Let's look at the set of the probabilities of having at least one occurrence of an event if we make $n$ attempts, where the probability of the event occurring is $1/n$.

For example, if we are looking at the probability of the occurrence of rolling a $4$ on a die ($1/6$ chance), we will actually make $6$ attempts. The probability of success in that case (at least one occurrence of a $4$ in the $6$ attempts) is $1-(5/6)^{6} = 0.6651$ (I have rounded the results.) If we make $15$ attempts at something with a probability of its happening (on each separate occasion) being $1/15$, then the probability of its occurring at least once in those $15$ attempts is $1-(14/15)^{15} = 0.6447$. If we use $n=1,000,000$ then we get $0.63212$.

So, we see that we have an asymptotic situation. I have always wondered if there was any other known mathematical significance to this specific approximate number/asymptote. Thanks in advance for reading and for all responses!

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    $\begingroup$ Just a remark: indeed, $1-((n-1)/n)^n\to (e-1)/e\approx 0.63212$. $\endgroup$ – anderstood Sep 7 '17 at 1:54
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    $\begingroup$ This inverse symbolic calculator is a tool that may be useful if you ever have a value that you have calculated to many decimal places and are wondering if there is a well-known closed form for it. $\endgroup$ – Zubin Mukerjee Sep 7 '17 at 2:49
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    $\begingroup$ Welcome to the site! Great first post! I hope to see you around on this site again. I recommend also that you take the tour to get a grasp on how this particular site is different from (perhaps better than?) others. ;) $\endgroup$ – Wildcard Sep 7 '17 at 4:33
  • $\begingroup$ Wait, this makes no sense. Are you saying that as we try more times, the probability of success goes down? Furthermore, I'd expect that the probability of not getting a gven value in a million die rolls to be pretty miniscule. $\endgroup$ – DeadMG Sep 7 '17 at 10:47
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    $\begingroup$ Ohh wait, the die sides and the number of rolls are both n at all times? $\endgroup$ – DeadMG Sep 7 '17 at 10:52

Roll an $n$-sided die numbered $1$ through $n$. Roll it $n$ times. Your question is equivalent to the following:

What is the probability that you roll at least one $1$, over $n$ throws of this $n$-sided die?

This is equal to $1$ minus the probability that you get no $1$s during the $n$ throws. The probability of not getting a $1$ on any given throw is $(n-1)/n$, so this gives your desired probability as:

$$1 - \left(\frac{n-1}{n}\right)^n$$

$$1 - \left(1- \frac{1}{n}\right)^n$$

Look at number $1$ in this list of characterizations of the exponential function:

Define $e^x$ by the limit

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

This definition applies to your problem, when we let $x=-1$.

In the limit as $n$ goes to infinity, we can write $$\lim_{n\to\infty} \left[\,1 - \left(1- \frac{1}{n}\right)^n\,\,\right] \,\,=\,\, 1 - \,\lim_{n\to\infty} \,\left(1+\frac{-1}{n}\right)^n\,\,=\,\, \boxed{1 - \frac{1}{e}\,} \,\,\approx\,\, 0.6321$$

  • $\begingroup$ This answer would be better if I hadn't had to click on the link to see the alternative characterization of e that you thought was interesting. Why not just include it in your answer (with a link if you want a reference for people to follow). Its a little annoying to read your answer and have to click out half way to be able to follow it. $\endgroup$ – Chris Sep 7 '17 at 11:03
  • $\begingroup$ @Chris What browser are you using? What OS? If it's Google Chrome then I have suggestions for you, if the OS is Windows I also have suggestions. Let me know :P $\endgroup$ – Zubin Mukerjee Sep 7 '17 at 11:27
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    $\begingroup$ I'm using chrome. The trouble with it is that if sites go down then the links stop working too making answers nonsensical. Any suggestions on how to fix this problem appreciated! ;-) Seriously though I know I can follow the link to see the info, I just think the answer would be better if it were in the answer. On StackOverflow at least we tend to frown on answers that rely on you following the link to get key information. Given the thing you have linked to is the most crucial part of your answer it seems strange not to actually have it in the answer. Your answer though, do what you want. ;-) $\endgroup$ – Chris Sep 7 '17 at 11:38
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    $\begingroup$ @Chris That makes sense. For convenience (and insurance against the possibility of links breaking), I've added an image snippet of what I had linked to. $\endgroup$ – Zubin Mukerjee Sep 7 '17 at 13:34
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    $\begingroup$ The jump from $e = \lim (1+1/n)^n$ to $1/e = \lim (1-1/n)^n$ is nontrivial, and needs an explanation if the target audience is people who didn't even know the first limit. $\endgroup$ – JiK Sep 11 '17 at 9:12

The point is that $$\lim_{n \to \infty} \left(1 - \frac{1}{n}\right)^n = \frac{1}{e}$$ where $e$ is the base of natural logarithms. This is approximately $0.3678794412$.

