Past exam of the year 1 of the computer science course of the University of Oxford / 2021 / Discrete mathematics, continuous mathematics, and probability / 9 / a / Solution

Condition on the fair coin choosing the initial chamber. The law of total probability gives $\mathbb{P}(\text{reach }k)=(p_1+p_2)/2$. With $r=(1-p)/p$, substituting the biased gambler's ruin probability givesBoth numerator and denominator in the first case have the same sign. Its limit as $r\to 1$ is the fair-walk value, since the ratio of their first derivatives is $3/(2k)$. The averaging is over initial locations, not over the transition probabilities.