1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Is it self-study ? If so, please add the tag. I just posted a hint ;) $\endgroup$
    – RUser4512
    Oct 2 '15 at 10:53
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.