Counting consecutive chain of independent Bernoulli random variables [duplicate]

Suppose a biased coin (probability of head being $p$) was flipped $n$ times. I would like to find the probability that the length of the longest run of heads, say $\ell_n$, exceeds a given number $m$, i.e. $\mathbb{P}(\ell_n > m)$.

It suffices to find the probability that length of any run of heads exceeds $m$. I was trying to approach the problem by fixing a run of $m+1$ heads, and counting the number of such configurations, but did not get anywhere.

It is easy to simulate it:

I would appreciate any advice on how to analytically solve this problem, i.e. express an answer in terms of a sum or an integral.

Thank you.

This problem was solved using generating functions by de Moivre in 1738. The formula you want is $$\mathbb{P}(\ell_n \geq m)=\sum_{j=1}^{\lfloor n/m\rfloor} (-1)^{j+1}\left(p+\left({n-jm+1\over j}\right)(1-p)\right){n-jm\choose j-1}p^{jm}(1-p)^{j-1}.$$

References

1. Section 14.1 Problems and Snapshots from the World of Probability by Blom, Holst, and Sandell

2. Chapter V, Section 3 Introduction to Mathematical Probability by Uspensky

3. Section 22.6 A History of Probability and Statistics and Their Applications before 1750 by Hald gives solutions by de Moivre (1738), Simpson (1740), Laplace (1812), and Todhunter (1865)

Added: The combinatorial class of all coin toss sequences without a run of $ m $ heads in a row is $$\sum_{k\geq 0}(\mbox{seq}_{< m }(H)\,T)^k \,\mbox{seq}_{< m }(H), $$ with corresponding counting generating function $$H(h,t)={\sum_{0\leq j< m }h^j\over 1-(\sum_{0\leq j< m }h^j)t}={1-h^ m \over 1-h-(1-h^ m )t}.$$ We introduce probability by replacing $h$ with $ps$ and $t$ by $qs$, where $q=1-p$: $$G(s)={1-p^ m s^ m \over1-s+p^ m s^{ m +1}q}.$$ The coefficient of $s^n$ in $G(s)$ is $\mathbb{P}(\ell_n<m).$

The function $1/(1-s(1-p^ m s^ m q ))$ can be rewritten as \begin{eqnarray*} \sum_{k\geq 0}s^k(1-p^ m s^ m q )^k &=&\sum_{k\geq 0}\sum_{j\geq 0} {k\choose j} (-p^ m q)^js^{k+j m }\\ %&=&\sum_{j\geq 0}\sum_{k\geq 0} {k\choose j} (-p^ m q )^js^{k+j m }. \end{eqnarray*} The coefficient of $s^n$ in this function is $c(n)=\sum_{j\geq 0}{n-j m \choose j}(-p^ m q)^j$. Therefore the coefficient of $s^n$ in $G(s)$ is $c(n)-p^ m c(n- m ).$ Finally, \begin{eqnarray*} \mathbb{P}(\ell_n\geq m)&=&1-\mathbb{P}(\ell_n<m)\\[8pt] &=&p^ m c(n- m )+1-c(n)\\[8pt] &=&p^ m \sum_{j\geq 0}(-1)^j{n-(j+1) m \choose j}(p^ m q)^j+\sum_{j\geq 1}(-1)^{j+1}{n-j m \choose j}(p^ m q)^j\\[8pt] &=&p^ m \sum_{j\geq 1}(-1)^{j-1}{n-j m \choose j-1}(p^m q)^{j-1}+\sum_{j\geq 1}(-1)^{j+1}{n-j m \choose j}(p^mq )^j\\[8pt] &=&\sum_{j\geq 1}(-1)^{j+1} \left[{n-j m \choose j-1}+{n-j m \choose j}q\right]p^{ jm } q^{j-1}\\[8pt] &=&\sum_{j\geq 1}(-1)^{j+1} \left[{n-j m \choose j-1}p+{n-j m \choose j-1}q+{n-j m \choose j}q\right]p^{ jm } q^{j-1}\\[8pt] &=&\sum_{j\geq 1}(-1)^{j+1} \left[{n-j m \choose j-1}p+{n-j m +1\choose j}q \right]p^{ jm} q^{j-1}\\[8pt] &=&\sum_{j\geq 1}(-1)^{j+1} \left[p+{n-j m +1\over j}\, q\right] {n-j m \choose j-1}\,p^{ jm} q^{j-1}. \end{eqnarray*}

Define a Markov chain with states $0, 1, \ldots m$ so that with probability $1$ the chain moves from $m$ to $m$ and for $i<m$ with probability $p$ the chain moves from $i$ to $i+1$ and with probability $1-p$ the chain moves from $i$ to $0$. If you look at the $n$th power of the transition matrix for this chain you can read off the probability that in $n$ flips you have a sequence of at least $m$ consecutive heads.

You can find a limiting distribution, otherwise it's a difficult problem and the closed form solution won't have much practical value. See this for an elementary approach. [Update] Previous link moved to this new address. "Longest Run of Heads", M.F.Schilling.

I don't think you'll get a simple analytic formula. The problem is essentially equivalent to this one, see my answer there: it involves the $n$-power of a $m \times m$ stochastic matrix (notice that there we are interested in the runs equal or greater than $m$), using a Markov chain (as suggested in the comments by Shawn). You can find also there some asymptotics.