Find the number of ways in which coin tosses can be made such that there are not three consecutive heads or tails.

Question

Find the number of ways in which coin tosses can be made such that there are not three consecutive heads or tails.

actually I think THT TTH HHT HTH possibilities and the recursive relation but I dont know how can I express them ??

Answer 1

Consider creating a string of H and T of length $k$ by adding one toss to the previous string of length $k-1$.

Let $P_k$ be the number of strings of length $k$ that contain no triplets and end in TH or HT.

Let $M_k$ be the number of strings of length $k$ that contain no triplets and end in TT or HH. Thus the number we desire is $N_k = P_k + M_k$.

But while either toss can be added to $P_{k-1}$ only one specific toss leads to a $P_k$ string; another toss leads to an $M_k$ string. And only one specific toss can be added to an $M_{k-1}$ string; it leades to a $P_k$ string. Thus

$$P_k = P_{k-1} + M_{k-1} \\ M_k = P_{k-1}$$

Then $$P_k = P_{k-1} + P_{k-2}$$

This is the equation determining the Fibonacci sequence; the starting conditions are that $P_2 = 2, P_3 = 6$. So

$$P_k = 2F_k$$ where $F_k$ is the $k$-th Fibonacci number. And then

$$ N_k = 2 F_{k+1}$$

Answer 2

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

We consider the set of binary words of length $n\geq 0$ and the set $B=\{HHH,TTT\}$ of bad words, which are not allowed to be part of the words we are looking for. We derive a generating function $f(s)$ with the coefficient of $s^n$ being the number of wanted words of length $n$.

According to the paper (p.7) from Goulden and Jackson the generating function $f(s)$ is \begin{align*} f(s)=\frac{1}{1-ds-\text{weight}(\mathcal{C})}\tag{1} \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}[HHH])+\text{weight}(\mathcal{C}[TTT]) \end{align*} We calculate according to the paper \begin{align*} \text{weight}(\mathcal{C}[HHH])&=\text{weight}(\mathcal{C}[TTT])\\ &=-\frac{s^3}{1+s+s^2}=-\frac{s^3(1-s)}{1-s^3}\\ \end{align*}

and

we obtain the generating function $f(s)$ for the binary words which do not contain the bad words \begin{align*} f(s)&=\frac{1}{1-2s+\frac{2s^3(1-s)}{1-s^3}}\\ &=\frac{2}{1-s-s^2}-1\\ &=1+2s+4s^2+6s^3+10s^4+16s^5+26s^6+42s^7+\cdots \end{align*}

Note, the result is essentially twice the generating function of the Fibonacci numbers \begin{align*} \frac{1}{1-s-s^2} \end{align*}

Answer 3

The DFA method (which is the naive approach) will produce the following transcript, matching the answer by @MarkusScheuer.

> GFNC([[0,0,0],[1,1,1]], 2, true);
                            [[1, 1, 1], [0, 0, 0]]

                                 Q[], 0, Q[0]

                                 Q[], 1, Q[1]

                               Q[0], 0, Q[0, 0]

                                 Q[0], 1, Q[1]

                            Q[0, 0], 0, Q[0, 0, 0]

                               Q[0, 0], 1, Q[1]

                           Q[0, 0, 0], 0, Q[0, 0, 0]

                           Q[0, 0, 0], 1, Q[0, 0, 0]

                                 Q[1], 0, Q[0]

                               Q[1], 1, Q[1, 1]

                               Q[1, 1], 0, Q[0]

                            Q[1, 1], 1, Q[1, 1, 1]

                           Q[1, 1, 1], 0, Q[1, 1, 1]

                           Q[1, 1, 1], 1, Q[1, 1, 1]

                                    2
                                   z  + z + 1
                                 - ----------
                                    2
                                   z  + z - 1

> series(%, z=0, 9);
                  2      3       4       5       6       7       8      9
     1 + 2 z + 4 z  + 6 z  + 10 z  + 16 z  + 26 z  + 42 z  + 68 z  + O(z )

We can also derive this from first principles, getting the generating function (use $z$ for heads and $w$ for tails):

$$f(z,w) = (1+z+z^2) \left(\sum_{q\ge 0} (w+w^2)^q (z+z^2)^q\right) (1+w+w^2).$$

This simplifies to

$$(1+z+z^2) \frac{1}{1-w(w+1)z(z+1)} (1+w+w^2).$$

Now we are only interested in the count, so we may drop the distinction between heads amd tails to get

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

This is the same generating function as before.

Answer 4

this is easy using recursion, suppose $f_n$ is the desired number. We have $f_1=2,f_2=4$. And for $n> 2$ we have:

The desired sequences of length $n$ can be separated into $2$ types:

The ones in which the last two flips are different ( there are $f_{n-1}$ of this type).

The ones in which the last three flips are all the same ( there are $f_{n-3}$ of this type).

We hence have the recursion $f_n=f_{n-1}+f_{n-2}$