63% chance of event happening over repeated attempts This question is of interest:

If there is a $1 / x$ chance of something happening, in $x$ attempts, for
  large numbers over $50$ or so, the likelihood of it happening is about
  $63\%$. If there's a $1$ in $10\,000$ chance of getting hit by a meteor if you
  go outside, if you go outside $10\,000$ times, you have a $63\%$ chance of
  getting hit with a meteor at some point. If there's a $1$ in a million
  chance of winning the lottery and you buy a million (random) lottery
  tickets, you have a $63\%$ chance of winning.

What is the main idea underlying this property?
 A: Since the probability of each individual event is small, and the events are independent, these scenarios are well-approximated by a Poisson distribution. If the expected number of times an event occurs in some sample is $\lambda$, then the probability that the number of events that occurs is $k$ is
$$P(k) \approx \frac{\lambda^k}{k!} e^{-\lambda} .$$
If the probability of an individual event occurring is $1 / N$ and our sample consists of $N$ trials, the expected number of events is $\lambda = N \cdot \frac{1}{N} = 1$, so the probability of $k$ events occurring among $N \gg k$ trials is
$$P(k) \approx \frac{1}{e \cdot k!} .$$
In particular, the probability of at least one event occurring is
$$P(\{1, 2, \ldots\}) = 1 - P(0) \approx 1 - \frac{1}{e} = 0.63212 \ldots .$$
As you've noticed, this approximation is good even for modest $N$: For $N = 10$, the actual value is $0.65132\ldots$, and for $N = 100$, it is $0.63396\ldots$.
We can write a higher-order approximation for $P(\{1, 2, \ldots\})$ by computing a power series for the exact probability, $P(\{1, 2, \ldots\}) = 1 - \left(1 - \tfrac{1}{N}\right)^N$, in $\frac{1}{N}$:
$$P(\{1, 2, \ldots\}) = \left(1 - \frac{1}{e}\right) + \frac{1}{2 e N} + \frac{5}{24 e N^2} + O\left(\frac{1}{N^3}\right) .$$
This also suggests a way of formulating and proving a precise estimate: We can readily check that $x \mapsto 1 - (1 - x)^{1 / x}$ is increasing, so for $N$ at least, e.g., the suggested value of $50$, we have that
$$0.63212\ldots < P(\{1, 2, \ldots\}) \leq 1 - \left(1 - \frac{1}{50}\right)^{50} = 0.63583\ldots,$$ giving a maximum relative error for $N \geq 50$ of $\approx 0.37\%$.
A: It's a limit problem.. the limit of 1-(x-1)/x)x as x approaches infinity is 1-(1/e) which is ~ 63%.
So yes, if you have a one in a million chance of winning the lotto, and you buy 1 million random tickets, you have a 63% chance of winning.
