I recently came across this deceptively simple question: how likely is a roulette player, who always bets $10\%$ of his current bankroll, to increase his holdings from $\$1,000$ to $\$10,000$ before going bankrupt?
He plays American roulette and bets on red, which wins with probability $9/19\approx47.368\%$. With each win he pockets twice his bet. The house has a minimum bet of $\$10$ so, for simplicity's sake, he will stop playing when he is effectively bankrupt with \$$100$ on hand and can no longer use his system.
This is reminiscent of many gambler's ruin problems, though those usually concern fixed bets. Intuitively, it hints at a negative binomial distribution. My attempt at a solution notes that he will require $25$ more wins than losses, and works out an expected value for the number of required trials. However, computing the probability from this expectation value does not account for the possibility of bankruptcy (which occurs with $22$ fewer wins than losses).
How does he compute his odds of cashing out with $\$10,000$ at the end of the day?