How do I calculate the probability of repetition of a series? This problem might sound trivial, but it is a question I have pondered on for hours without a clue how to crack it. Say, you have a series which resulted from a random process. The series is as follows: ABAAABBBBABAABBBAAABA.
How do you calculate the probability of repetition of this series?
EDIT: Each outcome is independent i.e. the events "A" and "B" are independent just as "heads" and "tails" are independent in separate coin flips.
 A: Assuming that "probability of repetition" is referring to the probability that two consecutive entries in the sequence are identical, and that our sequence of random variables $\left( X_n \right)_{n=1}^N$ are randomly generated i.i.d. such that $\mathbb{P}(X_n = A) = p$ for any integer $1 \le n \le N$, where $0 \le p \le 1$, and that $\mathbb{P}(X_n = B) = 1-p$.
We want to find $\mathbb{P}(X_n = X_{n+1})$. There are two cases:
Case 1: $X_n = A = X_{n+1}$.
As these events are independent, we have $$\mathbb{P}(X_n = A \land X_{n+1} = A) = \mathbb{P}(X_n = A)\mathbb{P}(X_{n+1} = A) = p^2$$
Case 2: $X_n = B = X_{n+1}$
Similarly, we have
$$\mathbb{P}(X_n = B \land X_{n+1} = B) = \mathbb{P}(X_n = B)\mathbb{P}(X_{n+1} = B) = (1-p)^2 = 1 - 2p + p^2$$
Thus, as these two cases are exclusive,
$$\mathbb{P}(X_n = X_{n+1}) = \mathbb{P}(X_n = A \land X_{n+1} = A) + \mathbb{P}(X_n = B \land X_{n+1} = B) = 1-2p+2p^2$$
So our probability of repetition is $1-2p+2p^2$, where $p$ can be approximated by $\widehat{p} = \frac M N$, where $M$ is the number of $A$ values in the sample, and $N$ is the total number of values.
In the sample above, $M=11$ and $N=21$. Thus, $\widehat{p} = \frac{11}{21}$, and so we can approximate the probability of repetition as $\mathbb{P}(X_n = X_{n+1}) \approx 0.5$.
