# Modified “gambler's ruin” probability

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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You may want to change the title to be more informative. –  Patrick Li Oct 30 '12 at 17:35
OP being unresponsive, I have edited the title. –  Gerry Myerson Dec 1 '12 at 5:09

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

i mean $h_k$ be the probability A ruin at the state when A has a capital of $k$ instead of a state. –  Mathematics Oct 30 '12 at 17:09
$q(1-\epsilon)$??? do you mean $q\epsilon$? –  Mathematics Oct 30 '12 at 18:01
do you mean $h_1=p h_2+\epsilon h_0$ –  Mathematics Oct 30 '12 at 18:07