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Consider $n$ coin tosses. In how many ways can we have at least $k$ consecutive heads? Call this number $f(n,k)$. Is there a general expression for it? Or at least tight upper and lower bounds?

For example take $n=5,k=2$. Then the possibilities are:

HHTTT, THHTT, TTHHT,TTTHH,

HHHTT, HHTHT, HHTTH, THHHT, THTHH, THHTH, HTHHT, TTHHH, HTTHH,

HHHHT, HHHTH, HHTHH, HTHHH, THHHH,

HHHHH.

that is 19 ways

Vincenzo Oliva
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Richard
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  • Is the sequence $THTHH$ valid as well for the case two consecutive heads? – callculus42 Apr 20 '18 at 15:09
  • @callculus Oops yes, you got what I meant.. I wonder if I missed anything else – Richard Apr 20 '18 at 15:10
  • I will say the answer is 19 by my method, but I cannot tell what is missing right now. – Hw Chu Apr 20 '18 at 15:16
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    I'd do it recursively. Count the "bad" sequences, i.e. those without a string of the form $H^k$. if a bad sequence has at least $k$ terms then it must end in one of ${T,TH,\cdots, TH^{k-1}}$. – lulu Apr 20 '18 at 15:19
  • @lulu So is it too hard to get a closed form? Or at least a couple of bounds – Richard Apr 20 '18 at 15:42
  • Well, it depends what you mean by a closed form. It's like the Fibonacci recursion (just longer). The same methods that are used there can get you a closed form solution. Working with the largest root of the characteristic polynomial, I'd expect you could get decent asymptotics as well. – lulu Apr 20 '18 at 15:44
  • Might want to read about the Pisot Numbers. I think (but did not check) that the largest root of the characteristic polynomial in this case is always a Pisot number. Asymptotically, then, the number of bad sequences would be a constant times powers of this root. Needs checking, of course. – lulu Apr 20 '18 at 15:51
  • @lulu Very interesting read – Richard Apr 23 '18 at 09:42

3 Answers3

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The "dual" question is easier to answer. Namely, let us define $g(n,k)$ as the number of combinations of $n$ coin tosses where at most $k$ consecutive heads can show up. By adding a $T$ in front of the sequence, it is equivalent to count a sequence of length $n+1$ composed by $k+1$ possible sequences $T$, $TH$, $THH$, ..., $TH\cdots H$ ($k$ $H$'s). So you have the recursive relation:

$$ g(n, k) = g(n-1, k) + g(n-2, k) + \cdots + g(n-k-1, k), $$

with the initial conditions $g(n, k) = 2^n$ for $1 \leq k \leq n$ and $g(n+1, k) = 2^{n+1}-1$. This gives you enough information for a general formula, though it will still be hard for a "pen and paper" calculation.

Hw Chu
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  • I need to leave now. lulu's idea is the same of mine, but I cannot elaborate, will edit this post later if needed. – Hw Chu Apr 20 '18 at 15:20
  • +1 So is it too hard to get a closed form? Or at least a couple of bounds – Richard Apr 20 '18 at 15:41
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We consider the binary alphabet $V=\{H,T\}$. We are looking for the number $g(n,k)$ of strings of length $n$ having runs of $H$ at most length $k-1$. The wanted number is $$f(n,k)=2^n-g(n,k)$$

Strings with no consecutive equal characters at all are called Smirnov or Carlitz words. See example III.24 Smirnov words from Analytic Combinatorics by Philippe Flajolet and Robert Sedgewick for more information.

A generating function for the number of Smirnov words over a binary alphabet is given by \begin{align*} \left(1-\frac{2z}{1+z}\right)^{-1}\tag{1} \end{align*}

Replacing occurrences of $H$ in a Smirnov word by one up to $k-1$ $H$ generates words having runs of $H$ with length less than $k$. \begin{align*}\ z\longrightarrow z+z^2+\cdots+z^{k-1}=\frac{z\left(1-z^{k-1}\right)}{1-z}\tag{2} \end{align*}

Since there are no restrictions to the length of runs of $T$'s we replace occurrences of $T$ in a Smirnov word by one or more $T$s. \begin{align*}\ z\longrightarrow z+z^2+\cdots=\frac{z}{1-z}\tag{3} \end{align*}

The resulting generating function is by substituting (2) and (3) in (1) \begin{align*} \left(1- \frac{\frac{z\left(1-z^{k-1}\right)}{1-z}}{1+\frac{z\left(1-z^{k-1}\right)}{1-z}}-\frac{\frac{z}{1-z}}{1+\frac{z}{1-z}}\right)^{-1} &=\frac{1-z^k}{1-2z+z^{k+1}} \end{align*}

Denoting with $[z^n]$ the coefficient of $z^n$ in a series we obtain the number of wanted words of length $n$ as \begin{align*} \color{blue}{f(n,k)}&=2^n-g(n,k)\\ &\color{blue}{=[z^n]\left(\frac{1}{1-2z}-\frac{1-z^k}{1-2z+z^{k+1}}\right)} \end{align*}

Example: Let's look at OPs example. We take $k=2$. We obtain with some help of Wolfram Alpha \begin{align*} \frac{1}{1-2z}-\frac{1-z^2}{1-2z+z^{3}}=z^2+3 z^3 + 8z^4+\color{blue}{19} z^5 + 43 z^6 +\cdots \end{align*}

The blue colored coefficient of $z^5$ shows there are $\color{blue}{19}$ words of length $5$ built from characters $\{H,T\}$ and runs of $H$ with length at least $k=2$ in accordance with OP's result.

Markus Scheuer
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This problem is better approached by looking first at the number $g(n,k)$ of $n$ coin toss outcomes that have less than $k$ consecutive heads. Call such outcome 'valid'. Each valid outcome will start with one of T, HT, HHT,... up to H...HT with $k-1$ H's at the beginning, and the number of valid outcomes in each case will be $g(n-1,k), g(n-2,k), \dots , g(n-k,k)$. This leads to the recurrence relation: $g(n,k) = g(n-1,k) + g(n-2,k) + \cdots + g(n-k,k)$. The initial values for that recurrence are $g(1,k) = 2, g(2,k) = 4,\dots, g(k-1,k) = 2^{k-1}, g(k,k) = 2^k - 1$. Solve the recurrence, and then the function you are looking for is $f(n,k) = 2^n - g(n,k)$. The particular case you mention with $n=5$, $k=2$ yields the recurrence $g(n,2) = g(n-1,2) + g(n-2,2)$, $g(1,2) = 2$, $g(2,2) =3$, so $g(n,k)$ is the $(n+2)$th Fibonacci number, hence $g(5,2) = F_7 = 13$, and $f(5,2) - 2^5 - 13 = 19$ (besides the 17 outcomes you wrote there are two more, namely HTTHH and THTHH).

mlerma54
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