One of my stats students stopped by this afternoon with an interesting question: she needed a new hard drive, and was looking at this one, which boasted an Annual Failure Rate of 0.34%. What she was wondering was how to calculate the chances that the hard drive would fail within, say, 5 or 10 years. Her first guess had been to multiply 0.34% by that number of years, but she didn’t think that was actually the way to go.
My first thought was to break it down into the chances that the hard drive would fail in one year [(0.0034)] and then add on the chances that it didn’t fail the first year but did the second [(0.9966)(0.0034)] and then add on the chances that it didn’t fail the first two years but did the third [(0.9966)2(0.0034)] etc., all the way up to adding on the chances that it didn’t fail the first (N-1) years but did in the Nth year [(0.9966)N-1(0.0034)].
Then I added them all up, and factored out the (0.0034), giving (0.0034)[1+(0.9966)+(0.9966)2+…+(0.9966)N-1]. But 1+x+x2+…+xN-1=(1-xN)/(1-x), so that equation simplified to (0.0034)[1-(0.9966)N]/[1-0.9966],
which further simplified to 1-(0.9966)N. And when I thought about it, that made sense: (0.9966)N is the probability that the hard drive does NOT fail anytime in the first N years, so the probability that it does fail would be 1-(0.9966)N.
Then we started looking at a few actual numbers. The likelihood that the hard drive fails within one year is 0.34%, that it fails within two years is 0.68%, that it fails within three years is 1.02%. Indeed, since 0.9966 is so close to 1 it has a very small effect so the student’s initial thought of N·(0.34%) turns out to be a pretty good estimate for quite a few years [at 10 years out the theoretical probability would be 3.35% compared to 3.4%, and 20 years out the theoretical probability is 6.59% compared to 6.8%.]
This lead me to wondering about another piece of data she’d pointed out: the Mean Time Before Failure (MTBT) was listed as 700,000 hours, which translates to about 79.85 years of continuous play. And where do they get THAT number from, I ask you? Certainly not experimental data. A later thought I had was that it was the number of years before the likelihood of disk failure was 50%, but 1-(0.9966)79.85≈0.238, so there should be less than a 25% chance that a given hard drive fails within 79.85 years. Indeed, 1-(0.9966)N=.5 when (0.9966)N=.5, which happens when N=ln(.5)/ln(0.9966)≈203 years. Hey, why aren’t stores advertising that data? Shouldn’t the Antiques Roadshow have featured a few Revolutionary War hard drives, bought for 25¢ at the neighborhood garage sale?
But then when I was showing a rough draft of this post to TwoPi, he pointed out that that’s the median, not the mean. The mean would be an expected value, which would be the infinite sum ΣN(0.0034)(0.9966)N-1, which is (0.0034)·ΣN(0.9966)N-1. Thinking back to Calc II, the geometric series ΣxN converges to 1/(1-x) for |x|<1; taking the derivative of both sides yields ΣNxN-1 converging to 1/(1-x)2 when |x|<1. This means that (0.0034)·ΣN(0.9966)N-1 just equals (0.0034)·1/(1-0.9966)2, or 1/(0.0034)≈294 years.
OK, all that work and it didn’t clear up anything except to indicate that George Washington could have gotten a hard drive for his birthday and it might still be working.
All in all, I think we got a good handle on the theory and absolutely none on the reality. She sent me a link later this afternoon to this article on a study by Carnegie-Mellon that indicated that the Mean Time to Failure Rates were greatly exaggerated by manufacturers (ya think?), up to 15 times as long as they should be, because they failed to take into account that the #1 cause of failure is age: the likelihood that a hard drive will fail in its 10th year is a lot higher than the likelihood that it will fail in its 2nd year.
In other words, it doesn’t really matter if they say the drive will last 80 years or 800 years, because it won’t. But it’s still worthwhile to buy a new hard drive when the old one goes belly up.
Thanks to Anya for the idea and for the links in this post!