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We have a gambler who at each step wins and loses $1$ dollar with probability $p$ and $1-p$ respectively. The game ends when he loses everything or wins $m$ dollar. Now starting with $i$ dollar the probability of winning $m$ dollar with $p = \frac{1}{2}$ comes out to be $\frac{i}{m}$. I want to know the intuitive argument behind this

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At present, 50+ of your questions received answers, none of which you accepted. What are your plans about this state of affairs? – Did May 25 '13 at 10:35
up vote 1 down vote accepted

Let $X_i$ be the outcome of game $i$. $$\mathbb P[X_i=1]=p, \space \mathbb P[X_i=-1]=(1-p)$$ When $$p=1-p=\frac12$$ we're dealing with a symmetric random walk problem. Let $$S_n = \sum_{i=1}^n X_i$$ be the sum of winnings up to the time $n$. We'll define a stopping time $$\tau = \inf\{n:S_0=0, S_n=M \vee S_n=-I\}$$ That is: starting at $S_0=0$ we stop either if we win M or if we lose I.

As you might notice, $S_n$ is a martingale: $$\mathbb{E}[S_n | \mathcal{F}_{n-1}] = S_{n-1}$$ and (most importantly) $$\mathbb{E}S_{\tau}=0$$

Now we'll proceed computing the expectation of winnings at the stopping time $\tau$ \begin{align} \mathbb{E}S_{\tau} =0 &= M \mathbb P[S_{\tau}=M] -I \mathbb P[S_{\tau}=-I] \\ &=M \mathbb P[S_{\tau}=M] -I \Bigg(1- \mathbb P[S_{\tau}=M] \Bigg)\\ \end{align} By the way, since there are only two ways to end this game (by the definition of $\tau$, then $$ \mathbb P[S_{\tau}=-I]= 1- \mathbb P[S_{\tau}=M]$$ must be true. Therefore, $$I=(M+I) \mathbb P[S_{\tau}=M]\\ \mathbb P[S_{\tau}=M] = \frac{I}{M+I}$$

Intuition: it's only the range $(-I,M)$ that matters. Symmetric random walk is basically a discrete version of Brownian motion, so you can try to visualize it like a particle moving randomly in the X-Y plane...winning if it reaches the upper bound, going broke if it hits the lower bound.

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