Consider this simple problem:
If an event has a $1/N$ chance of success, what's the probability of having at least one success after $N$ events?
Basically, what's the chance of getting a $20$ if you roll a D20 $20$ times?
That probability $P(N)$ can be expressed as: $$P(N) = 1 - \left(\frac{N-1}{N}\right)^{N}$$
I found that this seems to converge towards $1-\frac{1}{e}$. Why is that?
I'm having trouble understanding what this means. Why does it converge at all? If something has a one in a million chance of happening and you try a million times, shouldn't your chance to succeed be higher that $63\%$? And why this value specifically?
$\endgroup$ 52 Answers
$\begingroup$We have
$$P(N) = 1 - \left(\frac{N-1}{N}\right)^{N}=1 - \left(1-\frac{1}{N}\right)^{N}=1 -\frac1{ \left(1-\frac{1}{N}\right)^{-N}}\to 1-\frac1e$$
since
$$\lim_{x\to \pm \infty} \left(1+\frac1x\right)^x \to e$$
Refer to the related
$\endgroup$ $\begingroup$This is more generally related to the fact that the Binomial distribution on $N$ events with success probability $\lambda/N$ converges in distribution to a Poisson distribution with rate $\lambda$. In your case, $\lambda=1$, and for Poisson with rate $\lambda$, $1-P_\lambda(X=0)=1-\frac{1}{e^\lambda}=1-1/e$.
A proof can be found here.
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