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Assume initially there are $2$ player gamble,$A$ and $B$. $A$ has $k$ and $B$ has $a-k$ and each round they will bet $1$ and there is always a winner in every game. Also when $A$ have $1$ dollars and just lose it, then she have probability $\epsilon$ s.t $B$ will give her $1$ dollars. Also each game A has probability $p$ to win the game

My solution: for $k\ge 1$ let $h_k$ be the probability that A is ruined at the state A has a capital of $k$ dollars. Then $h_k=ph_{k+1}+qh_{k-1}$ where $p+q=1$ and and we also know $h_0=(\epsilon)h_1+(1-\epsilon)$ and $h_a=0$ and after solving the difference equation i got $h_k=\frac{\epsilon -1}{(\epsilon-1)a-\epsilon}(a-k)$

is the probability and the answer correct ???

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    OP being unresponsive, I have edited the title.2012-12-01

1 Answers 1

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I think it should be

$h_a=0$ $h_k=ph_{k+1}+qh_{k-1}$ $h_1=ph_2+qh_0$ $h_0=\epsilon h_1+(1-\epsilon)$

We have

$h_{a-1}=qh_{a-2}$ $h_{a-2}=ph_{a-1}+qh_{a-3}=pqh_{a-2}+qh_{a-3}$

So

$h_{a-2}=\frac{qh_{a-3}}{1-pq}$

$h_{a-3}=ph_{a-2}+qh_{a-4}$

So

$h_{a-3}=\frac{pqh_{a-3}}{1-pq}+qh_{a-4}$

$h_{a-3}=\frac{q(1-pq)}{1-2pq}h_{a-4}$

$h_{a-4}=ph_{a-3}+qh_{a-5}=p\frac{q(1-pq)}{1-2pq}h_{a-4} +qh_{a-5}$

So

$h_{a-4}=ph_{a-3}+qh_{a-5}=p\frac{q(1-pq)}{1-2pq}h_{a-4} +qh_{a-5}$ $h_{a-4}=\frac{q(1-2pq)}{1-3pq+(pq)^2} h_{a-5}$

TBC

Answer updated after comment.

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    so you get the same $h_k$? i mean the solution to the difference equation2012-10-30