# Gambler's Ruin variant: each bet is for 1/k dollars // probability of winning as k approaches infinity?

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.

• Is it self-study ? If so, please add the tag. I just posted a hint ;) Oct 2 '15 at 10:53

Why not looking at the problem this way: each player keeps playing 1 dollar, but the total amount of dollars is $Nk$ (and $A$ starts with $ki$ dollars) ?
My intuition is that the answer will solely depend on the ration $p/q$, since it becomes morally equivalent to giving an infinite fortune to each player.