If you model drive failure as a Poisson Process (each drive flips a coin each day with probability of failure x), then the measurements imply air and helium drives both have about x=2.8e-5. But drive failure rate may not be constant over time, especially if the failure is wear/aging related. The author should consider fitting the data to a https://en.m.wikipedia.org/wiki/Weibull_distribution with k>1.
In particular, k=2 is a model where failure rate increases linearly over time. To estimate the instrinsic drive failure rate for this model, we must look at the raw failure times (not enough info in blog's totals table) and compute the root-mean-square of time between failures. Then divide this by the total number of drives. Do this for the two classes (air vs helium) and see which is better.
> If you model drive failure as a Poisson Process (each drive flips a coin each day with probability of failure x)
IIRC the Poisson distribution applies when the rate is roughly constant per interval of time, which is not exactly the same as a probability of x per interval of time. (Failures are independent in the latter, not in the former).
In particular, k=2 is a model where failure rate increases linearly over time. To estimate the instrinsic drive failure rate for this model, we must look at the raw failure times (not enough info in blog's totals table) and compute the root-mean-square of time between failures. Then divide this by the total number of drives. Do this for the two classes (air vs helium) and see which is better.