What is the theorem that approximate a "success" 1 out N probability tried N times, 2N times, etc, for not succeeding at least once?

Question

That is, if the probability of success is 1/450, so failing means 449/450, then if we try it 450 times, then the chance of failing (not succeeding even one time) is:

$$ (449/450) ^ {450} = 0.367470307338961$$

and it can be approximated by

$$ e^{-1} = 0.367879441171442 $$

and if we try 900 times, it is:

$$ (449/450) ^ {900} = 0.13503442677579 $$

and since $ 900 $ is $ 450 \times 2 $, so we approximate using:

$$ e^{-2} = 0.135335283236613 $$

And it doesn't have to be $ 450 $. It can be $ 200, 700, 800, $ or $ 1200 $.

Is there a name of this theorem? And perhaps its origin of how it got found out?

Answer 1

This is already expressed in J. W. Tanner's comment.

By Taylor's theorem, $e^x = 1 + x + O(x^2)$. Therefore, for a fixed integer $k$, $$ \lim_n \left(1 - k/n \right)^n = \lim_n \left( e^{-k/n} + O((k/n)^2) \right)^{n} = e^{-k} + \lim_n \sum_{j = 1}^n O((k/n)^{2j}) e^{-(n-j)k/n} = e^{-k}. $$ Moreover, since $1 - k/n = (n - k)/n$, we can interpret $(1 - k/n)^n$ as the probability of not succeeding even once in an experiment with $n$ trials and success probability $k/n$.

The above suggests that when $n$ is large, $$\left(1 - k/n\right)^n \approx e^{-k}.$$

In your example, $k=1$, but you can take $k=2$ or $k=3$ (etc.) to arrive at $e^{-2}$ or $e^{-3}$.

Answer 2

The short answer is that it is Bernoulli Trails. If the probability of succeeding per trial is $1/n$, then not succeeding once after $n$ trails is $e^{-1}$. So using this fact, if you try it $2n$ times and not succeeding, then it is

$$ e^{-1} \times e^{-1} = \frac 1 e \times \frac 1 e = \frac 1 {e^2} = e^{-2} $$

and you can use similar calculations if it is $3n$, $4n$ times, and so forth.

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If you are interested in how it can be derived:

First, the original definition of Euler's number $e$ is:

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

by Jacob Bernoulli in 1683 (although he didn't use the letter $e$). Euler started to use the letter $e$ for the constant in 1727 or 1728.

And it can be derived, or some reference even states that this is the definition of $e^x$:

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

Now we can let $x = -1$ and it becomes $$\begin{align} e^{-1} & = \lim_{n \to \infty} \left(1+\frac {-1} n\right)^n \\ & = \lim_{n \to \infty} \left(1-\frac {1} n\right)^n \\ \end{align}$$

So if we use a somewhat large number $n$, such as $450$:

$$\begin{align} e^{-1} & \approx \left(1-\frac {1} {450}\right)^{450} \\ & = \left(\frac {450} {450} -\frac {1} {450}\right)^{450} \\ & = \left(\frac {450 - 1} {450}\right)^{450}\\ & = \left(\frac {449} {450}\right)^{450}\\ \end{align}$$

Now note the above is exactly: the probably of not succeeding even once out of 450 trials, when the probability of succeeding in one try is $\frac 1 {450}$.

What about trying 900 times? Then since

$$ \left(\frac {449} {450}\right)^{450} \approx e^{-1} $$

If we square both sides:

$$\begin{align} \left( \left(\frac {449} {450}\right)^{450}\right)^2 & \approx \left(e^{-1}\right)^2\\ \left(\frac {449} {450}\right)^{900} & \approx e^{-2}\\ & = \frac 1 {e^2}\\ \end{align}$$

and similar calculations can be done for $450 \times 3 = 1350$ tries: $$ e^{-3} \text{ or } \frac 1 {e^3} $$ or $450 \times 4 = 1800 $ tries: $$ e^{-4} \text { or } \frac 1 {e^4}$$

In fact, one of the applications of Euler's number, besides the original study of Compound Interest at the time done by Bernoulli, is Bernoulli Trails related to probability. If the probability of succeeding per trial is $1/n$, then not succeeding once after $n$ trails is $e^{-1}$. So using this fact, if you try it $2n$ times and not succeeding, then it is

$$ e^{-1} \times e^{-1} = \frac 1 e \times \frac 1 e = \frac 1 {e^2} = e^{-2} $$