1/1000 chance of a reaction. If you do the action 1000 times, whats the new chance the reaction occurs? A hypothetical example:

You have a 1/1000 chance of being hit by a bus when crossing the street.
  However, if you perform the action of crossing the street 1000 times, then your chance of being hit by a bus increases to about 60% because every time you do the action, the probability of it happening again increases.

What is the math behind this to support this? Just curious.
 A: \begin{align}
P(\text{hit by bus in 1000 crossings}) & = 1-P(\text{not hit by bus in 1000 crossing}) \\ & = 1-(999/1000)^{1000} \\ & \approx 0.63
\end{align}
A: You can approximate this very well using a Poisson distribution: If over a large number $N$ of trials an event occurs an average of $\lambda \ll N$ times, the probability that it occurs $k$ times in a set of $N$ trials is $$P(k) \approx \frac{\lambda^k e^{-\lambda}}{k!}.$$
In our case, over $1000$ trials we expect our event with per-trial probability of $1/1000$ to occur an average of 1 time, so
$$P(k) \approx \frac{e^{-1}}{k!}.$$
In particular the probability of the event occurring at least once is
$$1 - P(0) \approx 1 - \frac{1}{e} = 0.63212\ldots$$
which is very close to the actual value of
$$1 - P(0) = 1 - \left(\frac{999}{1000}\right)^{1000} = 0.63230\ldots$$
as E W H Lee gives in his good answer.
This method shows two additional interesting facts:


*

*The approximate probability $1 - \frac{1}{e}$ is actually universal, and not particular to the case $N = 1000$ (in fact, this quantity is the limit of the probability as $N \to +\infty$).

*Over $N \gg 1$ trials the probability than an event with a per-event probability $1/N$ does not occur is very close to the probability that it occurs exactly once: both of these probabilities are $\approx \frac{1}{e}$.

A: Calling $p$ the probability of being hit on one crossing and $q=1-p$ the probability to be safe in one crossing you can compute $p_{1000}$ (the probability of being hit at least once in $1000$ crossings) like this:
$$p_{1000} = p + qp + qqp +...+q^{999}p$$
This mean: you get hit at the first attempt, or you escape the first and get hit at the second, or you escape the first and the second and get hit at the third...
Now we can solve the geometric series:
$$p_{1000} = p \sum_{n=0}^{999} q^n = p \frac{1-q^{1000}}{1-q} = p \frac{1-q^{1000}}{p} =1-q^{1000} \approx 0.63$$
A: For such questions, with every attempt, your chance to win increases. The chance is never 100% (as far as I know) but generally, you could write the probability out as being:
1-((1–1/x)^n)
Where x is your denominator (out of what is your chance i.e 1/1000) and n is the number of tries. So for your example, it would be 1-((1–1/1000)^1000) which equals 1-(0.999^1000), which turns out to be about 0.63230457, or 63.230457%
There is a lot of confusion about this topic, as intuitively, you would think that if the odds are 1/1000 playing 1000 times would guarantee a win. I was taught this method, and it has always worked for my purposes.
P.S.
Turns out, the lottery is even worse than you thought.
