Setup:
Suppose I am interested in a simple binomial random process with two possible outcomes: $A$ with probability $p$, and $B$ with probability $1-p$
Each event of the random process is independent.
Problem:
For a given number of trials $N$, define the set of lengths $l$ as the lengths of repeated segments of $A$ outcomes. From this, I can define a distribution $P(l)$ that a given length takes the value $l$.
An example with 13 events:
$AAABABAABBAAA$ has four lengths $l$: 3, 1, 2, 3.
Hence $P(l=3) = 1/2$, $P(l=1)=1/4$ and $P(l=2) = 1/4$.
The question is simple: what is this (normalised) distribution as $N\to\infty$?
Why I'm stuck:
Though the probability of having $l$ events in a row all be $A$ is trivial to compute ($p^l$), computing the probability that a given `chain' of $A$ events is of length $l$ is confusing me! Naively I would say that a chain of length $l$ has weight $p^l$, and hence $P(l) = p^l/\sum_lp^l$ - but I am worried I might be thinking too simplistically...