Gambling probability A gambler plays a fair game where he can win or lose $\$1$ in each round. His initial stock is $\$200$. He decides _a priori_ to stop gambling at the moment when he either has $\$500$ or $\$0$ in his stock. Time is counted by the number of rounds played. 1. Show that the probability that he will never stop gambling is zero. 2. Compute the probability that at the time when he stops gambling he has $\$500$ and the probability that he has $\$0$.

i) I do not know a great way to do, but I'll try to make a proof. He will eventually get to $0$ or $500$ since he moves up one dollar or down one dollar without a pattern that restricts the $0$ or $500$ option.

ii) The probability is $\boxed{\frac25}.$ The only thing I can think of is states. First, try $0$ or $400$ and they go to either one with probability $\frac12.$ Then when at $400,$ it can go to $300$ or $500$ with equal probability too. At $300,$ it can go to $100$ or $500.$ At $100,$ it can go to $0$ or $200.$ This has five variables and is much simpler to solve. Specifically, find $b$ in the system of equations $b=0.5d,d=0.5+0.5c,c=0.5a+0.5,a=0.5b.$ Solving this gives us $b=\frac25.$

To clarify, in each multiple on $100,$ we try either a) The amount of money needed to reach $500$, if possible, or b) If a isn't possible, then double or nothing. Since it has equal probability in each step, the states in the better idea are possible.