2 players, A and B, play a series of games Question:-
2 players, A and B, play a series of games where player A has a 60% chance of winning each time.  The winner of the series is the first player to win 3 games in a row.  What is the probability of player A winning the series?
Further question: 2 players, A and B, play a series of games where player A has a 60% chance of winning each time.  The winner of the series is the first player to win 2 more games than their opponent.  What is the probability of player A winning the series? 
MyApproach:-
I've tried so many ways to solve this problem but it does not seem to resemble any other "probability game" I've encountered. So here are my thoughts:
Suppose there are n games. The result of last 4 games is determined: BAAA (representing winner). Starting from there, we fill in A or B back without placing three in a row.
I am stuck here and tried solving the problem for when n=4,5,6..... But it does not seem to work out, where I find no pattern. 
For the second part, I got:
Suppose there are n games, so A wins n/2+1 and B wins n/2-1. Therefore the probability is 0.6^(n/2+1) x 0.4^(n/2-1). 
 A: For the contest where winning the series requires winning $3$ consecutive games . . .

For $-2 \le n \le 2$, define state $n$ as follows . . .


*

*If $n > 0$, player $A$ has accumulated $n$ consecutive (immediately prior) winning games.$\\[4pt]$

*If $n < 0$, player $B$ has accumulated $|n|$ consecutive (immediately prior) winning games.$\\[4pt]$

*If $n = 0$, no games have yet been played.


Let $p[n]$ be the probability that player $A$ wins the series, assuming the current state is $n$.

Then we have the following equations
\begin{align*}
p[-2]&=\bigl({\small{\frac{3}{5}}}\bigr)p[1]\\[4pt]
p[-1]&=\bigl({\small{\frac{3}{5}}}\bigr)p[1]+\bigl({\small{\frac{2}{5}}}\bigr)p[-2]\\[4pt]
p[0]&=\bigl({\small{\frac{3}{5}}}\bigr)p[1]+\bigl({\small{\frac{2}{5}}}\bigr)p[-1]\\[4pt]
p[1]&=\bigl({\small{\frac{3}{5}}}\bigr)p[2]+\bigl({\small{\frac{2}{5}}}\bigr)p[-1]\\[4pt]
p[2]&=\bigl({\small{\frac{3}{5}}}\bigr)+\bigl({\small{\frac{2}{5}}}\bigr)p[-1]\\[4pt]
\end{align*}
So $5$ linear equations in $5$ unknowns. 

Solving the system yields $p[0]={\large{\frac{1053}{1445}}}$.

For the contest where winning the series requires having won $2$ games more than the opponent . . .

Define states $1,0,-1$ as follows . . .


*

*If $n =1$, player $A$ has won $1$ more game than player $B$.$\\[4pt]$

*If $n = 0$, the players have won the same number of games.

*If $n = -1$, player $B$ has won $1$ more game than player $A$.$\\[4pt]$


Let $p[n]$ be the probability that player $A$ wins the series, assuming the current state is $n$.

Then we have the following equations
\begin{align*}
p[-1]&=\bigl({\small{\frac{3}{5}}}\bigr)p[0]\\[4pt]
p[0]&=\bigl({\small{\frac{3}{5}}}\bigr)p[1]+\bigl({\small{\frac{2}{5}}}\bigr)p[-1]\\[4pt]
p[1]&=\bigl({\small{\frac{3}{5}}}\bigr)+\bigl({\small{\frac{2}{5}}}\bigr)p[0]]\\[4pt]
\end{align*}
So $3$ linear equations in $3$ unknowns. 

Solving the system yields $p[0]={\large{\frac{9}{13}}}$.
