A token sits at position 1 among the positions 0, 1, 2, 3. Each move goes up by 1 with probability 2/3 and down by 1 with probability 1/3, independently. Positions 0 and 3 are absorbing. What is the probability that the token is absorbed at 3?

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Biased walk on $4$ points

ProbabilityMedium

A token sits at position $1$ among the positions $0$ , $1$ , $2$ , $3$ . Each move goes up by $1$ with probability $\frac{2}{3}$ and down by $1$ with probability $\frac{1}{3}$ , independently. Positions $0$ and $3$ are absorbing. What is the probability that the token is absorbed at $3$ ?

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Hint

Write one equation for each interior position for the probability of ending at the top, then solve the pair.

Solution

Let $h (k)$ be the probability of absorption at $3$ from position $k$ , with $h (0) = 0$ and $h (3) = 1$ .
From position $1$ : $h (1) = (\frac{2}{3}) \times h (2)$ .
From position $2$ : $h (2) = \frac{2}{3} + (\frac{1}{3}) \times h (1)$ .
Substituting gives $h (1) = \frac{4}{9} + (\frac{2}{9}) \times h (1)$ , so $(\frac{7}{9}) \times h (1) = \frac{4}{9}$ .
Therefore the probability is $\frac{4}{7}$ .

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

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