Winning exactly once before going broke

Each round of a game costs $10$ to enter; you win the round with probability $0.1$, in which case you are handed $80$, and otherwise you get nothing. You begin with $30$ and keep playing until you can no longer afford a round. What is the probability, to three decimal places, that you win exactly one round before running out of money?

About $0.105$. One win brings total funds to $30+80=110$, i.e. exactly $11$ rounds, so the path must have $1$ win and $10$ losses. The win must come within the first three rounds (before the initial $30$ runs out), giving $3\times0.1\times0.9^{10}\approx0.105$.