For some fixed $n$, let $p_n$ be the probability that a random infinite binary string contains a run of consecutive $1$s, containing $n$ more $1$s than the total number appearing before the run.

For example, if $n=2$, the binary string $010010111100...$ is of the sort we're looking for, but the string $01001011100...$ is not (yet). Call a run of $1$s that is sufficiently long, like the four $1$s in the first string, a satisfying run.

Is there any reasonably nice way to compute $p_n$?

Here's what I know:

  1. We can compute the expected number of satisfying runs in a string. Suppose a satisfying run begins after $k$ preliminary bits. If $k=0$, this occurs with probability $2^{-n}$; otherwise, it occurs with probability $\sum_{i=0}^{k-1} {k-1 \choose i} 2^{-n-k-i}=2^{-n-k}\left(\frac{3}{2}\right)^k$. Summing over all $k$, we get that the expected number of satisfying runs is $3(2^{-n})$, and so $p_n<3(2^{-n})$. You could extend this argument to compute $p_n$ by inclusion-exclusion, but it looks to me like it'd be incredibly ugly. In fact, $3(2^{-n})$ is a pretty good estimate for $p_n$, since the probability of a given string having multiple satisfying runs is so low.

  2. By conditioning on the location of the first $0$ in the string, we can get a recurrence relation: $p_n=2^{-n+1}+\sum_{i=1}^{n-1} 2^{-i+1}p_{n+i}$. This is not sufficient to compute $p_n$ on its own, even given some initial values -- the largest index that appears in it is $p_{2n-1}$, so if you try to use it recursively you'll never learn anything about $p_n$ for $n$ even. You might be able to get something out of it by repeatedly substituting it into itself (in the form I've given it in, with the smallest coefficient singled out), and using the bound from 1. to make some kind of limit argument, but this also looks nasty to me.


(This originally comes from this Magic: the Gathering scenario, but I hope I've managed to successfully de-Magic it.)


Slight trivial change in notation: Let $P_n$ be the probability that every run of consecutive 1s in our string does not exceed $n$ plus the previous acumulated runs.

Then, the following linear recursion (equivalent to that of your second approach) holds:

$$P_n = \sum_{k=1}^n P_{n+k} \; 2^{-k} \hspace{12mm} n=1, 2 \ldots \tag{1} $$

(this is obtained by summing over all the possible lengths of the first run; because, after the first run of length $k$, we have the equivant problem with $n$ replaced by $k+n$)

This recursion is rather tricky to solve, because our "initial value" would be $P_{\infty}=1$, and the recursion is not time-reversible. Here's my approach (perhaps there are more elegant procedures, I'm not really familiar with this).

I try a solution of the form:

$$P_n = 1 + a_1 2^{-n} + a_2 2^{-2n} + a_3 2^{-3n} + \cdots$$

Actually, we'll only keep the terms with exponents of the form $-(2^p-1)n$. That is, we set $a_i=0$ unless $i = 2^p-1$ for some integer $p$. Calling $Q(r, n) = 2^{-r \, n}$, we have:

$$P_n = Q(0,n) + a_1 \, Q(1 , n) + a_3 \, Q(3,n) + a_7 \, Q(7,n) + a_{15} \, Q(15,n) +\cdots$$

Now, it's straighfroward to check (geometric sum) that:

$$\sum_{k=1}^n Q(r,n+k) \; 2^{-k} = \frac{1}{2^{r+1}-1} \bigl( Q(r,n) - Q(2r+1,n) \bigr) $$

Hence, the recursion (1) induces the transformations $Q_0 \to Q_0 -Q_1$, $Q_1 \to \frac{1}{3}(Q_1 -Q_3)$, $Q_3 \to \frac{1}{7}(Q_3 -Q_7)$, etc, and thus the coefficients can be obtained:

$$\begin{align} a_1 &= \frac{a_1}{3} - 1 \Rightarrow a_1 = -\frac{3}{2} \\ a_3 &= \frac{a_3}{7} - \frac{a_1}{3} \Rightarrow a_3 = \frac{1}{3} \frac{7}{6} \frac{3}{2}=\frac{7}{6 \times 2}\\ a_7 &= \frac{a_7}{15} - \frac{a_3}{7} \Rightarrow a_7 = - \frac{1}{7} \frac{15}{14} \frac{1}{3} \frac{7}{6} \frac{3}{2} =-\frac{15}{14 \times 6 \times 2} \\ a_{15} &= \cdots = \frac{31}{30 \times 14 \times 6 \times 2}\\ a_{2^p-1} &= \cdots = (-1)^p \frac{2^{p-1}-1}{(2^{p-1}-2) \times (2^{p-2}-2) \cdots \times 14 \times 6 \times 2}\\ \end{align} $$

Or, more compact (perhaps not more illuminating) :

$$ a_{2^p-1} = \frac{2-2^{-p}}{(2;2)_p}$$

where $(a;q)_n$ is the q-Pochhammer_symbol.

