Inspired by This topic:
Let there be $N$ balls in a straight line. We colour them with $m$ $(c_1, c_2,\ldots , c_m)$ different colours.
Determine the probability there exists at least one streak of monochromatic balls of at least length $1\leq k\leq N$.
The real question here is, how do we count how many of the colourings are desirable to us and how many there are in total.
For instance, pick $N=5, m=2$ (equal probability), $k=3$. With two colours, there should be $2^5 = 32$ different colourings.
EDIT: Thanks Arthur
$5$ ways for three consecutive to occur, $2$ ways for four consecutives and $1$ way of five consecutives. By symmetry, there should be $16$ desirable choices.
With some modifications we can also work out the solution if the colourings had different probabilities.
Question: How do we count desirable choices when $N=100, m=2$ (equal probability) and $k = 5$? It's not very sensible to count each case. Is there a way to work this problem out with pen and paper?