Suppose we have a symmetric case of Gambler's Ruin, i.e. we start with £$k$, $k\in\{1,2,...,n-1\}$, each play has a probability $p=\frac{1}{2}$ of winning £$1$, and an equal probability of losing £$1$. The game ends when we have either £$0$ or £$n$.
I have shown that the probability of starting with £$k$ and winning the game with £$n$ is $p_k=\frac{k}{n}$, and that $\mathbb{P}(\text{wins on 1st play}\;|\;\text{winning game with £n})=\frac{k+1}{2k}$
The next part of the question asks to find: $$\mathbb{P}(\text{we have £1 total at some point}\;|\;\text{winning game with £n)}$$
I am very unsure how to proceed with finding this probability. I have tried to express it in terms of more standard probabilities involved but I can't seem to make any progress. If I could get some pointers in the right direction that would be great! Thank you
What is the probability that you have $1$ unit at some time, given that you win the game?
Clearly, first your money must decrease from $k$ to $1$; this is then followed by the process of winning (= going from $1$ to $n$). These two processes are independent, hence the overall probability is the product of the probabilities of these two steps.
For the first step, we simply reduce the initial amount $k$, the winning amount $n$ and the unit amount $1$ by $1$. We can then use the formula for the chance of losing and obtain $$P = \frac {n-k}{n-1}$$
For the second step, we simply set $k = 1$ and use the formula for winning: $$P = \frac {1}{n}$$
Multiplying these two results yields $$P = \frac {n-k} {n(n-1)}$$