If you play a game where you flip a coin if it lands heads you win £1 and tails you lose £1.If you start with $£K$ what is the probability that you are bankrupt after $n$ games?
MY ATTEMPT
I have just started learning about Short Random Walks, but I am not sure how to work this out while ruling out the possibility that the walk does not go below zero (play after bankruptcy) before the nth turn, any help would be appreciated.
Suppose that we keep on playing even if we go bankrupt (possibly ending in negative numbers as if ending with debt), the money you have after $n$ coin-flips would be distributed like $X = K-n + 2 Bin(0.5,n)$ where $Bin(p,n)$ denotes a binomial distribution.
An important observation is that the probability of reaching $0$ pounds somewhere during gambling and ending with $k$ pounds is the same as the probability of ending with $-k$ pounds. You can see this by flipping the last part of the random walk after reaching zero. Therefore, the probabilty of going bankrupt is $2P\{X<0\} + P\{X=0\}$.
This is $2P\{Bin(0.5,n)<\frac{n-k}{2}\} + P\{Bin(0.5,n)=\frac{n-k}{2}\}$ which you can calculate exactly or you can approximate this using the central limit theorem.