Probability for the length of the longest run in nn Bernoulli trials

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 n, exceeds a given number m, i.e. P(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:

Distribution of the length of the longest run of head in a sequence of 1000 Bernoulli trials with 60% change of getting a head

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.

Answer

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*}

Attribution
Source : Link , Question Author : Sasha , Answer Author : Community

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