Question on probability - A and B toss a pair of coins alternatively A and B toss a pair of coins alternatively. One who gets two heads together will win the game. If A starts the game, find probability of B winning the game.
Can anyone guide me how to approach this problem?
 A: 
One who gets two heads together will win the game.

I going to assume this means that Player A takes a turn, flips twice and has to get $HH$ to win. Otherwise, it would be Player B's turn, and they alternate like that. Let $\mathcal A = \{\text{A wins in first turn}\}$, and $\mathcal B = \{\text{B wins in first turn}\}$. For $\mathcal B$, it can happen in three disjoint ways, $\{HTHH, THHH, TTHH\}$. Therefore
$$P(\mathcal B) = P(HTHH)+P(THHH)+P(TTHH) = \frac{3}{16}.$$Then using craps principle, 
$$P(\text{B wins}) = \frac{P(\mathcal B)}{P(\mathcal B)+P(\mathcal A)} = \frac{3/16}{3/16+1/4} = \frac{3}{7}. $$
A: Here's a simple recursive method:  Let $p$ be the probability that $A$ wins this game (so the answer you want is $1-p$.)
Consider the first toss:  Either $A$ wins (probability $\frac 14$) or the turn passes to $B$, at which point $A$ will clearly have a $1-p$ probability of victory.  Thus we get the recursion:  $$p=\frac 14\;1+\frac 34\;(1-p)\;\;\implies\;\;p=1-\frac 34\;p\;\;\implies\;\;p=\frac 47$$  
Hence the answer you want is $\frac 37$
A: Can we also approach and solve it as follows?
P(A wins in first toss) = 1/4
P(A does not win in First)*P(B does not win in second)*P(A wins in third)
=3/4*3/4*1/4
...
Then sum of the infinite series?
