Ruin game with a favourable two-thirds edge

Player M starts with $1 and player N with $2. In each round M takes $1 from N with probability $\tfrac23$, otherwise N takes $1 from M. The game stops as soon as either player has no money. What is the probability that M ends up with all $3?

$4/7$. With $p=\tfrac23$, $q=\tfrac13$ the gambler's-ruin formula gives $\dfrac{1-(q/p)^1}{1-(q/p)^3}=\dfrac{1-\tfrac12}{1-\tfrac18}=\tfrac47$. Equivalently, set up the Markov chain on M's wealth $0,1,2,3$ and solve the two linear equations.