Hitting it Big in Vegas

This is a problem where it is very difficult to think about all the different orders in which things could happen. It could take a very long time in order for you to ever reach $\$N$ or run out money. However, we can leverage linearity of expectation and the fact that each game is fair. Because in each gamble, the expected value of how much money you'll make is 0, playing any number of times doesn't change the fact that on average, you neither gain nor lose money.

Thus, we know that the expected value of the total gain/loss is 0. With probability $\dfrac{1}{3}$ we will lose $100, and with probability $\frac{2}{3}$ we will gain some amount of money $n$ . Therefore, we can write the following equation

$\text{expected gain } = 0 = (-100)\frac{1}{3} + \frac{2}{3}n \implies n = 50$

If we stand to gain $50, we must be waiting until we have a value of $N = 100 + n = \boxed{\$ 150}$ to stop playing.

This problem requires a bit of creative thinking to simplify, but this makes the problem much easier to handle. Letting the probability of losing all your money when you have $n be ${ P }_{ n }$ , a recurrence relation can be constructed, where ${ P }_{ n }=\cfrac { 1 }{ 2 } \left( { P }_{ n-1 }+{ P }_{ n+1 } \right)$ . This comes from the fact that the probability of reaching $n from $(n-1) and $(n+1) by the next game are both $\cfrac { 1 }{ 2 }$ . Rearranging this relation gives ${ P }_{ n+1 }=2{ P }_{ n }-{ P }_{ n-1 }$ .

Solving this relation yields ${ P }_{ n }=A-Bn$ , where A and B are arbitrary constants. Solving for A and B using the conditions ${ P }_{ 0 }=1$ and ${ P }_{ N }=0$ , the relation becomes ${ P }_{ n }=1-\cfrac { n }{ N }$ . Given that the probability of losing all your money at $100 is $\cfrac { 1 }{ 3 }$ , ${ P }_{ 100 }=1-\cfrac { 100 }{ N } =\cfrac { 1 }{ 3 }$ . Solving this gives $N=150$ .