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Given an unbalanced coin that emits H at 2/5 and emits T at 3/5.
Two players, A and B, each choose a different sequence of 3 coin flips, then toss the coin together again and again until they get one of the sequences. The winning player is the player whose sequence was first received. Player A chooses the sequence T T T. Is there a sequence B can choose so that its probability of winning will be greater than half? If so, find this sequence and calculate the probabilities of each player to win. Otherwise, explain why there is no such sequence.

I have made a simulation on my PC trying to figure out the answer first, and found out T H T has a probability greater than half to win (~0.516). But I need help finding a mathematical way to show that...

Choosing $HTT$ gives $B$ a $frac {98}{125}=.784$ chance of winning.

Indeed, unless $A$ sees $TTT$ initially it is impossible for $A$ to win (as some Head must be tossed and the first instance of $TTT$ will inevitably be preceded by $HTT$).

Pf: Suppose $TTT$ first occurs starting in toss $n>1$. Then the $(n-1)^{st}$ toss must have been $H$, else we'd have had $TTT$ starting from slot $n-1$. But in that case the block starting from toss $n-1$ is $HTT$.

Of course, $TTT$ occurs initially with probability $left( frac 35right)^3=frac {27}{125}$ so $B$ wins with probability $1-frac {27}{125}=frac {98}{125}$.

Note: $TTT$ is a terrible choice.

Note 2: the calculation from your choice is harder, but perfectly possible. Work with states according to how much of either string has been assembled. Thus we have the states $emptyset, T, TH, TT, TTT, THT$ and look at the various transitions to write down linear equations. A little tedious, but not difficult. the calculation shows that, with this choice, $B$ wins with probability $frac {16}{31}$. Letting $P_S$ denote the probability that $B$ wins given that we start from state $S$, the relevant equations are $$P_{emptyset}=frac 35times P_T+frac 25P_{emptyset}implies P_{emptyset}=P_T$$ $$P_T=frac 35P_{TT}+frac 25P_{TH}$$ $$P_{TH}=frac 35times 1+frac 25P_Tquad &quad P_{TT}=frac 35times 0 +frac 25P_{TH}$$ This is easily solved to get the desired answer $$P_{emptyset}=frac {16}{31}$$