The Drunk on the Bridge

A drunk man stands at the 17th metre of a 100-metre bridge. Each step he staggers one metre forward or one metre back, each with probability $\tfrac{1}{2}$ , until he either falls off the near end (metre $0$ ) or reaches the far end (metre $100$ ). What is the probability he reaches the far end, and what is the expected number of steps he takes before leaving the bridge?

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#A fair walk between two cliffs

His position is a symmetric random walk on $\{0, 1, \dots, 100\}$ , moving up or down one with equal chance, absorbing at both $0$ and $100$ .

#The hitting probability is linear

Let $h(x)$ be the probability of reaching $100$ before $0$ from position $x$ . It satisfies $h(0) = 0$ , $h(100) = 1$ , and the averaging rule $h(x) = \tfrac{1}{2}h(x+1) + \tfrac{1}{2}h(x-1)$ , which forces $h$ to be linear. Hence $h(x) = \tfrac{x}{100}$ and

\PP(\text{reaches the far end}) = h(17) = \frac{17}{100}. \tag{1}

A fair walk makes the chance of reaching the far end linear in the start, so from the 17th meter of a 100-meter bridge the walker reaches the end with probability 17/100.

#The expected wandering time

Let $T(x)$ be the expected number of steps to absorption. It satisfies $T(0) = T(100) = 0$ and $T(x) = 1 + \tfrac{1}{2}T(x+1) + \tfrac{1}{2}T(x-1)$ , that is $T(x+1) - 2T(x) + T(x-1) = -2$ . The quadratic $T(x) = x(100 - x)$ has constant second difference $-2$ and vanishes at both ends, so

\E[\text{steps}] = T(17) = 17 \cdot 83 = 1411. \tag{2}

#Slim odds, long wander

He reaches the far end only $17$ times in $100$ , since the head start of $17$ metres against a fall is small. Yet he is expected to wander $1411$ steps first, because a fair walk drifts nowhere and dawdles near the middle for a long time before either edge claims him.