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!
Best Answer
Roll an $n$-sided die numbered $1$ through $n$. Roll it $n$ times. Your question is equivalent to the following:
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:
This definition applies to your problem, when we let $x=-1$.
In the limit as $n$ goes to infinity, we can write $$\lim_{n\to\infty} \left[\,1 - \left(1- \frac{1}{n}\right)^n\,\,\right] \,\,=\,\, 1 - \,\lim_{n\to\infty} \,\left(1+\frac{-1}{n}\right)^n\,\,=\,\, \boxed{1 - \frac{1}{e}\,} \,\,\approx\,\, 0.6321$$