Consecutive Heads

Let H deontes the head, T the tail. $*$ Any of the head or tail
$P(H) = P(T) = \dfrac{1}{2} , P(*) = 1$

Cases :

$HHHH*** = \left(\dfrac{1}{4}\right)^4 \times 1 = \dfrac{1}{16}$
$THHHH** = \left(\dfrac{1}{4}\right)^5 \times 1 = \dfrac{1}{32}$
$*THHHH* = \left(\dfrac{1}{4}\right)^5 \times 1 = \dfrac{1}{32}$
$**THHHH = \left(\dfrac{1}{4}\right)^5 \times 1 = \dfrac{1}{32}$

$\therefore$ The required probability = $\dfrac{5}{32}$

We can consider multiple cases for this problem.

Case 1: 4 heads in a row.

For the 4 heads in a row involving the first or last flips, the number of events is $2\times2^{2}=8$ .

For the 4 heads in a row not involving the first or last flips, the number of events is $2\times2=4$ .

Hence, the total number of possibilities for this case is 12.

Case 2: 5 heads in a row.

For the 5 heads in a row involving the first or last flips, the number of events is $2\times2=4$ .

For the 5 heads in a row not involving the first or last flips, the number of events is 1.

Hence, the total number of possibilities for this case is 5.

Case 3: 6 heads in a row.

The number of possibilities for this case is just 2.

Case 4: 7 in a row. (1 possibility)

Hence, the probability is therefore $\dfrac{12+5+2+1}{2^{7}}=\dfrac{5}{32}$