Probability of 4 consecutive heads in 10 coin tosses

Question

I am trying to compute the probability of having 4 (or more) consecutive heads in 10 coin tosses.

I tried using recursion but it led to a complicated expression so i think i did not quite manage.

I saw similar questions asked here that were solved with difficult approaches, but this problem looks like it could be solved in a couple of lines so I must be doing something wrong.

If anyone could help me understand it or propose a different approach for solving the problem i would be very grateful.

Have a nice day!

Answer 1

In general, this question has been answered here several times, eg here.

For these particular small numbers, you might group the total count by the position of the first successful run

So $N = N_1 + N_2 + \cdots N_7$

$N_1 = 2^6$ [HHHH******]

$N_2 = 2^5$ [THHHH*****]

$N_3 = 2^5$ [*THHHH****]

$N_4 = 2^5$ [**THHHH***]

$N_5 = 2^5$ [***THHHH**]

$N_6 = 2^5 -2$ [****THHHH*] minus [HHHHTHHHH*]

$N_7 = 2^5 -3$ [*****THHHH] minus ([HHHHTTHHHH] , [THHHHTHHHH] , [HHHHHTHHHH])

Which gives $N=64 + 6 \times 32 - 5 = 251$

So the probability is $N/2^{10} = 251/1024$

Answer 2

Here we are looking for the number $f_N$ of binary strings of length $N$ which do not contain the substring $HHHH$. The probability $p$ is then $$p=1-\frac{f_{10}}{2^{10}}$$

The so-called Goulden-Jackson Cluster Method is a convenient technique to derive a generating function for problems of this kind.

We consider words of length $N\geq 0$ built from an alphabet $$\mathcal{V}=\{H,T\}$$ and the set $\mathcal{B}=\{HHHH\}$ of bad words which are not allowed to be part of the words we are looking for.

We derive a function $F(x)$ with the coefficient of $x^N$ being the number of wanted words of length $N$. According to the paper (p.7) the generating function $F(x)$ is \begin{align*} F(x)=\frac{1}{1-dx-\text{weight}(\mathcal{C})} \end{align*} with $d=|\mathcal{V}|=2$, the size of the alphabet and with the weight-numerator $\mathcal{C}$ with \begin{align*} \text{weight}(\mathcal{C})=\text{weight}(\mathcal{C}[HHHH]) \end{align*}

We calculate according to the paper \begin{align*} \text{weight}(\mathcal{C}[HHHH])&=-x^4-\text{weight}(\mathcal{C}[HHHH])\left(x+x^2+x^3\right) \end{align*}

It follows: A generating function $F(x)$ for the number of words built from $\{H,T\}$ which do not contain the subword $HHHH$ is \begin{align*} F(x)&=\frac{1}{1-dx-\text{weight}(\mathcal{C})}\\ &=\frac{1}{1-2x+\frac{x^4}{1+x+x^2+x^3}}\\ &=\frac{1+x+x^2+x^3}{1-x-x^2-x^3-x^4} \end{align*}

Since the generating function counting the number $2^N$ of all binary strings of length $N$ is \begin{align*} \frac{1}{1-2x}=1+2x+4x^2+\cdots \end{align*}

A generating function for the number binary strings of length $N$ which contains the string $HHHH$ is

\begin{align*} \frac{1}{1-2x}-F(x)&=\frac{1}{1-2x}-\frac{1+x+x^2+x^3}{1-x-x^2-x^3-x^4}\\ &=\frac{x^4}{(1-2x)(1-x-x^2-x^3-x^4)}\\ &=x^4+3x^5+8x^6+\color{green}{20}x^7+48x^8+111x^9+\color{blue}{251}x^{10}\\ &\qquad 558x^{11}+1224x^{12}+2656x^{13}+5713x^{14}+12199x^{15}+\cdots\tag{1} \end{align*}

The last line was calculated with the help of Wolfram Alpha and we see there are $\color{blue}{251}$ strings of length $10$ which contain the subword $HHHH$.

For example the $20$ strings of length $7$ containing the substring $HHHH$ are

\begin{array}{lllll} \color{green}{HHHH}HHH\quad&\quad \color{green}{HHHH}HHT\quad&\quad \color{green}{HHHH}HTH\quad&\quad \color{green}{HHHH}HTT\\ \color{green}{HHHH}THH\quad&\quad \color{green}{HHHH}THT\quad&\quad \color{green}{HHHH}TTH\quad&\quad \color{green}{HHHH}TTT\\ HHT\color{green}{HHHH}\quad&\quad HT\color{green}{HHHH}H\quad&\quad HT\color{green}{HHHH}T\quad&\quad HTT\color{green}{HHHH}\\ T\color{green}{HHHH}H\quad&\quad T\color{green}{HHHH}HT\quad&\quad T\color{green}{HHHH}TH\quad&\quad T\color{green}{HHHH}TT\\ THT\color{green}{HHHH}\quad&\quad TTH\color{green}{HHHH}\quad&\quad TT\color{green}{HHHH}T\quad&\quad TTT\color{green}{HHHH}\\ \end{array}

We finally conclude from (1): The probability of $4$ heads in $10$ coin tosses is $$\frac{251}{2^{10}}\doteq 0.2451$$