Some numerical values:

$ a_1=-1.50$, $ a_3\approx0.58333$, $a_7\approx-0.08929 $, $a_{15}\approx 0.00615$, $a_{31}\approx-0.0002$

So, finally

$$ P_n = 1 - \frac{3}{2} 2^{-n} + \frac{7}{12} 2^{-3 n } + \cdots =\\ = 1 +\sum_{p=1}^\infty \frac{2-2^{-p}}{(2;2)_p}2^{-(2^p-1)n}$$

This converges quite quickly, specially for big $n$.

To get, according to the original question, the probability that some run is greater or equal than $m$, we'd evaluate

$$1 - P_{m-1} = 3 \, 2^{-m} - \frac{14}{3} 2^{-3 m} + \cdots $$

which in the first approximation coincides with the OP's estimate (using the expectation); it is quite a good approximation for, say, $n>4$.

        n        1         2         3         4         5        6          7     
 p(n) exact   0.32222   0.63411   0.81364   0.90639   0.95314   0.97656   0.98828  
 p(n) approx  0.25000   0.62500   0.81250   0.90625   0.95312   0.97656   0.98828  

Added Rigurous bounds for $P(n)$ can be obtained by noting that, with my definition, it's equivalent to the probability that a sequence of iid geometric variables $x_i = 1, 2 \cdots$ attaining values in $x_1 \le n$,$x_2 \le n + x_1 $,$x_3 \le n + x_1 + x_2$ ...


$$P(n) = \sum_{k_1}^{n} 2^{-k_1} \sum_{k_2}^{n+k_1} 2^{-k_2} \sum_{k_3}^{n+k_1+k_2} 2^{-k_3} \cdots$$

In particular, a lower bound for the $m$-th terms in this infinite nested product is obtained by setting the outer values in $k_1=1$, $k_2=1$..., so

$$P(n) \le \sum_{k_1}^{n} 2^{-k_1} \sum_{k_2}^{n+1} 2^{-k_2} \sum_{k_3}^{n+2} 2^{-k_3} \cdots = (1 - 2^{-n})(1 - 2^{-n-1})(1 - 2^{-n-2}) = \\ = \prod_{i=0}^\infty (1 - 2^{-n} 2^{-i})= S(2^{-n},2^{-1}) $$ where $S(a,q)=\prod_{i=0}^\infty (1 - a q^{i})$ is the q-Pochamer symbol or (q-series) QPochhammer[a,q]

We can check that this tends to 1 when $a\to 0$ by using the bound $0<-\log(1-x)<2x$ for $0<x \le 1/2$, so $\log( S(2^{-n},2^{-1})$ is bounded between $0$ and $-2^{-n+1}$. Hence $\lim_{n\to\infty} P(n)=1$.

Added 2: Regarding uniqueness: We must prove that the recursion (1), with the boundary condition $P_{\infty}=1$ does not admit more than one solution (with $0 \le P_n \le 1)$. This is equivalent to prove that the recursion does not have a solution with $P_{\infty}=0$ except for the trivial one: $P_n=0$. 

Now, suppose $P_{n_1}>0$. Then, the equation (1) implies that there exists at least one $n_2$ in $n_1 +1 \cdots 2 n_1$ with $P_{n_2} > P_{n_1}$. Applying the same to $n_2$, this implies that we can find an increasing subsequence of $P(n)$; hence its limit cannot be zero.

  • $\begingroup$ Very nice, thanks! My only worry is about uniqueness; how do we know that $P_\infty=1$ and the recursion are sufficient to uniquely identify $\{P_n\}$? One possibility is to say something like "we could make an inclusion-exclusion argument, so $P_n$ must have the given power series form with some set of coefficients", at which point your argument certainly shows that those are the only possible coefficients, but it'd be neat if there were something a little slicker... $\endgroup$ – Micah May 30 '12 at 21:11
  • $\begingroup$ @Micah Your concerns are totally justified. $\endgroup$ – leonbloy May 31 '12 at 1:22
  • 1
    $\begingroup$ @Micah: I added a justification for $P_{\infty}=1$ $\endgroup$ – leonbloy May 31 '12 at 12:51
  • $\begingroup$ Great, thanks a bunch. That was the thing I was really concerned about; I'm now entirely satisfied. $\endgroup$ – Micah Jun 3 '12 at 3:45
  • $\begingroup$ it is a great answer and great effort. $\endgroup$ – mez Jan 24 '13 at 21:43

This answer has a "motivational" stage, and afterwards a "calculations" stage. Even if the deduction of the formulas is not as pleasant as the formulas provided by the OP and leonbloy, I think that my answer qualifies as "nice", at least because I obtain a decreasing recurrence and because of its "constructive" flavor. Please prepare yourself to see a lot of summation symbols, but do not get impatient, just don't forget the moral of the reasoning, and feel free to skip the easy steps.


