This problem is a classic example of a Gambler's Ruin scenario, where we aim to calculate the probability that player $A$ reaches a specific target amount before being ruined (i.e., their capital decreases to a set amount). Here’s a step-by-step approach to solve the problem:

1. Problem Setup:

   • Let $a$ be the initial capital of player $A$, and $b$ be the initial capital of player $B$.
   • The probability of player $A$ winning $1 in each game is $\frac{1}{3}$, while the probability of losing $1 is $\frac{2}{3}$.
   • The target for player $A$ is to reach a capital of $a + b - 4$, and the ruin threshold is when $A$'s capital drops to $5$.

2. Objective:

   • To find the probability that player $A$ reaches $a + b - 4$ before reaching 5 for the given cases:
     • i) $a = 7$ and $b = 7$
     • ii) $a = 9$ and $b = 15$

3. Solving the Gambler's Ruin Problem with Biased Probability:

   • For a biased random walk (where the probability of moving up is different from the probability of moving down), we use the formula: $P(\text{reaching target before ruin}) = \frac{1 - \left( \frac{q}{p} \right)^s}{1 - \left( \frac{q}{p} \right)^N}$ where:
     • $p = \frac{1}{3}$ (probability of winning)
     • $q = \frac{2}{3}$ (probability of losing)
     • $s = a - 5$ (current distance from ruin)
     • $N = (a + b - 4) - 5$ (distance between ruin and target)

Case (i): $a = 7$, $b = 7$

1. Substitute values:

   • $s = 7 - 5 = 2$
   • $N = (7 + 7 - 4) - 5 = 5$

2. Calculate $\frac{q}{p} = \frac{2/3}{1/3} = 2$.

3. Plug into the formula: $P(\text{reaching target}) = \frac{1 - 2^2}{1 - 2^5} = \frac{1 - 4}{1 - 32} = \frac{-3}{-31} = \frac{3}{31}$

Case (ii): $a = 9$, $b = 15$

1. Substitute values:

   • $s = 9 - 5 = 4$
   • $N = (9 + 15 - 4) - 5 = 15$

2. Using $\frac{q}{p} = 2$:

3. Plug into the formula: $P(\text{reaching target}) = \frac{1 - 2^4}{1 - 2^{15}} = \frac{1 - 16}{1 - 32768} = \frac{-15}{-32767} = \frac{15}{32767}$

Thus, the probability for each case is:

• Case (i): $\frac{3}{31}$
• Case (ii): $\frac{15}{32767}$

Would you like a detailed breakdown of each calculation or further clarification on the approach?

Further Questions:

1. How would the probability change if the probabilities of winning and losing were equal?
2. What would be the effect on the probability if $a$ and $b$ values were significantly increased?
3. Can this approach be applied if the target is further away, e.g., $a + b$ instead of $a + b - 4$?
4. How would the probability formula adjust if the ruin threshold were different from $5?
5. What if the probability of winning increased slightly—how would that impact the results?

Tip:

In Gambler's Ruin problems, understanding the relationship between winning and losing probabilities and calculating distances between thresholds are key to finding the solution accurately.