Probability question on tossing a coin Two people, A and B,  toss a coin which has two possible outcomes, T and H. The probability to get H when A tosses the coin is $\;p_A\;$, and the probability to get H when B tosses the coin is $\;p_B\;$. The appearance of either H or T in each toss of both players A and B is independent of the other tosses' outcomes.
In each person's turn he tosses the coin until T appears and then it is the other player's turn, and the winner of the game is the first player who gets in his turn at least two H's.
I am asked to calculate the probability A wins the game.
I am very confused: first, why do they say the winner is the one who gets " at least " two H's in his turn? I would say that it is exactly at the second appearance of H in one person's turn that the game ends and that person wins, and this can happen only in the second toss, as otherwise there already appeared T and the turn ended.
Second: by independency, I calculated that the probability A wins in his turn equals the probability he gets H exactly in that turn's first two tosses (otherwise he gets T and his turn is over!), so the probability is $\;p_b^2\;$ ...but I can't figure out what happens if A doesn't win in his first turn: then it must be that B does not win in his first turn, so that he either gets T in his first toss ( probability: $\;1-p_B\;$), or in his second one ( probability: $\;(1-p_B)p_B\;$), as any player that completes two tosses and has not lost his turn has already win...is this right? Anyway, I thought the answer could be $\;p_A^2\;$ as it wouldn't matter what happened before...but I really am not sure at all.
Any input will be appreciated.

P($A$ wins on first round) $=p_A^2$
P($A$ does not win on first round and $B$ wins on second round)$=(1-p_A^2)(p_B^2)$
Odds in favor of $A = \dfrac{p_A^2}{(1-p_A^2)(p_B^2)}$
[Subsequent cycles of $2$ rounds will only add some common multiplier, odds won't change]
Thus P(A wins) = $\dfrac{p_A^2}{p_A^2 + (1-p_A^2)(p_B^2)}$ 
