Fighting Game 2 inspired by Borut Levart

Let $M_n=\Pr(\text{Melody wins} | \text{Melody has } n \text{ points})$ probablity of winning if Melody has $n$ points and $p$ be the probability of she winning a single round. For winning score of k=3, we dissect it into 3 cases $M_0,M_1,M_2$ .

Whenever anyone loses a round, it just reverts into the initial state for all players i.e. $\Pr(\text{Melody Wins} | \text{Melody loses previous round}) = M_0$ . For the problem, in event where Melody has $n\ge 1$ points but loses the round ,the winner of that round has to battle Melody again before declared winner. Denote $W^+$ and $W^-$ the probability that the winner wins and loses the round respectively.

Note that for every even round, Melody is playing so its possible for her to gain a point and she cannot lose a round if someone else has 2 points. Then,

$\begin{aligned} M_2 &= p + (1-p)(W^- M_0 + W^+W^-M_1)\\ M_1 &= pM_2 + (1-p)(W^- M_0 + W^+W^-M_1)\\ M_0 &= pM_1 + (1-p)(W^- M_0) \end{aligned}$ Plugging in variables and solve for M_0 results in $\boxed{M_0=8/27=0.296...}$