A surprising appearance of e

Here’s a simple probability problem that should be accessible to high school students who have learned the Multiplication Rule:

Suppose that you play the lottery every day for about 20 years. Each time you play, the chance that you win is 1 chance in 1000. What is the probability that, after playing  1000 times, you never win?

This is a straightforward application of the Multiplication Rule from probability. The chance of not winning on any one play is 0.999. Therefore, the chance of not winning 1000 consecutive times is (0.999)^{1000}, which we can approximate with a calculator.


Well, that was easy enough. Now, just for the fun of it, let’s find the reciprocal of this answer.


Hmmm. Two point seven one. Where have I seen that before? Hmmm… Nah, it couldn’t be that.

What if we changed the number 1000 in the above problem to 1,000,000? Then the probability would be (0.999999)^{1000000}.


There’s no denying it now… it looks like the reciprocal is approximately e, so that the probability of never winning for both problems is approximately 1/e.

Why is this happening? I offer a thought bubble if you’d like to think about this before proceeding to the answer.

green_speech_bubbleThe above calculations are numerical examples that demonstrate the limit

\displaystyle \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n = e^x

In particular, for the special case when n = -1, we find

\displaystyle \lim_{n \to \infty} \left(1 - \frac{1}{n}\right)^n = e^{-1} = \displaystyle \frac{1}{e}

The first limit can be proved using L’Hopital’s Rule. By continuity of the function f(x) = \ln x, we have

\ln \left[ \displaystyle \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n \right] = \displaystyle \lim_{n \to \infty} \ln \left[ \left(1 + \frac{x}{n}\right)^n \right]

\ln \left[ \displaystyle \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n \right] = \displaystyle \lim_{n \to \infty} n \ln \left(1 + \frac{x}{n}\right)

\ln \left[ \displaystyle \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n \right] = \displaystyle \lim_{n \to \infty} \frac{ \displaystyle \ln \left(1 + \frac{x}{n}\right)}{\displaystyle \frac{1}{n}}

The right-hand side has the form \infty/\infty as n \to \infty, and so we may use L’Hopital’s rule, differentiating both the numerator and the denominator with respect to n.

\ln \left[ \displaystyle \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n \right] = \displaystyle \lim_{n \to \infty} \frac{ \displaystyle \frac{1}{1 + \frac{x}{n}} \cdot \frac{-x}{n^2} }{\displaystyle \frac{-1}{n^2}}

\ln \left[ \displaystyle \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n \right] = \displaystyle \lim_{n \to \infty} \displaystyle \frac{x}{1 + \frac{x}{n}}

\ln \left[ \displaystyle \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n \right] = \displaystyle \frac{x}{1 + 0}

\ln \left[ \displaystyle \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n \right] = x

Applying the exponential function to both sides, we conclude that

\displaystyle \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n= e^x

green lineIn an undergraduate probability class, the problem can be viewed as a special case of a Poisson distribution approximating a binomial distribution if there’s a large number of trials and a small probability of success.

The above calculation also justifies (in Algebra II and Precalculus) how the formula for continuous compound interest A = Pe^{rt} can be derived from the formula for discrete compound interest A = P \displaystyle \left( 1 + \frac{r}{n} \right)^{nt}

All this to say, Euler knew what he was doing when he decided that e was so important that it deserved to be named.

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