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


The probability of no 3 consecutive letters in a word of length $n$ is $$\frac{(1-p)^2}{a-b}\,\left(\frac{a^{n-1}}{1-a}-\frac{b^{n-1}}{1-b}\right),$$ where $$a=\frac{p+\sqrt{p(4-3p)}}2,\quad b=\frac{p-\sqrt{p(4-3p)}}2,\quad p=1-\frac1{26}.$$ In particular, when $n\to\infty$, the probability of no 3 consecutive letters in a word of length $n$ is equivalent to $$\frac{13}{25\sqrt{29}}(5+\sqrt{29})\,\left(\frac{5}{52}(5+\sqrt{29})\right)^n\approx1.00281\times(0.99857)^n.$$
For $n=100$, $n=500$ and $n=1000$, this predicts approximate probabilities of 3 consecutive letters in a word of length $n$ of $13\%$, $51\%$ and $76\%$ respectively, to be compared to the exact values in @Byron's answer. Probabilities for higher values of $n$ are direct with our formula and become difficult to evaluate using summation formulas.
