How does probability of unlikely event increase with time? If I flip a coin repeatedly, the chance of getting
m heads in a row is
$$P(E \text{ m times }) =  \frac{1}{2^m}$$
For a large $m$, eg $10$, we would expect to almost never see it, but as we keep flipping the coin, at a certain point it becomes unlikely for it to not show up, How does the probability of an "unlikely event" (i.e., $m$ heads in a row, given $n$ tosses) grow with respect to $n$?
Is it
$$P(E^C \text{ n times }) = \left(1-\frac{1}{2^m}\right)^n$$
 A: This reddit post has the answer in the comments. https://www.reddit.com/r/math/comments/4kj27s/probability_of_getting_n_heads_in_a_row_after_m/
You need to think of it as one minus the probability of never getting a run of $m$. The number of possible strings of 'H' and 'T' that achieve this follows the recurrence relation $a_{n+m}=a_{n+m-1}+a_{n+m-2}+ ... + a_{n}$.
This is because you can divide it into cases based on the end of the string.
For a string of length $n$ which doesn't have such a run:
If it ends in T, the remaining symbols can be any string of length $n-1$ without HHH.
If it ends in TH, the remaining symbols can be any string of length $n-2$ without HHH.
If it ends in THH, the remaining symbols can be any string of length $n-3$ without HHH.
etc. up to ending with $m-1$ heads after a T
Tthese are the only possibilities and they are mutually exclusive. So $a_{n+m}$ is just the sum of the number of possibilities for each. And the probability you get of NOT having a run of m heads is after $n$ flips is this divided by the total number of combinations, $2^{n}$
