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

 A: 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.
A: 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
$$p_n=1-{B(n)\over 26^n}\qquad(n\geq3)\ .$$
A: Use the binomial cumulative distribution function with probability of $\frac 1{26}$ to find the probability of having at most a sequence of two duplicate letters out of $n$. You are looking for $1-P(x \leq 2)$
A: The number of $n$ strings over the alphabet $\{a,b,c,\dots,z\}$ with no runs of 3 or more consecutive letters the same is $$26\sum_{j\geq 0}  25^{n-j-1}{n-j\choose j}.\tag1$$
The probability of no such "triple" runs  is (1) divided by  $26^n,$
and subtracting this from 1 gives you what you want.  

Added: $\ $ A derivation of formula (1). Every such string of length $n$ consists of 
 $n-j$ "runs"of $n-2j$ single letters and $j$ double letters for some $j\geq 0$. 
There are 26 choices of letter for the first run, and 25 choices for each run 
thereafter, since you cannot choose the letter most recently used. 

Here's a table of some  values.
$$\begin{array}{cc}
n&P(n)\\
1&0.000000\\
2&0.000000\\
3&0.001480\\
4&0.002902\\
5&0.004324\\
6&0.005744\\
7&0.007163\\
8&0.008579\\
9&0.009993\\
100& 0.130593\\
200& 0.246248\\
300& 0.346518\\
400& 0.433449\\
500& 0.508816\\
600& 0.574157\\
700& 0.630805\\
800& 0.679918\\
900& 0.722498\\
1000& 0.759413
\end{array}$$
