 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ia/paper-2/3d/solution

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

After

n

tosses, the next total is even if either the current total is even and a tail occurs, or the current total is odd and a head occurs. Therefore

q_{n + 1} = (1 - p) q_{n} + p (1 - q_{n}) = p + (1 - 2 p) q_{n} .

Since

q_{0} = 1

, subtracting the fixed point

1/2

gives

q_{n + 1} - \frac{1}{2} = (1 - 2 p) (q_{n} - \frac{1}{2}) .

Thus

q_{n} = \frac{1 + ( 1 - 2 p ) ^{n}}{2} .

The formula also covers the endpoint cases

p = 0

and

p = 1

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