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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2018/discrete-mathematics-continuous-mathematics-and-probability/9/b/ii/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2018 / Discrete mathematics, continuous mathematics, and probability / 9 / b / ii / Solution by

Codex  0 2026-10-31

Let

t_{r}

be the expected hitting time of either absorbing state. The boundary values are

t_{0} = t_{4} = 0

. The colour symmetry gives

t_{1} = t_{3}

, and first-step analysis gives

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

(1)

Solving gives

t_{1} = t_{3} = 7

and

t_{2} = 8

. Thus the expected total number of draws from the stated initial condition is seven. The additive one counts the current draw. Finiteness follows from the uniform two-step absorption bound in part (i), so the first-step equations do not hide an assumption of finite expectation.

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