Winning Exactly Once Before Ruin

Measure money in units of $10. She starts with $3$ units. Each loss changes her bankroll by $-1$ and each win by $+7$. Every amount is a multiple of $10, so being ruined means her bankroll is exactly $0$.

Counting the games. Suppose she wins exactly once and is then ruined after $L$ losses. Her final bankroll is

so $L = 10$. She therefore plays exactly $11$ games: one win and ten losses, with the last game a loss.

Placing the win. Suppose $k$ losses come before the win. Her bankroll must stay positive until the win, so $3-k\ge 1$, which means $k\in\{0,1,2\}$. The win is therefore game $1$, $2$ or $3$.

After the win her bankroll is $10-k$. The remaining $10-k$ games are all losses, and they take the bankroll down by one unit each, reaching $0$ only at the last game. So she is not ruined early, and each of these three placements is a valid path.

Probability. Each valid path has probability $(0.1)(0.9)^{10}$, and the three paths are distinct sequences. Hence

Key idea. Because the net changes are fixed, a single win pins down the total number of games. The only constraint left is that the win comes before the bankroll first reaches zero.