Probability that a word contains at least 3 same consecutive letters? Assume we have a word of length $n$ and an alphabet of length $26$ (the small letters a through z, if you want so.
How likely is it that this word contains at least $k := 3$ consecutive letters of any type?
Examples that match:
aaabababab 
aoeuuuuuuu
aaaaaaaaaa

Examples that do not match:
ababababab
banananana
abcdefghij


Call a word bad if it does not contain a run of $3$ equal letters.
Denote by $D(n)$ the number of bad words of length $n$ whose last two letters are different, and by $E(n)$ the number of such words whose last two letters are equal. Then
$$D(2)=26^2-26=650, \qquad E(2)=26\ .$$
Furthermore we have the rercursion
$$D(n+1)=25 D(n) + 25 E(n),\qquad E(n+1)=D(n)\qquad(n\geq2)\ .$$
This leads to the linear difference equation
$$D(n+2)-25D(n+1)-25D(n)=0\ ,$$
which can be solved by the "Master Theorem".
The number of bad words of length $n$ is then given by $B(n)=D(n)+D(n-1)$, so that the probability $p_n$ in question comes to
$$p_n=1-{B(n)\over 26^n}\qquad(n\geq3)\ .$$
