Penney's game with a biased coin A coin is biased so that the probability of obtaining a head is $p$, where $0 < p < 1$. If the sequence
$HHH$ occurs first then Player $A$ wins, and if the sequence $HTH$ occurs first then player $B$ wins. The
coin is tossed until one player wins. For what value of $p$ is the game fair? i.e. Probability of Player $A$ and Player $B$ winning are equal.
I've attempted to research on Penney's game with biased coins, but there doesn't appear to be any information on it. Personally, I'm not well versed with probability, just know the basics, so any help would be useful. I've seen ideas of using Markov Chains but I've never used them myself.
 A: Flip the coin, obtaining some number of $\color{blue} T$'s until you see the first $\color{blue} H$. Once this happens, it becomes possible for Player A or B to win in the next two flips. Then flip the coin twice, obtaining a sequence of two letters. If we see $\color{red}{HH}$, Player A wins. If we see $\color{red}{TH}$, Player B wins. If we see $\color{red}{TT}$, then neither player can win in the next flip, so we start this process over. If we see $\color{red}{HT}$, then flip once more. Player B wins if we see $\color{green}{H}$; otherwise, we start over. For example,
$$ \color{blue}{TTTTTTH} \color{red}{TH} $$
results in Player B winning, but
$$ \color{blue}{TTTTTTTH} \color{red}{HT} \color{green}{T}$$
results in us starting over. The probability that Player A wins in one of these rounds is the probability of seeing $\color{red}{HH}$, which is $p^2$. The probability that Player B wins is the probability of seeing $\color{red}{TH}$ or $\color{red}{HT} \color{green}{H}$, which is $(1-p)p + p(1-p)p$.  Therefore, if we repeat this until some player wins, the probability Player A wins is
$$ \frac{p^2}{p^2 + (1-p)p + p(1-p)p} .$$
