# Probability for the length of the longest run in $n$ 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 $\ell_n$, 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.

• You could have a Markov chain with states $0, 1, 2, \ldots l$ where the probability you move from $i$ to $i+1$ is $p$ and from $i$ to $0$ is $1-p$ for $i<l$ and the probability you move from $l$ to $l$ is $1$. Then you could look at $n$th power of the transition matrix for this chain and read off the answer. – ShawnD Aug 25 '11 at 17:29
• @Shawn What you suggest is a Markov chain that would give rise to geometric probability for the length of a single run, yet in $n$ simulations there might be more runs that that. The number of runs is random itself, so distribution I expect to get is not geometric. Besides your idea does not take into account the number of coin flipping. – Sasha Aug 25 '11 at 17:52
• I believe my "l" is your "m." The number of flips you make is accounted for by the power of the transition matrix you look at. – ShawnD Aug 25 '11 at 18:09
• @Sasha: Both of your concerns are taken into account in Shawn's proposal. The fact that the transition from $m$ to $m$ has probability $1$ takes care of the possibility of multiple runs of length $m$; the system stays in state $m$ after the first such run. If all you need is the result for a particular $m$, you can use Shawn's approach. Another question is whether the structure of the transition matrix allows one to write the probability in closed form for general $m$ and $n$. – joriki Aug 25 '11 at 18:26
• This paper may be useful: The Longest Run of Heads, M.K. Schilling – Luis Mendo Jul 23 '14 at 13:36

## 5 Answers

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*}

• Well damn, didn't see that coming. – anon Aug 25 '11 at 19:17
• @Byron Wow, thank you! This is really beyond my expectations. I will check those books out. – Sasha Aug 25 '11 at 19:19
• Great answer, Byron. – Mike Spivey Aug 25 '11 at 23:11
• I found that Byron's answer doesn't give the right values. I don't have any of the books he mentions, so I figured it out from scratch. Big fun! I think the floor value for the upper limit of $j$ should be $\frac{n+1}{m+1}$. The term, ${{n-jm} \choose {j-1}}$ should multiply $p$ inside the big bracket. The term, $\frac{n-jm+1}{j}$ should be a term ${{n - jm + 1} \choose j}$. Can anybody tell me whether I got this right? – user92236 Aug 29 '13 at 17:11
• @DaveNeville I have added a derivation of the formula. Hope you find it useful! – user940 Mar 8 '14 at 18:56

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.

• Shawn Thanks for posting your question (+1). I think I should accept Byron's answer as the one that provides an analytic answer. But your answer has proven a valuable lesson in Markov chains, so thank you again very much for that. – Sasha Aug 25 '11 at 19:24
• Yes, his answer is better. – ShawnD Aug 25 '11 at 20:03
• @Shawn: The Markov should have $(m+1)(m+2)/2$ states instead of $m+1$. Each state should be an ordered pair $(R,r)$, where $0\le R\le m$ is the length (capped at $m$) of the longest run so far, and $0\le r\le R$ is the length of the current run. Then $(R,r)$ should go to $(R,0)$ with probability $1-p$, and go to state $(R+1,r+1)$ (or itself if $R=m$) with probability $p$. – user1551 Aug 25 '11 at 21:00
• @user1551 Shawn's proposal will do just fine, because as I said, the probability equals to probability that length of any run exceeds $m$. Indeed $\max(x_1, x_2, \ldots, x_k) > m$ is logically equivalent to $x_1>m \land x_2 > m \land \ldots \land x_k>m$. – Sasha Aug 25 '11 at 21:07
• There was an error in my comment above. State $(R,r)$ should go, with probability $p$, to $(R, r+1)$ if $r<R$, or $(R+1,r+1)$ if $r=R<m$, or itself if $r=R=m$. – user1551 Aug 25 '11 at 21:27

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.

• Thank you for the link. In regards to closed form comment, see Byron's answer above. – Sasha Aug 25 '11 at 19:39
• That's a nice reference. Thanks! – user940 Aug 25 '11 at 19:58
• @karakfa Think the link got broken. – Kai Sikorski Apr 13 '14 at 20:42
• @KaiSikorski updated the link, thanks for the warning. – karakfa Apr 14 '14 at 14:46

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.

Remark. The following answer uses the same methodology as the accepted leading answer. It is self-contained, includes Maple code and some asymptotics. Perhaps it can serve a didactic purpose and show that the basic computation is very simple.

With $z$ for successes and $w$ for failures we get the generating function

$$\left(1+z+\cdots+u\frac{z^m}{1-z}\right) \left(\sum_{q\ge 0} \frac{w^q}{(1-w)^q} \left(z+\cdots+u\frac{z^m}{1-z}\right)^q \right)\frac{1}{1-w}.$$

Now as a sanity check when we put $w=z$ and $u=1$ we should get all binary strings. Doing the calculation we find

$$\frac{1}{1-z} \frac{1}{1-wz/(1-w)/(1-z)} \frac{1}{1-w} \\ = \frac{1}{(1-w)(1-z)-wz} = \frac{1}{1-w-z}.$$

Putting $z=w$ yields

$$\frac{1}{(1-z)^2} \frac{1}{1-z^2/(1-z)^2} = \frac{1}{(1-z)^2-z^2} = \frac{1}{1-2z}$$

and the sanity check goes through.

