Gambler's ruin problem (conditioned on reaching a certain state before winning)

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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?

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Let's work on the simpler problem where

  • Gambler starts with some fortune $k$
  • He wins 1 with probability 1/3 and loses 1 with probability 2/3
  • and the game ends if he loses all the money or if he wins $15.

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:

  • With probability 1/3, he wins and has fortune $k+1$, or
  • with probability 2/3, he loses and has fortune $k-1$.

This lets us set up the recurrence $$p_k = \frac{1}{3} p_{k+1} + \frac{2}{3} p_{k-1}.$$ Moreover,

  • $p_{15} = 1$ (since we win at that point), and
  • $p_0 = 0$ (since we lost all the money).

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:

  • Gambler starts with some fortune $k$
  • He wins 1 with probability 1/3 and loses 1 with probability 2/3
  • and the game ends if he loses all the money or if he reaches 15.

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 $k$,

  • With probability 1/3, he wins and has fortune $k+1$, or
  • with probability 2/3, he loses and has fortune $k-1$.

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:

  • $q_3 = p_3 = \frac{2^3-1}{2^{15}-1}$, since once we hit 3, we only need to make sure the gambler will eventually win.
  • $q_{15} = 0$, since we already finished the game (by reaching 15), but never got to 3.

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 $q_9 = 64/304265$.