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?


If all you want to know is his chance of ruin, where he plays indefinitely without ever stopping, then it is $1$ because the amount he has multiplies by $\frac{11}{10}$ on winning a bet and multiplies by $\frac{9}{10}$ on losing a bet. If we consider the natural logarithm of the amount, it increases by less than $0.096$ with probability $\frac{9}{19}$ and decreases by more than $0.105$ with probability $\frac{10}{19}$. Clearly it is biased downwards and hence will decrease to bankruptcy with probability $1$.

However, your title asks for the probability of getting to a large amount first before going bankrupt. Then all we need is the probability of getting from $1000$ to $10000$, since the probability of getting from $10000$ to $100$ is $1$. I doubt there will be a simple form because the upward and downward movements do not form a rational ratio and hence the bankruptcy threshold creates a very 'jagged' boundary. I suppose you can estimate it using a sufficiently close rational approximation of the ratio and the standard method for handling 1-dimensional random walks with endpoints (one-step analysis), but even that would require roots of high-degree polynomials and which are not solvable.

So I would be very interested to see if anyone can derive a simple form, even if it uses summations!

  • $\begingroup$ My bad, I edited the last line to reflect that I'm looking for his chances of a pay day and not chance of ruin-- was thinking one and typed the other! As for a simple form, that's what I'm after. I will post my work toward an approximate solution tomorrow if I have time to write it up. Initial simulations show that my solution is close, but I know it's just approximate. $\endgroup$ – Qalnut Jun 4 '15 at 2:59

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