A trader holds 1 chip and plays rounds. In each round, independently, the chip count rises by 1 with probability 2/3 and falls by 1 with probability 1/3. Play stops when the count reaches 0 or 3. What is the probability that play stops at 3? Give a fraction a/b in lowest terms.

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A trader holds $1$ chip and plays rounds. In each round, independently, the chip count rises by $1$ with probability $\frac{2}{3}$ and falls by $1$ with probability $\frac{1}{3}$ . Play stops when the count reaches $0$ or $3$ . What is the probability that play stops at $3$ ? Give a fraction a/b in lowest terms.

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Hint

Write the success probability from each of the counts $1$ and $2$ in terms of the other, then solve.

Solution

Let $h 1$ and $h 2$ be the probabilities of ending at $3$ starting from counts $1$ and $2$ .
From count $1$ , a loss ends the game at $0$ , so $h 1 = (\frac{2}{3}) h 2$ .
From count $2$ , a win ends the game at $3$ , so $h 2 = \frac{2}{3} + (\frac{1}{3}) h 1$ .
Substituting gives $h 1 = \frac{4}{9} + (\frac{2}{9}) h 1$ , so $(\frac{7}{9}) h 1 = \frac{4}{9}$ .
Therefore the probability is $\frac{4}{7}$ .

Answer: $\frac{4}{7}$

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