Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 2 / 11F (Probability)

Let $h_k$ be the hitting probability of the right endpoint before the left endpoint, starting at $k$. A surviving vase corresponds to that right-endpoint event. Conditioning on the first step of the simple symmetric random walk givesThe Markov property makes the remaining probability depend only on the position after that step. These absorbing endpoint values distinguish the event of interest from its complement.

Let $m_k$ be the expected remaining hitting time of either endpoint, measured in minutes. The first step consumes one minute, after which first-step analysis givesThe expectation is finite: from any interior state, a block of twenty leftward steps guarantees absorption and has probability $2^{-20}$. Thus the survival probability over successive twenty-step blocks has a geometric upper bound. This justifies using finite expectations in the recurrence.

The homogeneous linear recurrence relation has characteristic polynomial $(\lambda -1{)}^2$, hence general solution $h_k=A+Bk$. For the expected-time recurrence, rearrangement gives $m_{k+1}-2m_k+m_{k-1}=-2$, for which $-k^2$ is a particular solution. ThusImposing the absorbing boundary conditions from the preceding parts gives the gambler's ruin formulas

The initial position is $k=17$. Applying the gambler's ruin hitting probability and mean hitting time givesFor comparison, the original central starting position $k=10$ would give probability $1/2$ and mean time $100$ minutes.

Now the event is to hit $20$ before $1$, starting at $3$. Up to the first of those two hitting times, translate the interval by one unit; its endpoints become $0,19$ and the starting point becomes $2$. The gambler's ruin hitting probability is thereforeStopping at $1$ is essential: using the original absorbing endpoint $0$ would answer a different event.