Drunk Man

Solution

Answer: 250

Let \(E_i\) denote the expected remaining number of steps needed to reach position \(50\) when the drunk man is currently at position \(i\). At each step he moves \[ +1 \text{ with probability } 0.6, \qquad -1 \text{ with probability } 0.4. \] From position \(i\), one step occurs immediately and then the process continues from either \(i+1\) or \(i-1\). Therefore the expectation satisfies \[ E_i = 1 + 0.6E_{i+1} + 0.4E_{i-1}. \] \[ 0.6E_{i+1} - E_i + 0.4E_{i-1} = -1. \] We try a linear solution of the form \[ E_i = ai + b. \] So: \[ 0.6[a(i+1)+b] - [ai+b] + 0.4[a(i-1)+b]. \] \[ 0.6ai + 0.6a + 0.6b - ai - b + 0.4ai - 0.4a + 0.4b. \] This simplifies to \[ 0.2a. \] The recurrence requires this to equal \(-1\), so \[ 0.2a = -1 \] \[ a = -5. \] Thus \[ E_i = -5i + b. \] When the drunk man reaches position \(50\), the process stops immediately, so \[ E_{50} = 0. \] Substituting, \[ 0 = -5(50) + b \] \[ b = 250. \] Therefore \[ E_i = 250 - 5i. \] He starts at position \(0\), so \[ E_0 = 250. \] Another way to see this intuitively is to note the average drift per step: \[ 0.6(1) + 0.4(-1) = 0.2. \] On average he gains \(0.2\) steps of progress each move. To gain \(50\) units of position, the expected number of steps is approximately \[ \frac{50}{0.2} = 250. \] Thus the expected number of steps needed to get \(50\) steps ahead is \[ 250. \]