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.


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*Show that the probability that he will never stop gambling is zero.

*Compute the probability that at the time when he stops gambling he has $\$500$ and the probability that he has $\$0$.
 A: 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.
A: This is a standard gambling problem.
To prove that the game stops, you could note that it stops when a sequence of 300 wins occur. Now, you could try to link this action to a geometric distribution and you find that it will stop eventually.
To calculate the probability, Ross Milikan made a nice comment. Note that, we know the expectation of his end capital and note that his end capital can either be $0$ or $500$. The calculation of the probability should now be easy.
If you have any questions, feel free to ask them.
