Probability of winning tennis game (gambler's ruin?) You and an opponent are playing tennis - first to get 2 wins in a row wins. The probability of you getting a win is .6. The probability of him getting a win is .4. What's the probability of you winning the game?
I think this can be modeled as a Markov chain with 5 states (2 Losses, 1 Loss, 0 net, 1 Win, 2 Wins). Therefore, I think I could write out some equations to solve this. I'm trying out a different approach; does this make sense:
P(you win right off the bat) = (.6)(.6) = .36
P(he wins right off the bat) = (.4)(.4) = .16
P(you win) = .36/(.36+.16)
Edit: additionally, how can I relate this problem to the equations of gamblers ruin, if at all?
Other edit: Online, someone said the answer is .91, using the following argument. Why am I wrong? I even checked my answer using equations with the Markov chain.

 A: Say $J$ and $M$ are playing a tennis match ($J$ and $M$). Whoever wins two consecutive games wins the match. We have to find the probability of two consecutive $J (JJ)$ before two consecutive $M (MM)$.
Probability of J winning a single game $= p = \dfrac{3}{5}$
Probability of M winning a game $= q = 1-p = \dfrac{2}{5}$
There are four possibilities after two matches - $JJ, JM, MJ, MM$
Probability of $JJ = p^2$ and you win.
Probability of $JM = pq$ and say your probability of win from there is $P(W|JM)$
Probability of $MJ = pq$ and say your probability of win from there is $P(W|MJ)$
Probability of $MM = q^2$ (you lose)
$i)$ If it is $JJ$, you win with probability $p^2$.
$ii)$ From $JM$, you can get to $MJ$ with probability $p$ and from there you can win with probability $P(W|MJ)$.
$P(W|JM) = p \times P(W|MJ)$
$iii)$ From $MJ$, you can win with probability $p$ or get to $JM$ with probability $q$ and then win with probability $P(W|JM)$ from there.
$P(W|MJ) = p + q \times P(W|JM)$
Solving $ii)$ and $iii)$,
$P(W|MJ) = \dfrac {p}{1-pq}$
$P(W|JM) = \dfrac {p^2}{1-pq}$
Probability of you winning
$
\begin {align*}
P(W) &= P(JJ) + pq \times P(W|JM) + pq \times P(W|MJ) \\ 
&= p^2 + pq \times \dfrac {p^2}{1-pq} + pq \times \dfrac {p}{1-pq} \\
&= \dfrac {p^2(2-p)}{1-p(1-p)} \space \text {, replacing } q \text { with } (1-p)\\
\end {align*}
$
As $p = \dfrac{3}{5}$, $P(W) = \dfrac{63}{95}$
A: Your approach is wrong. What you calculated is $P(you\ win | the\ games\ lasted\ 2\ points)$ which is clearly different than $P(you\ win)$.
As for your edit: once you find the correct $P(you\ win)$ you can relate the problems since the tennis match deduced to a Bernoulli trial with $p = P(you\ win)$.
A: Every sequence of games where you win overall ends in a $WW$, and either starts with $WL...$ or $LW....$, so the probability you eventually win is $$P(WW) + P(LWW) + P(WLWW) + P(LWLWW) + P(WLWLWW) + ... = \frac{P(WW) + P(LWW)}{1 - P(W)P(L)} = \frac{0.36 + 0.144}{1 - 0.24} = 0.663215789... = \frac{63}{95}.$$