  • $\begingroup$ There was a typo (missing parentheses) in the original OP, the actual limit is different. $\endgroup$ – anderstood Sep 7 '17 at 1:58
  • $\begingroup$ @anderstood I think the idea is that $1/e$ is the number you subtract from $1$ to get the limit in the question. But it might have been clearer if that had been stated explicitly. $\endgroup$ – David K Sep 7 '17 at 2:28
  • $\begingroup$ @anderstood only by complementation, the point remains taht $\lim\limits_{n\to\infty}(1-{(\frac{n-1}{n})}^n) = 1-\tfrac 1e$. $\endgroup$ – Graham Kemp Sep 7 '17 at 2:29

Others have already pointed out that the probability of at least one success in $n$ independent trials, each succeeding with probability $1/n$, tends to $1-1/e \approx 0.63212$ as $n$ increases. However, it might be instructive to work out how you could determine this limit yourself.

You already know that the exact probability of observing at least one success in $n$ trials equals $1$ minus the probability of observing no success in $n$ trials, which in turn equals the probability of a single trial failing raised to the $n$-th power. Given the assumption that a single trial succeeds with probability $1/n$, it fails with probability $1 - 1/n$, and thus the probability $p_n$ of observing at least one success in $n$ trials is: $$p_n = 1 - \left( 1 - \frac1n \right)^n.$$

In this formula, $n$ appears twice: once in the denominator and once in the exponent. You've already experimentally observed that, as $n$ increases, $p_n$ seems to tend to a finite limit $0 < p_\infty < 1$ that is independent of $n$. Thus we can expect that, in the limit, those $n$'s should somehow cancel out. Our task, then, is to somehow manipulate the expression for $p_\infty = \lim_{n\to\infty} p_n$ into a form where we can algebraically cancel out the $n$'s and obtain a closed-form expression for $p_\infty$.

To do that, we somehow need to bring the $n$'s together. We can't get the $n$ in the exponent down, but we can raise the other $n$ into the exponent too, by replacing $1 - 1/n$ with the equivalent expression $e^{\log_e(1 - 1/n)}$ and applying the rule $(x^a)^b = x^{ab}$ to get: $$p_n = 1 - e^{\textstyle n \log_e(1 - 1/n)}.$$

(Strictly speaking, this reformulation is only valid for $n > 1$, since the expression $1 - 1/n$ needs to be positive for its logarithm to be a well defined real number. But since we're concerned with the limit $n \to \infty$, we can safely ignore any special cases involving small values of $n$.)

Of course, we could've just as well picked any other base for the logarithm instead of $e$ above and obtained an equally valid expression for $p_n$. However, what's special about the natural logarithm $\log_e$ is that the derivative of $\log_e(x)$ at $x=1$ equals $1$. In particular, this means that, for values of $x$ close to $0$, we have the approximation $\log_e(1 + x) = x + o(x)$, where $o(x)$ denotes additional higher-order terms such that: $$\lim_{x\to0} \frac{o(x)}{x} = 0.$$

Letting $x = -1/n$, we get $\log_e(1-1/n) = -1/n - o(1/n)$, which we can substitute into the expression for $p_n$ above to get: $$p_n = 1 - e^{\textstyle n\left(-\tfrac1n - o\left(\tfrac1n\right)\right)} = 1 - e^{\textstyle -1 - n \cdot o\left(\tfrac1n\right)}.$$

As $n$ increases, the $n \cdot o(1/n)$ term in the exponent tends to zero, and therefore: $$p_\infty = \lim_{n\to\infty} p_n = 1 - e^{\textstyle -1} = 1 - \frac1e.$$

The general trick here is that, for $x$ close to $0$, $\log_e(1+x) \approx x$ (and, conversely, $e^x \approx 1+x$). Thus, in particular, $(1+x)^n \approx (e^x)^n = e^{nx}$ when $x \approx 0$.

If $nx$ is also small, we can even further approximate $(1+x)^n \approx 1+nx$. This has an intuitive probabilistic interpretation: if an experiment succeeds with a very small probability $p \approx 0$, then repeating it $n$ times yields a success probability of $1-(1-p)^n \approx 1-e^{-np} \approx np$, i.e. approximately $n$ times higher than for a single experiment.

For larger $n$, the second approximation breaks down as the likelihood of multiple successes becomes non-negligible, but the first approximation $1-(1-p)^n \approx 1-e^{-np}$, which only depends on $p$ being small, still holds. In particular, if $n$ and $p$ are inversely proportional, such that $np = c$, then the probability of at least one success tends to $1-e^{-c}$ as $n$ get larger and $p$ gets smaller.

Another way of looking at this is that we're effectively approximating the binomial distribution of the number of successful trials out of $n$ using a Poisson distribution with the event rate parameter $\lambda = np$. This approximation, which gets better as $n$ gets larger and $p$ gets smaller, then yields a probability of $1 - e^{-\lambda}$ of at least one event occurring.

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    $\begingroup$ +1 for the final paragraphs, which are the first here that do not just determine the limit -- which, pardon everyone, is a basic and easy calculus exercise -- but give some context in probability theory. $\endgroup$ – Torsten Schoeneberg Sep 7 '17 at 19:04

$$1- \left(\frac{n-1}{n}\right)^n=1-\left(1 - \frac{1}{n} \right)^n$$

As we take limit $n \to \infty$,

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

Hence that is why you observe the number that approximate $1-\exp(-1)$

  • $\begingroup$ I'm not sure why this answer is so much less upvoted than the others... $\endgroup$ – anderstood Sep 7 '17 at 14:50

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