Proof of Law of Total Probability Two players take turns flipping, independently, a fair coin, where first player starts. Game ends as second heads comes up. The player who flips the second heads wins the game. Define the event: A = "Player 1 wins game" 
Use the Law of Total Probability to prove that Pr(A) = 4/9 
  I am completely stuck here, can't even get where to start. Any help will be greatly appreciate. 
 A: Let $H$ be the count of heads that have already been thrown.   We want to find $\mathsf P(\mathcal A\mid H=0)$
Then the first two turns can result in, 2 heads (player 2 wins), a head and a tail, or two tails.  The Law of Total Probability then sais:
$$\mathsf P(\mathcal A\mid H=0)
 = \frac \Box \Box\cdot 0
 + \frac \Box \Box \mathsf P(\mathcal A\mid H=1)
 + \frac \Box \Box \mathsf P(\mathcal A\mid H=0)$$
Now starting with one tail thrown, the next two turns can result in a head (immediate win for player 1), or a tail then either a head (player 2 wins) or another tail.
$$\mathsf P(\mathcal A\mid H=1)
 = \frac \Box \Box \cdot 1
 + \frac \Box \Box \cdot 0
 + \frac \Box \Box \mathsf P(\mathcal A\mid H=1)$$
Supply the probabilities and solve these simultaneous equations.
A: Here is a tedious computational approach:
Note that the winning sequences are of the form $T^{k_1}H T^{k_2}H$, where
$k_1+k_2$ is odd if player 1 wins. 
The probability of this sequence occurring is
${1 \over 2}^{k_1+k_2+2}$.
It is straightforward to see that the probability of a sequence not being of this form is zero, so the probability of neither player winning is zero.
Let $p_k$ be the probability that player $k$ wins. We have $p_1+p_2 = 1$.
We also have
$p_1 = \sum_{n=0}^\infty \sum_{k=0}^{2n+1} {1 \over 2}^{2n+3} = \sum_{n=0}^\infty (2n+2) {1 \over 2}^{2n+3}$, and similarly,
$p_2 = \sum_{n=0}^\infty (2n+1) {1 \over 2}^{2n+2}$.
Hence $p_1 = {1 \over 2} p_2 + \sum_{n=0}^\infty {1 \over 2}^{2n+3} = {1 \over 2} p_2 +{1 \over 6}$, from which we obtain the desired result.
Alternative: 
I think one has to know which player has thrown the first head.
Let $F_k$ be the event that player $k$ throws the first head. Considerations above show that $P(F_1 \cup F_2) = 1$ and the events are clearly disjoint. 
Hence 
$PA = P(A \cap F_1) + P(A \cap F_2) = P(A | F_1)P F_1 + P(A | F_2) P F_2$.
We have $F_1 = \{ T^{2n}H \}_{n \ge 0}$, $F_2 = \{ T^{2n+1}H \}_{n \ge 0}$,
and so $PF_2 = {1 \over 2} P F_1$, which lets us compute $P F_k$.
Now show $P(A | F_1) = P F_2$ and $P(A | F_2) = P F_1$ which gives
$PA = 2 P F_1 P F_2$.
