probability of winning matches consecutively England and Australiya 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.
What I've tried:
$A:$ event in which England wins $P(A)=\frac{1}{2}$
$B:$ event in which Australiya wins $P(B)=\frac{1}{2}$
probability that England wins at least $3$ consecutive matches
$WWWLLLLL$ or $()WWWLLLL$ or $()()WWWLLL$ or $()()()WWWLL$ or $()()()()WWWL$ or $()()()()()WWW$
total probability $$\sum_{r=3}^{8}\frac{1}{2^r}$$
but is is wrong how do i solve it help me
 A: Let's count all the possibilties where
L = lost, W=win and * = lost or win


*

*We have only $1$ string for winning all games and $2$ for winning 6 games.

*Winning 5 games in a row: 


$WWWWWL*\;\;$ $2$ strings $\times 2$ (simmetry)
$LWWWWWL\;\;$ $1$ string
so we have $5$ strings, for 5 games in a row.


*

*To win 4 games in a row: 


$WWWWL**\;\;\;$ $4$ stings $\times 2$ (simmetry)
$LWWWWL*\;\;\;$ $2$ stings $\times 2$ (simmetry)
so we have $12$ strings, for 4 games in a row. 


*

*To win 3 games in a row: 


$WWWL***\;\;\;$ $8$ stings $\times 2$ (simmetry)$-1$
$LWWWL**\;\;\;$ $4$ stings $\times 2$ (simmetry)
$*LWWWL*\;\;\;$ $4$ stings 
so we have $27$ strings, for 3 games in a row. 
So we have $1+2+5+12+27=47$ good strings (among 128) and thus the answer is $$P = {47\over 128}$$
A: Look at the sequence of $W$ and $L$ incrementally until $WWW$ occurs.  Except the first case, in other cases, $WWW$ is prepended by one $L$.  Decompose the required event according to the first occurence of $WWW$.
\begin{align}
r=3 \quad &\verb+WWW+ \\
\vdots \quad &\verb+LWWW+ \\
&\verb+?LWWW+ \\
&\verb+??LWWW+ \\
r=7 \quad &\verb+???LWWW+
\end{align}
You can take each $?$ to be either $W$ or $L$, except when $r=7$, the case $WWWLWWW$ has to be excluded to avoid double-counting it with $WWW$ in the case $r=3$.
$$
\begin{aligned}
&\text{Total probability} \\
=& \left(\frac12\right)^3 + \left(\frac12\right)^4 \sum_{k=0}^3 2^k \cdot \left(\frac12\right)^k - \left(\frac12\right)^7 \\
=& \frac18 + \frac{1}{16} \cdot 4 - \frac{1}{128} \\
=& \frac{47}{128}
\end{aligned}
$$
A: I though about it in reverse. Let's define $P(x, y)$ as the probability of wining 3 matches consecutively after you've won $x$ matches consecutively (think about it as your streak) and still have $y$ games to play. We want to know what is $P(0,7)$. In that case, to start our base case, we have that:
$$P(0,1)=P(1,1) = 0$$
$$P(2,1)=\frac12$$
That is to say, it is impossible to complete a 3 win streak if you haven't at least two games streak. Okay, now everything we are missing is a recursive formula to get us back to $P(0,7)$.
If we are in a $0$ streak, we can either win a game and get us to a $1$ streak, or continue in our $0$. Similarly, if we are at a $1$ streak, our next victory will bump us to $2$, or get us back to zero. Finally, in a $2$ streak you are either done with your goal in the next match, or reset your counting. Translating that to probabilities:
$$P(0,x) = \frac12P(1,x-1) + \frac12P(0,x-1)$$
$$P(1,x) = \frac12P(2,x-1) + \frac12P(0,x-1)$$
$$P(2,x) = \frac12 + \frac12P(0,x-1)$$
I suppose we could do some more analysis to get a fancy schmancy formula for P(x,y), but honestly, I didn't bother to do that, and just plugged numbers in a matrix until I got to a result
$$P(0,7)=\frac{47}{128}$$
