A coin is flipped sequentially. The game finishes when the sequence TTH is formed(player X wins) or the sequence HTT is formed(player Y wins).
I can find the expected time until absorption by X or Y and the probability of absorption by X or Y using markov chain. How can I find the the probability that the game will finish at the $n$th flip?
X can only win on step $n$ if it there were $(n-1)$ tails followed by a head on the last step. If a head precedes any sequence of two or more tails, Y would have won then. Find the probability of $n-1$ tails followed by 1 head.
Y can only win on step $n$ if the last three flips were a head, tail, and tail, and if a pair of tails had not occurred previously. Find the probability of their being no two adjacent tails in $n-3$ flips times the probability of HTT on the last three flips.
These are mutually exclusive conditions, so just add these probabilities together.
Note: Are you sure about those winning sequences? The outcome of this game is rather biased towards Y; it's entirely decided by the first two flips. If the first two flips are both tails, X will ultimately win, otherwise Y will ultimately win.