Now for the probabilities we must subtract the value for $u=0$ from the generating function and then set $u=1.$ We obtain for the zero term

$$\frac{1-z^m}{1-z} \left(\sum_{q\ge 0} \frac{w^q}{(1-w)^q} z^q \frac{(1-z^{m-1})^q}{(1-z)^q}\right) \frac{1}{1-w} \\ = \frac{1-z^m}{1-z} \frac{1}{1-wz(1-z^{m-1})/(1-w)/(1-z)} \frac{1}{1-w} = \frac{1-z^m}{(1-w)(1-z)-wz(1-z^{m-1})} = \frac{1-z^m}{1-w-z+wz^{m}}.$$

To obtain the probabilities set $z=pv$ and $w=(1-p)v$ to get for the one term

$$\frac{1}{1-w-z} = \frac{1}{1-(1-p)v-pv} = \frac{1}{1-v}$$

so that $$[v^n] \frac{1}{1-v} = 1.$$

The subtracted contribution from the zero term is

$$\frac{1- p^m v^m}{1-v+(1-p)p^m v^{m+1}}.$$

Now for coefficient extraction we write

$$[v^Q] \frac{1}{1-v+(1-p)p^m v^{m+1}} \\ = [v^Q] \sum_{q\ge 0} (v-(1-p)p^m v^{m+1})^q \\ = [v^Q] \sum_{q\ge 0} \sum_{r=0}^q {q\choose r} v^{q-r} (-1)^r (1-p)^r p^{mr} v^{(m+1)r}. \\ = [v^Q] \sum_{q\ge 0} \sum_{r=0}^q {q\choose r} v^{mr+q} (-1)^r (1-p)^r p^{mr} \\ = \sum_{r=0}^{\lfloor Q/m\rfloor} {Q-mr\choose r} (-1)^r (1-p)^r p^{mr}.$$

We thus get the closed form

$$\bbox[5px,border:2px solid #00A000]{ 1 - a_N + p^m a_{N-m} \quad \text{where} \quad a_Q = \sum_{r=0}^{\lfloor Q/m\rfloor} {Q-mr\choose r} (-1)^r (1-p)^r p^{mr}.}$$

Note that for $N\lt m$ we get

$$1- {N\choose 0} \times 1 \times 1\times 1 = 0$$

which is the correct result. Also for $N=m$ we obtain

$$1 - 1 - {0\choose 1} \times -1 \times (1-p) \times p^m + p^m \times 1 = p^m$$

which is correct as well.

Now for an asymptotic we note that the root $\rho$ with the smalltest modulus of $1-v+(1-p)p^m v^{m+1}$ is the one close to $v_0=1$ and we get from Newton-Raphson the first approximation

$$v_1 = v_0 - \frac{1-v_0 + (1-p)p^m v_0^{m+1}}{-1+(m+1)(1-p)p^m v_0^{m}}.$$

This works out to

$$1+ \frac{(1-p)p^m}{1-(m+1)(1-p)p^m} = \frac{1-m(1-p)p^m}{1-(m+1)(1-p)p^m} = \rho.$$

The corresponding term from the partial fraction decomposition is

$$\frac{1}{v-\rho} \mathrm{Res}_{v=\rho} \frac{1}{1-v+(1-p)p^m v^{m+1}} \\ = \frac{1}{v-\rho} \times \left.\frac{1}{-1+(m+1)(1-p)p^m v^{m}}\right|_{v=\rho} \\ = - \frac{1}{\rho} \frac{1}{1-v/\rho} \frac{1}{-1+(m+1)(1-p)p^m \rho^{m}}.$$

We thus obtain

$$\bbox[5px,border:2px solid #00A000]{ a_Q \sim \frac{1}{\rho^{Q+1}} \frac{1}{1-(m+1)(1-p)p^m \rho^{m}}.}$$

We may replace $\rho$ by a better approximation from Newton-Raphson in a setting where we require numerics, which is done in the sample code which now follows.

MRUNPROB :=
proc(N, m)
option remember;
local ind, d, pos, cur, run, runs, prob,
zcnt, wcnt;

prob := 0;

for ind from 2^N to 2*2^N-1 do
d := convert(ind, base, 2);

cur := -1; pos := 1;
run := []; runs := [];

while pos <= N do
if d[pos] <> cur then
if nops(run) > 0 then
runs :=
[op(runs), [run, nops(run)]];
fi;

cur := d[pos];
run := [cur];
else
run := [op(run), cur];
fi;

pos := pos + 1;
od;

runs := [op(runs), [run, nops(run)]];

if nops(select(r -> (r = 1 and r >= m), runs)) > 0
then
wcnt := add(if(r = 0, r, 0), r in runs);
zcnt := add(if(r = 1, r, 0), r in runs);

prob := prob + p^zcnt * (1-p)^wcnt;
fi;
od;

expand(prob);
end;

V1 :=
proc(N, m)
option remember;
local gf;

gf := (1-p^m*v^m)/(1-v+(1-p)*p^m*v^(m+1));
expand(1-coeftayl(gf, v=0, N));
end;

a := (Q, m) ->
add(binomial(Q-m*r,r)*(-1)^r*(1-p)^r*p^(m*r),
r = 0 .. floor(Q/m));

V2 :=
proc(N, m)
option remember;
expand(1-a(N,m)+p^m*a(N-m,m));
end;

a2 :=
proc(Q, m)
local rho;

rho := (1-m*(1-p)*p^m)/(1-(m+1)*(1-p)*p^m);

1/rho^(Q+1)*1/(1-(m+1)*(1-p)*p^m*rho^m);
end;

a3 :=
proc(Q, m, p)
local rho;

rho :=
sort([fsolve(1-v + (1-p)*p^m*v^(m+1), v)],
(r1, r2) -> abs(r1-1) < abs(r2-1));
rho := op(1, rho);

1/rho^(Q+1)*1/(1-(m+1)*(1-p)*p^m*rho^m);
end;