 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/paper-2/12/b/iii/solution

The equality must be a first return time: an order starting

GR

RG

would already have stopped at draw two. The only successful four-draw orders are therefore

GGRR

and

RRGG

\frac{g}{N} \frac{g - 1}{N - 1} \frac{r}{N - 2} \frac{r - 1}{N - 3} + \frac{r}{N} \frac{r - 1}{N - 1} \frac{g}{N - 2} \frac{g - 1}{N - 3} = \frac{2 g ( g - 1 ) r ( r - 1 )}{N ( N - 1 ) ( N - 2 ) ( N - 3 )} .

Here

N = g + r

. Counting all balanced four-draw orders would include games that have already stopped.

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