A gambler starts with an initial fortune of $\$9$. He wins $\$1$ with $p=\dfrac13$ and loses $\$1$ with $q=\dfrac23$. The game ends when the gambler loses all of his money or has a total fortune of $\$15$. What is the probability that the gambler will win the game, but that he also reached $\$3$ in certain point of time?
2026-03-25 01:36:12.1774402572
Gambler's ruin problem (conditioned on reaching a certain state before winning)
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Let's work on the simpler problem where
Let $p_k$ be the probability that starting with fortune $k$, the gambler will win. (Here $0 \leq k \leq 15$) When he has fortune $k$, there are two possibilities:
This lets us set up the recurrence $$p_k = \frac{1}{3} p_{k+1} + \frac{2}{3} p_{k-1}.$$ Moreover,
Usual methods to solve recurrence allows one to find $p_k = \frac{2^k - 1}{2^{15}-1}$ for $0 \leq k \leq 15$.
For the given problem, we can formulate a similar set up as above:
Let $q_k$ be the probability that starting with fortune $k$, the gambler will win and reach 3 at some point of time.
We can again try to formulate a recurrence relation. For $3 \leq k \leq 15$, with gambler starting with fortune $k$,
Thus we have the same recurrence $$q_{k} = \frac{1}{3}q_{k+1} + \frac{2}{3} q_{k-1}.$$ However, the boundary conditions changed:
Solving the recurrence, we have $$q_k = \frac{(2^3-1)(2^{15} - 2^k)}{(2^{15}-1)(2^{15}-2^3)}$$ Substituting $k = 9$ gives $q_9 = 64/304265$.