# Time until a consecutive sequence of ones in a random bit sequence

This a reformulation of a practical problem I encountered.

Say we have an infinite sequence of random, i.i.d bits. For each bit $X_i$, $P(X_i=1)=p$.

What is the expected time until we get a sequence of $n$ 1 bits?

Thanks!

• Though closely related, are you interested only in strings of 1's or the first time you see any string (0's or 1's) of length $n$? – cardinal Mar 19 '11 at 19:20
• Only 1's. Thanks – R S Mar 19 '11 at 21:43

There is a lot of literature on such questions concerning the mean time for patterns. For your particular problem a solution can be found on page 156 of Introduction to Probability Models (10th edition) by Sheldon Ross. The formula is $$E[T]=1/p+1/p^2+\cdots+1/p^n={(p^{-n}-1)/(1-p)}.$$

As expected, this is a decreasing function of $p$ for fixed $n$: it takes longer to see rarer events. As $p$ goes from 0 to 1, $E[T]$ decreases from infinity to $n$.

Let $T$ be the random variable that records the first time we see $n$ ones in a row. Let's also define the random variable $L$ to be the position of the first zero bit in the sequence.

Looking at the first $n$ bits there are, roughly speaking, two possibilities: either I get the desired pattern of $n$ ones or I got a zero bit at time $k$ and the whole problem starts over.

More formally, conditioning on the value of $L$ we get \begin{eqnarray*} E[T] &=& \sum_{k=1}^{n} E[T \ |\ L=k]\ P(L=k) + E[T\ |\ L> n]\ P(L>n)\cr &=& \sum_{k=1}^{n} (k+E[T])\ P(L=k) + n P(L > n)\cr &=& \sum_{k=1}^{n} (k+E[T])\ p^{k-1}(1-p) + n p^n. \end{eqnarray*}

Solving this equation for $E[T]$ gives the formula.

There are many generalizations of this problem and variations on the above proof that use, for instance, Markov chains, or martingales, or generating functions, etc. In addition to Ross's book mentioned above, you may like to look at

1. Section 8.4 of Concrete Mathematics by Graham, Knuth, and Patashnik
2. Chapter 14 of Problems and Snapshots from the World of Probability by Blom, Holst, and Sandell
3. Section XIII 7 of An Introduction to Probability Theory and Its Applications by Feller
• Thanks. As I don't have access to the book, can you direct me to some pointers to the material and terms involved? – R S Mar 19 '11 at 22:26
• @RS I have added some new material and references. Happy hunting! – user940 Mar 20 '11 at 1:33
• I have posted a generating function approach to this result in an addendum to my answer to a similar question. Perhaps I should expand and move the addendum here as an answer. – robjohn Sep 16 '12 at 15:38

Here is a generating function approach to this problem.

Preliminaries

For $0\le k<n$, let an atom $a_k$ be a sequence of $k$ $1$ bits followed by a $0$ bit. Let a terminal atom $a_n$ be a sequence of $n$ $1$ bits. Let a complete $n$ sequence be a binary sequence which ends at the first subsequence of $n$ consecutive $1$ bits. Any complete $n$ sequence can be written as a unique sequence of atoms followed by a terminal atom. The probability of an atom occurring is $p^k(1-p)$ and the length of an atom is $k+1$. The probability of a terminal atom occurring is $p^n$ and the length of a terminal atom is $n$.

Therefore, for each atom $a_k$, define the monomial $\phi_x(a_k)$ as $$\phi_x(a_k)=\left\{\begin{array}{ll}p^k(1-p)x^{k+1}&\text{if }0\le k< n\\p^nx^n&\text{if }k=n\end{array}\right.\tag{1}$$ and for the concatenation of two binary sequences, $s_1$ and $s_2$ $$\phi_x(s_1s_2)=\phi_x(s_1)\phi_x(s_2)\tag{2}$$ The probability of any sequence of atoms $s$ occurring is the coefficient of the monomial $\phi_x(s)$ and the length of $s$ is the exponent of $x$ in $\phi_x(s)$.

