Why I Say 1/e About This Roller Coaster

The Leap-The-Dips at Lakemont Park, Altoona, Pennsylvania, as photographed by Joseph Nebus in July 2013 from the edge of the launch platform.

So in my head I worked out an estimate of about one in three that any particular board would have remained from the Leap-The-Dips’ original, 1902, configuration, even though I didn’t really believe it. Here’s how I got that figure.

First, you have to take a guess as to how likely it is that any board is going to be replaced in any particular stretch of time. Guessing that one percent of boards need replacing per year sounded plausible, what with how neatly a chance of one-in-a-hundred fits with our base ten numbering system, and how it’s been about a hundred years in operation. So any particular board would have about a 99 percent chance of making it through any particular year. If we suppose that the chance of a board making it through the year is independent — it doesn’t change with the board’s age, or the condition of neighboring boards, or anything but the fact that a year has passed — then the chance of any particular board lasting a hundred years is going to be 0.99^{100} . That takes a little thought to work out if you haven’t got a calculator on hand.

Squaring 0.99 is easy; it’s 0.98, near enough. For that matter, the first couple powers of 0.99 are easy: 0.99 to the third power is about 0.97, to the fourth about 0.96, and so on. It’s only around the twentieth power that this approximation really breaks down — 0.99 to the twentieth is about 0.82 — and even then, it breaks only gradually. But 0.99 to the hundredth power, well, we need a better approximation than just “zero”.

Happily that better approximation is right on hand. One really good formula for evaluating e, roughly 2.71828, and that anyone who really gets into mathematics memorizes, is that it’s \lim_{N \rightarrow \infty} \left(1 + \frac{1}{N}\right)^{N} , which roughly means that as N gets larger and larger, the difference between the number you get by evaluating \left(1 + \frac{1}{N}\right)^{N} and the number e gets as small as you need it to be. You’ll get a pretty good approximation to e if you just evaluate \left(1 + \frac{1}{N}\right)^{N} with a big enough value of N. Using an N of 20 gives you 2.65; by the time N reaches 50, the non-limit version gets you to 2.69, and after that you’re just fiddling with decimal points some. An N of 100 gets you to 2.70.

This would be fine, if we wanted to figure out a number a little bigger than 1 raised to a pretty big power, but careful examination shows that 0.99 is in fact a little smaller than 1. No problem, though: if you get a little bit more into mathematics you get used to the formula \lim_{N \rightarrow \infty} \left(1 + \frac{a}{N}\right)^{N} = e^{a} , which I’m sure you’ll trust me on rather than insisting on seeing proved.

Since 0.99 is 1 - \frac{1}{100} , or if you prefer, 1 + \frac{-1}{100} , then, 0.99 to the hundredth power is going to be tolerably close to e^{-1}, and that’s 1/e. Since e is a little less than three, 1/e is a little more than one-third. (It’s about 0.37, or tolerably close to 0.4.)

This sort of problem — and this 1/e — turns up a lot in mathematics problems where the chance of whatever you’re interested in has a pretty low shot at happening any one chance, but a lot of chances to happen.

Next up: why I don’t actually believe this estimate.