 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ia/paper-2/10e/b/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

For hitting zero before four, the success probabilities are

h_{i} = \frac{4 - i}{4} .

Conditioning each transition on eventual success gives

state 123 down 2/3 3/4 1 up 1/3 1/4 0.

Thus the conditional expected hitting times satisfy

e_{1} = 1 + \frac{1}{3} e_{2}, e_{2} = 1 + \frac{3}{4} e_{1} + \frac{1}{4} e_{3}, e_{3} = 1 + e_{2} .

Solving this linear system gives

E (T ∣ B) = e_{1} = \frac{7}{3} .

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