First to the sequence HT between two players Two players, A and B, alternatively toss a fair coin (A tosses first and then B). The sequence of heads and tails is recorded. If there is a head followed by a tail (HT subsequence), the game ends and the person who tosses the tail wins. What is the probability that A wins the game?
The solutions states that $P(A) = 1/2*P(A|H) + 1/2*P(A|T)$ and takes $P(A|T) = 1-P(A)$ which makes sense. However, I can't understand why it takes $P(A|H) = 1/2*(1-P(A|H))$. I know the part $(1-P(A|H))$ has to do with the case when B gets a H, but I don't understand the logic behind this term.
Thank You
Thank You

PLEASE REVIEW: 
Can someone please help me spot the error in following approach. Let $A$ be the event when player A wins and $B$ be the event when player B wins. And $H_A$ represents Heads drawn by A and $H_B$ represents Heads drawn by B (and likewise Tails).
So, $P(A) = \frac 12P(A|H_A) +  \frac 12P(A|T_A)$
$\implies P(A) = \frac 12P(A|H_A) +  \frac 12(1 - P(A))$ [$\because$ by symmetry, $P(A|T_A) = P(B) = 1 - P(A)$]
$\implies 3P(A) = 1 + P(A|H_A)$                   ... (Eq. 1)
$\implies 3P(A) = 1 + (\frac12 P(A|H_AH_B) + \frac12 P(A|H_AT_B))$ [$\because$ B's turn to flip]
$\implies 3P(A) = 1 + (\frac12 P(A|H_AH_B) + 0)$ [$\because$ B has won in event of $H_AT_B$ ]
$\implies 3P(A) = 1 + \frac12 (\frac12 P(A|H_AH_BT_A) + \frac12 P(A|H_AH_BH_A))$  [$\because$ A's turn to flip]
$\implies 3P(A) = 1 + \frac12 (\frac12 *1 + \frac12 P(A|H_AH_BH_A))$ [$\because$ A has won in event of $H_AH_BT_A$ ]
$\implies 3P(A) = 1 + \frac14 (1 + P(A|H_A))$ [$\because P(A|H_AH_BH_A) = P(A|H_A)$]
Using eq.1, $3P(A) = \frac 54 + \frac14 (3P(A) - 1)$
$\implies \frac{9}{4}P(A) = 1$ 
$\implies P(A) = \frac49$ 
