Hi,This teaser is from a quant interview prepare book. The original question states as below:Two players, A and B, alternatively toss a fair coin (A tosses the coin first, then B tosses the coin, then A, then B...) If there is a head followed by a tail, the game ends and the person who tosses the tail wins. What's the probability that A wins the game?The answer given uses law of total probability. Prob of A win the game = P(A) = 1/2 P(A|H) + 1/2 P(A|T) where P(A|H), P(A|T) are probs of A wins the game given his first toss is H or T respectively.It further says P(A|H) = 1/2 * 0 + 1/2 * (1- P(A|H)), where 1/2 *0 is due to B tosses a T after A's H. So far I understand. The part I don't understand is about the 1/2 * (1 - P(A|H)) part. If after A's first H, B tosses a H as well, why/how do we get the prob of (1-P(A|H))? Also, it says P(A|T) = P(B) where P(B) is the prob of B wining. I understand if A's first toss is a T, the game essentially starts again with B is the first one to toss. But why do we have the prob P(A|T) = P(B), i.e. given A's first toss of T, A's prob of wining is the same as B's wining? I don't quite understand this equation here...Thanks for some insights.