 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ia/paper-2/11f/d/solution

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

Let

E_{j}

be the expected additional number of tosses when the current terminal run contains

j

consecutive heads. Then

E_{k} = 0

, and for

0 \leq j < k

E_{j} = 1 + p E_{j + 1} + (1 - p) E_{0} .

For

0 < p < 1

, iterating this recurrence from

j = k - 1

down to

0

gives

E_{0} = (1 + (1 - p) E_{0}) \frac{1 - p ^{k}}{1 - p} .

Solving,

E_{0} = \frac{1 - p ^{k}}{( 1 - p ) p ^{k}} .

When

p = 1

, the waiting time is deterministically

k

, which is also the continuous limit of this formula as

p \to 1

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