probability to win 3 consecutive matches :Alternate approach England and Australia play a 7 one day cricket matches and each team has equal probability of winning a match. No match end in a draw. Then find probability that England win at least 3 consecutive matches.
I know this question has been asked already  but i want to know where i went wrong
My approach:let us consider cases
case 1:england win $3$ matches in total
thus it will be an arrangment of (WWW),L,L,L,L which can be arranged in $\frac{5!}{4!}$ ways
P($3$)=$5$${(\frac{1}{2^7})}$
case2  england win $4$ matches in total
thus it will be an arrangment of (WWW),W,L,L,L which can be arranged in $\frac{5!}{3!}$
ie P($4$)=$\frac{5!}{3!}$${(\frac{1}{2^7})}$
similarly i considered cases upto P(7)
and the total probability came out to be
$\frac{1}{2^7}$${(5+\frac{5!}{3!}+\frac{5!}{{2!}{2!}}+\frac{5!}{3!}+5)}$
which came out to be $\frac{60}{128}$ but the answer is $\frac{47}{128}$
please tell me where i went wrong and how to get to the answer using this method
 A: According to your approach, this is how it should be:
Case 1: England win 3 matches in total $\{(WWW),L,L,L,L) \}$
Since all other four are $L$, it is a straightforward case.
$P(3) = \dfrac{5}{2^7}$
Case 2: England win $4$ matches in total $\{(WWW),W,L,L,L) \}$
Please note $W,L,L,L$ can be arranged in $4$ ways and for each arrangement of the same, $(W,W,W)$ can be placed in only $4$ places (not $5$).
$P(4) = \dfrac {4 \times 4}{2^7} = \dfrac {16}{2^7}$
Case 3: England win $5$ matches in total $\{(WWW),W,W,L,L) \}$
Please note $W,W,L,L$ can be arranged in $6$ ways and for each arrangement of the same, $(W,W,W)$ can be placed in only $3$ places.
$P(5) = \dfrac {6 \times 3}{2^7} = \dfrac {18}{2^7}$
Case 4: England win $6$ matches in total $\{(WWW),W,W,W,L) \}$
Please note $W,W,W,L$ can be arranged in $4$ ways and for each arrangement of the same, $(W,W,W)$ can be placed in only $2$ places. However $(W,W,W),L,W,W,W$ is same as $W,W,W,L,(W,W,W)$. So we need to subtract one arrangement.
$P(6) = \dfrac {4 \times 2 - 1}{2^7} = \dfrac {7}{2^7}$
Case 5: England win $7$ matches in total $\{(WWW),W,W,W,W) \}$
There is only one possible arrangement.
$P(7) = \dfrac {1}{2^7}$
