A position starts at 1. At each step it moves up 1 with chance 3/4 and down 1 with chance 1/4, independently, and it stops on reaching 0 or 3. What is the chance it reaches 3 first? Give a fraction in lowest terms.

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A position starts at $1$ . At each step it moves up $1$ with chance $\frac{3}{4}$ and down $1$ with chance $\frac{1}{4}$ , independently, and it stops on reaching $0$ or $3$ . What is the chance it reaches $3$ first? Give a fraction in lowest terms.

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Hint

Write the chance from each level in terms of the levels one step away.

Solution

Let $h_{1}$ and $h_{2}$ be the chances from levels $1$ and $2$ . First-step conditioning gives $h_{1} = \frac{3}{4} \times h_{2}$ and $h_{2} = \frac{3}{4} + \frac{1}{4} \times h_{1}$ .
Substituting gives $h_{1} = \frac{9}{16} + \frac{3}{16} \times h_{1}$ .
So $h_{1} = \frac{\frac{9}{16}}{\frac{13}{16}} = \frac{9}{13}$ .

Answer: $\frac{9}{13}$

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