I always try to solve problems concerning random binary strings in a constructive manner. Then we can ask: how to "construct" an "admissible" string?

This is how I proceed: let us say that you want that the desired condition be fulfilled at the $r$-th run of $1$s (shortly, $1$-run) and not before. Then you construct the previous $1$-runs with prescribed lengths $k_1,k_2,\dots,k_{r-1}\geq1$. Now we must interleave $0$-runs between our $1$-runs. The first $0$-run, which we will be put before the first $1$-run, can have any length $i_1$ including zero (because your sequence may or not start with $1$). On the other hand, the other dividing $0$-runs must have positive length (because they must separate our $1$-runs).

Therefore the lengths $i_1,\dots,i_r$ of the $0$-runs satisfy $i_1\geq0$ and $i_2,\dots,i_r\geq1$. Finally, the $r$-th $1$-run must have length greater or equal than $n+k_1+\cdots+k_{r-1}$. Actually, we can suppose that this length is exactly equal to $n+k_1+\cdots+k_{r-1}$, because in this case it does not matter how the rest of the sequence is.

Of course, the desired condition is not fulfilled before the $r$-th step if and only if $k_j\leq s_{j-1}$ for $j=1,\dots,r-1$, being $s_j=n-1+k_1+\cdots+k_j$ ($s_0=n-1$).

The set of sequences constructed in the way described above, with the values $k_i$ and $i_j\ $ fixed has probability

$$\underbrace{2^{-(i_1+\cdots+i_r)}}_{0\text{-}\style{font-family:inherit;}{\text{runs}}}\ \underbrace{2^{-(k_1+\cdots+k_{r-1})}}_{\style{font-family:inherit;}{\text{faulty}}\ 1\text{-}\style{font-family:inherit;}{\text{runs}}}\ \underbrace{2^{-(n+k_1+\cdots+k_{r-1})}}_{\style{font-family:inherit;}{\text{successful}}\ 1\text{-}\style{font-family:inherit;}{\text{run}}}\,.$$

It remains to sum this value over all admissible values of $r,i_1,\dots,i_r,k_1,\dots,k_{r-1}$. In other words, the desired probability is equal to

$$\sum_{r=1}^\infty\Biggl[\,\sum_{i_1=0}^\infty\sum_{i_2=1}^\infty\cdots\sum_{i_r=1}^\infty\Biggr]\sum_ {k_1=1}^{s_0}\sum_{k_2=1}^{s_1}\sum_{k_3=1}^{s_2}\cdots\sum_{k_{r-1}=1}^{s_{r-2}}2^{-(i_1+\cdots+i_r)}\,2^{- (k_1+\cdots+k_{r-1})}\,2^{-(n+k_1+\cdots+k_{r-1})}\,.$$

The part of the sum involving the $i_j$ can be easily solved: recall that $\sum_{i=0}^\infty 2^{-i}=2$ and $ \sum_{i=1}^\infty2^{-i}=1$, so our sum simplifies to

$$2^{1-n}\sum_{r=1}^\infty\ \sum_{k_1=1}^{s_0}\sum_{k_2=1}^{s_1}\sum_{k_3=1}^{s_2}\cdots\sum_{k_{r-1}=1}^{s_{r-2}}4^{-(k_1+\cdots+k_{r-1})}\,.$$

Now we concentrate on the inner sum, that is, with $r$ fixed. We have


Since $\sum_{k=1}^s4^{-k}=\frac13(1-4^{-s})$, our sum becomes




Now I will try to convince you that this is indeed a decreasing recurrence, unlike OP and leonbloy's approach (which of course are very clever and illuminating). This is a good time to make a generalization of the problem and introduce some convenient notation:


Let $p\in(0,1)$ and let $q=1-p$. Now we suppose that the probability of $1$ in each place of a binary string is equal to $p$ (so the probability of $0$ is $q$). In the case of interest we have $p=q=1/2$, but the general case is not harder than this particular case.

