You play a game with a coin. You may place a bet, if Heads if flipped then you receive your bet plus the same in winnings. If tails is flipped, then you lose your bet. You have \$ 10 and you want to turn this into \$ 100 by continuously betting \$ 1 at a time, walking away when you either have a total of \$ 100 or are bankrupt. What is the probability that you are bankrupt?
I was asked this question in a test today
I understand that its a 1D random walk, similar to the gambler's ruin problem but I'm not sure if the answer is 0.1 or 0.9
Our starting point is 10, bankrupt is -10 from the starting point and winning 100 dollars is +90 from the starting point
Now my confusion is should it be $$ \frac {10} {10+90} $$ OR $$ \frac {90} {10+90} $$
Since it's a fair game, the expected value of your wealth at any point in time is \$10. This includes random times (such as the first time you hit \$0 or \$100), as long as those random times are stopping times, which means that they don't look into the future. This is to rule out things like "The time when your wealth is the greatest out of the first 10 tosses."
Since the first time you hit \$0 or \$100 is indeed a stopping time, your expected value at that point is \$10. If we call $p$ the probability of bankruptcy, we can compute the expected value at that time to be $0 \cdot p + 100 \cdot (1-p) = 100 \cdot (1-p)$. Setting that equal to $10$ and solving for $p$ gives $p = \frac{90}{100} = 0.9$.