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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.

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    $\begingroup$ You might want to check the book "How to Gamble If You Must: Inequalities for Stochastic Processes", by Lester Dubins. If I remember correctly the answer given there is not as simple as $x/2A$. $\endgroup$
    – Ruy
    Dec 30, 2020 at 3:15
  • $\begingroup$ Does it help that $g(z):=f(z+A)-0.5$ is an odd function? $\endgroup$ Dec 30, 2020 at 4:26
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    $\begingroup$ @Ruy could you elaborate? The martingale method pretty conclusively gives the chance of winning as $x/2A$, if the gambler plays "as bold as possible"? $\endgroup$
    – jlammy
    Dec 30, 2020 at 14:51
  • $\begingroup$ Sorry, @jlammy but I cannot elaborate any further. All I can say is that many many years ago I took a class in Berkeley by Dubins in which this problem was extensively studied and I seem to remember that the function you are looking for was shown to be an increasing nowhere differentiable function. However memory often fails miserably so the only way forward is to go see what is in Dubins book which I think I still own but I must have misplaced since I could not find it yesterday. $\endgroup$
    – Ruy
    Dec 30, 2020 at 15:01

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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. $$P(achieving \ 2A \ capital)=x/2A$$

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