 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ia/paper-2/9e/b/ii/solution

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

Call the coins

1, 2, 3

in the order given. Coin 1 beats the constant score of coin 2 exactly when it shows a head, so coin 1 wins with probability

3/5

. Coin 2 beats coin 3 exactly when coin 3 shows a head and scores three, again with probability

3/5

Coin 3 beats coin 1 whenever coin 3 shows a tail, or when coin 3 shows a head and coin 1 shows a tail. Its winning probability is therefore

\frac{2}{5} + \frac{3}{5} \frac{2}{5} = \frac{16}{25} > \frac{1}{2} .

Thus the preferences form a nontransitive cycle:

1 beats 2, 2 beats 3, 3 beats 1.

The second chooser can always select a coin that has winning probability greater than one half against the first choice. Therefore

it is preferable to choose second .

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