The Generating Function

Thus, the probability of a complete $n$ sequence with length $k$ coming from the concatenation of $j$ atoms and a terminal atom is the coefficient of $x^k$ in \begin{align} &(\phi_x(a_0)+\phi_x(a_1)+\phi_x(a_2)+\dots+\phi_x(a_{n-1}))^j\phi(a_n)\\ &=p^nx^n((1-p)x)^j\left(1+px+p^2x^2+\dots+p^{n-1}x^{n-1}\right)^j\\ &=p^nx^n\left((1-p)x\frac{1-p^nx^n}{1-px}\right)^j\tag{3} \end{align} Summing over all $j$, the probability of a complete $n$ sequence with length $k$ is the coefficient of $x^k$ in \begin{align} f_n(x) &=\sum_{j=0}^\infty p^nx^n\left((1-p)x\frac{1-p^nx^n}{1-px}\right)^j\\ &=\frac{p^nx^n}{1-(1-p)x\frac{1-p^nx^n}{1-px}}\tag{4} \end{align} That is, $f_n(x)$ is the generating function for the probability of a complete $n$ sequence with a given length.

Since $f_n(1)=1$, the probability that a complete $n$ sequence occurs eventually is $1$.

Expected Length $$f_n(x)=\sum_{k=0}^\infty c_kx^k\tag{5}$$ where $c_k$ is the probability of a complete $n$ sequence with length $k$. Therefore, $xf_n^\prime(x)$ evaluated at $x=1$ is $$\sum_{k=0}^\infty kc_k\tag{6}$$ which is the expected length of a complete $n$ sequence.

$$xf'(x)=p^nx^n\frac{n\frac{1-x}{1-px}+x\frac{1-p}{1-px}\frac{1-p^nx^n}{1-px}}{\left(1-(1-p)x\frac{1-p^nx^n}{1-px}\right)^2}\tag{7}$$ Evaluating $(7)$ at $x=1$ yields that the expected length of a complete $n$ sequence is $$\mathrm{E}(k)=\frac{1-p^n}{p^n(1-p)}\tag{8}$$

Expected Variance

Using the same idea that was used to get $(6)$, we "take the derivative and multiply by $x$" twice to see that $xf_n^\prime(x)+x^2f_n^{\prime\prime}(x)$ evaluated at $x=1$ is $$\sum_{k=0}^\infty k^2c_k\tag{9}$$ which is the expected square of the length of a complete $n$ sequence. \begin{align} xf_n^\prime(x)+x^2f_n^{\prime\prime}(x) &=\frac{p^nx^n}{(1-px)^3\left(1-(1-p)x\left(\frac{1-p^nx^n}{1-px}\right)\right)^3}\\ &\times\left( \begin{aligned} &n^2(1-x)(1-px)\left(1-x-(1-p)xp^nx^n\right)\\ &+2n(1-p)x\left(1-x-(2-x-px)p^nx^n\right)\\ &+(1-p)x(1-p^nx^n)\left(1+x-(1-p)xp^nx^n\right)\\ \end{aligned} \right)\tag{10} \end{align} Evaluating $(10)$ at $x=1$ yields that the expected square of the length of a complete $n$ sequence is $$\mathrm{E}(k^2)=\frac{2(1-p^n)}{p^{2n}(1-p)^2}-\frac{1+2n-p^n}{p^n(1-p)}\tag{11}$$ Combining $(8)$ and $(11)$, we get that the expected variance of the length of a complete $n$ sequence is \begin{align} \mathrm{Var}(k) &=\mathrm{E}(k^2)-\mathrm{E}(k)^2\\ &=\frac{1-p^{2n+1}}{p^{2n}(1-p)^2}-\frac{2n+1}{p^n(1-p)}\tag{12} \end{align}