Reasoning as in the previous stage, the probability of the set of desired sequences with the numbers $r,i_1,\dots,i_r,k_1,\dots,k_{r-1}$ fixed is equal to

$$\underbrace{q^{i_1+\cdots+i_r}}_{0\text{-}\style{font-family:inherit;}{\text{runs}}}\ \underbrace{p^{k_1+\cdots+k_{r-1}}}_{\style{font-family:inherit;}{\text{faulty}}\ 1\text{-}\style{font-family:inherit;}{\text{runs}}}\ \underbrace{p^{n+k_1+\cdots+k_{r-1}}}_{\style{font-family:inherit;}{\text{successful}}\ 1\text{-}\style{font-family:inherit;}{\text{run}}}\,,$$

It is important to distinguish the case $r=1$ of the probability above: in this case there is no choice of numbers $k_1,\dots,k_{r-1}$, we only choose $i_1\geq0$, so in this case the probability is equal to $p^nq^{i_1}$. Thus, the total probability is equal to

$$\sum_{i_1=0}^\infty p^nq^{i_1}+\sum_{r=2}^\infty\Biggl[\,\sum_{i_1=0}^\infty\sum_{i_2=1}^\infty\cdots\sum_{i_r=1}^\infty\Biggr]\sum_ {k_1=1}^{s_0}\sum_{k_2=1}^{s_1}\sum_{k_3=1}^{s_2}\cdots\sum_{k_{r-1}=1}^{s_{r-2}}q^{i_1+\cdots+i_r}\,p^{k_1+\cdots+k_{r-1}}\,p^{n+k_1+\cdots+k_{r-1}}\,.$$

$$=\frac{p^n}{1-q}+\frac{p^n}{1-q}\sum_{r=2}^\infty\biggl(\frac{q}{1-q}\biggr)^{r-1}\sum_ {k_1=1}^{s_0}\sum_{k_2=1}^{s_1}\sum_{k_3=1}^{s_2}\cdots\sum_{k_{r-1}=1}^{s_{r-2}}\bigl(p^2\bigr)^{k_1+\cdots+k_{r-1}}\,.$$


$$S(\alpha,1)=1;\ S(\alpha,r)=\sum_ {k_1=1}^{s_0}\sum_{k_2=1}^{s_1}\sum_{k_3=1}^{s_2}\cdots\sum_{k_{r-1}=1}^{s_{r-2}}\alpha^{k_1+\cdots+k_{r-1}}\,,\ \style{font-family:inherit;}{\text{for}}\ r\geq2\,.$$

Then our probability can be written as


We have $S(\alpha,1)=1$, and for $r\geq2$:

$$S(\alpha,r)=\sum_ {k_1=1}^{s_0}\sum_{k_2=1}^{s_1}\sum_{k_3=1}^{s_2}\cdots\sum_{k_{r-2}=1}^{s_{r-3}}\alpha^{k_1+\cdots+k_{r-2}}\sum_{k_{r-1}=1}^{s_{r-2}}\alpha^{k_{r-1}}$$

$$=\sum_ {k_1=1}^{s_0}\sum_{k_2=1}^{s_1}\sum_{k_3=1}^{s_2}\cdots\sum_{k_{r-2}=1}^{s_{r-3}}\alpha^{k_1+\cdots+k_{r-2}}\biggl[\frac{\alpha}{1-\alpha}\,(1-\alpha^{s_{r-2}})\biggr]$$

$$=\frac{\alpha}{1-\alpha}\sum_ {k_1=1}^{s_0}\sum_{k_2=1}^{s_1}\sum_{k_3=1}^{s_2}\cdots\sum_{k_{r-2}=1}^{s_{r-3}}\alpha^{k_1+\cdots+k_{r-2}}-\frac{\alpha}{1-\alpha}\sum_ {k_1=1}^{s_0}\sum_{k_2=1}^{s_1}\sum_{k_3=1}^{s_2}\cdots\sum_{k_{r-2}=1}^{s_{r-3}}\alpha^{k_1+\cdots+k_{r-2}+s_{r-2}}$$

$$=\frac{\alpha}{1-\alpha}S(\alpha,r-1)-\frac{\alpha}{1-\alpha}\,\alpha^{n-1}\sum_ {k_1=1}^{s_0}\sum_{k_2=1}^{s_1}\sum_{k_3=1}^{s_2}\cdots\sum_{k_{r-2}=1}^{s_{r-3}}\alpha^{2(k_1+\cdots+k_{r-2})}$$


Changing $\alpha$ by $\alpha^j$, we obtain, for $r\geq2$:


Why the hell I did this? because defining

$$b_j=\frac{\alpha^j}{1-\alpha^j},\ c_j=-\,\frac{\alpha^{jn}}{1-\alpha^j}\quad \style{font-family:inherit;}{\text{and}}\quad T(j,r)=S(\alpha^j,r)$$

the recurrence becomes

$$T(j,1)=1;\quad T(j,r)=b_j\,T(j,r-1)+c_j\,T(2j,r-1),\ \style{font-family:inherit;}{\text{for}}\ r\geq2\,.$$

OK, I agree that this is not a genuine decreasing recurrence, as index $j$ is increasing; fortunately, the index $r$ decreases. In general, these recurrences can be explicitly solved, either by bare hands or using computer algebra systems. Later I will try to solve it step by step, for the benefit of those who believed (or disbelieved?) me and endured up this point.


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