A gambler plays an even odds game, where they either win or lose the amount they staked, each with probability $1/2$. The gambler starts with a capital of $x$ dollars, where $0<x<2A$, and adopts the following strategy:
- If the gambler's capital is $y\leq A$, then they stake all their money.
- If they have $y>A$, then they bet $2A-y$.
The gambler stops playing once they have exactly $2A$ dollars. What's the probability that they reach their aim without going bankrupt, as a function of $x$?
If $S_n$ denotes the fortune of the gambler after round $n$, then $S_n$ is martingale, and $\mathbb E[S_n]=x$. It can be seen that the stopping time $T$ for the end of the game satisfies the conditions for the optional stopping theorem, so $$x=\mathbb E[S_0]=\mathbb E[S_T]=2A\cdot\mathbb P(S_T=2A).$$ Hence the probability of winning is $\frac{x}{2A}$.
This problem feels too way too cute to be nuked by martingales. I'm interested in a more elementary solution. Here's some headway I've made.
Let the probability be $f(x)$. Then $f$ satisfies $$f(x)=\begin{cases}\frac{1}{2}f(2x)&\text{if }x\in[0,A] \\ \frac{1}{2}[1+f(2x-2A)]&\text{if }x\in[A,2A],\end{cases}$$and of course $f(0)=0$, $f(2A)=1$. I'm not sure how to go from here to establishing that $f$ is linear.
I can already tell, before doing any calculus at all, since the probability of winning is 1/2 at every turn and the probability of losing it's the same 1/2 at every turn, at you can either lose or gain the same amount that you can bet the average expectancy, that is the probability of achieving 2A before going bankrupt is going to be. $$P(achieving \ 2A \ capital)=x/2A$$