particular interval). You can still get a lower bound by e.g. ensuring none of the 56 intervals contain more than 1 failure, or by adding more intervals and ensuring none of them have more than 2 failures etc. Their binomial calculation contains the same mistake. A quick improved lower bound can be obtained by calculating the probability that any failure is followed by (at least) 3 other failures within 156 hours. For one failure this probability is given by the Poisson distribution and is Pc = 1 -\sum_{k<3} e^-λ λ^k / k! = 5.18413e-10. Now we get into some trouble because the failures and the probability of a 'catastrophic' failure are dependent, however the probability that any particular failure turns catastrophic is constant, so the expected number of catastrophic failures can't be greater than the expected number of failures times that constant, this gives a lower bound of Pc (365·24·λ) = 6.63154e-9 this is a lower bound, but that's still three fewer nines left than their claim. Anyway let's just hope their data centres are more reliable than their statistics. Edit: This last calculation can be justified by noting that the probability that 1 critical failure starts in a particular time interval is Pc times the probability of 1 failure in that interval plus some constant times the probability of more than one interval. Similarly the probability of more than one critical failure is at most the probability of more than one failure. Now the probability of more than one failure in a time interval is dominated by the length of the interval, therefore if you calculate the density those parts fall away and you're left with a density of Pc λ critical failures per hour. This seems to be an exact expression for the expected number of critical failures, and not just a lower bound. Although it is still a lower bound for the probability of a critical failure, albeit a fairly tight one.