I am trying to a solve a variant on the Gambler's Ruin problem, in which two gamblers $A$ and $B$ make a series of bets until one of the gamblers goes bankrupt. $A$ starts out with $i$ dollars, B with $N-i$ dollars. The probability of A winning a bet is given by $p$, with $0 < p < 1$. Each bet is for $\frac{1}{k}$ dollars, with $k$ a positive integer.

The problem asks us to find the probability that $A$ wins the game, and to determine what happens to this as $k \rightarrow \infty$.

I know that the probability of $A$ winning in the normal gambler's ruin problem (i.e. when $k=1$) if $A$ starts out with $i$ dollars is $\frac{1-(\frac{q}{p})^i}{1-(\frac{q}{p})^n}$. My intuition is that the probability that $A$ wins the game approaches $0$ as $k \rightarrow \infty$ in this particular problem, but I am unsure of how to show this algebraically.

  • 3
    $\begingroup$ You first convert the currency into $1/k$-dollars (dollaritos). Then it becomes the classical gambler's ruin problem where $A$ starts with $ik$ dollaritos and $B$, with $(N-i)k$. You then plug this into the formula and let $k\to\infty$. $\endgroup$ – zhoraster Oct 2 '15 at 5:01

Your expression $\frac{1-(\frac{q}{p})^i}{1-(\frac{q}{p})^n}$ is fine for $k=1$ when $q=1-p \not= p$.

But when $q=p=\frac12$, that expression becomes an unhelpful $\frac00$. Instead it is $\frac{i}{n}$; one way to show this is to consider A's expected wealth at the beginning of the game, during the game, and at the end.

As zhoraster said in comments, for larger integer $k$ (i.e. smaller bets) changes this probability to $\frac{1-(\frac{q}{p})^{ik}}{1-(\frac{q}{p})^{nk}}$ for $q\not= p$ but still $\frac{i}{n}$ when $q=p$. The expected time to finish the game will increase when $k$ increases

In the limit as $k \to \infty$ the probability will tend to $0$ when $p \lt \frac12\lt q$ [the denominator will be almost $(\frac{q}{p})^{(n-i)k}$ times the numerator for large $k$], and will tend to $1$ when $p \gt \frac12\gt q$ [both numerator and denominator will be close to $1$ for large $k$]. When $p = \frac12= q$ then it will stay at $\frac{i}{n}$


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