1. **Problem Statement:** You start with 3 tokens and want to reach 5 tokens before hitting 0. Each turn, you bet the maximum tokens possible but not more than needed to reach 5 if you win. The probability of winning each bet is $\frac{3}{4}$. We want to find the probability of reaching 5 tokens before 0 using this bold betting strategy. 2. **Understanding the Strategy:** At each state $k$ tokens, you bet $\min(k, 5-k)$ tokens. This means: - From 3 tokens, bet $\min(3, 2) = 2$ tokens. - From 4 tokens, bet $\min(4, 1) = 1$ token. 3. **States and Transitions:** - From 3 tokens: - Win (prob $\frac{3}{4}$): move to $3 + 2 = 5$ tokens (goal reached). - Lose (prob $\frac{1}{4}$): move to $3 - 2 = 1$ token. - From 1 token: - Bet $\min(1, 4) = 1$ token. - Win (prob $\frac{3}{4}$): move to $1 + 1 = 2$ tokens. - Lose (prob $\frac{1}{4}$): move to $1 - 1 = 0$ tokens (ruin). - From 2 tokens: - Bet $\min(2, 3) = 2$ tokens. - Win (prob $\frac{3}{4}$): move to $2 + 2 = 4$ tokens. - Lose (prob $\frac{1}{4}$): move to $2 - 2 = 0$ tokens. - From 4 tokens: - Bet $\min(4, 1) = 1$ token. - Win (prob $\frac{3}{4}$): move to $4 + 1 = 5$ tokens (goal). - Lose (prob $\frac{1}{4}$): move to $4 - 1 = 3$ tokens. 4. **Define $p(k)$ as the probability of reaching 5 tokens before 0 starting with $k$ tokens.** Boundary conditions: $$p(0) = 0, \quad p(5) = 1$$ From the transitions: $$p(3) = \frac{3}{4} \times p(5) + \frac{1}{4} \times p(1) = \frac{3}{4} \times 1 + \frac{1}{4} p(1) = \frac{3}{4} + \frac{1}{4} p(1)$$ $$p(1) = \frac{3}{4} p(2) + \frac{1}{4} \times 0 = \frac{3}{4} p(2)$$ $$p(2) = \frac{3}{4} p(4) + \frac{1}{4} \times 0 = \frac{3}{4} p(4)$$ $$p(4) = \frac{3}{4} \times 1 + \frac{1}{4} p(3) = \frac{3}{4} + \frac{1}{4} p(3)$$ 5. **Solve the system:** Substitute $p(1)$ and $p(2)$: $$p(1) = \frac{3}{4} p(2) = \frac{3}{4} \times \frac{3}{4} p(4) = \frac{9}{16} p(4)$$ Substitute $p(1)$ into $p(3)$: $$p(3) = \frac{3}{4} + \frac{1}{4} \times \frac{9}{16} p(4) = \frac{3}{4} + \frac{9}{64} p(4)$$ Substitute $p(3)$ into $p(4)$: $$p(4) = \frac{3}{4} + \frac{1}{4} p(3) = \frac{3}{4} + \frac{1}{4} \left( \frac{3}{4} + \frac{9}{64} p(4) \right) = \frac{3}{4} + \frac{3}{16} + \frac{9}{256} p(4) = \frac{15}{16} + \frac{9}{256} p(4)$$ Rearrange to isolate $p(4)$: $$p(4) - \frac{9}{256} p(4) = \frac{15}{16}$$ $$p(4) \left(1 - \frac{9}{256} \right) = \frac{15}{16}$$ $$p(4) \times \frac{247}{256} = \frac{15}{16}$$ $$p(4) = \frac{15}{16} \times \frac{256}{247} = \frac{15 \times 16}{247} = \frac{240}{247}$$ 6. **Find $p(3)$:** $$p(3) = \frac{3}{4} + \frac{9}{64} \times \frac{240}{247} = \frac{3}{4} + \frac{2160}{15808} = \frac{3}{4} + \frac{135}{988}$$ Convert $\frac{3}{4} = \frac{741}{988}$: $$p(3) = \frac{741}{988} + \frac{135}{988} = \frac{876}{988} = \frac{219}{247}$$ **Final answer:** $$\boxed{\text{The probability of reaching 5 tokens before 0 starting from 3 tokens is } p(3) = \frac{219}{247} \approx 0.8866}$$