Question

All the face cards are removed from a pack of 52 playing cards, and hence the remaining pack consists of four aces, four two's and four ten's of four different suits. A matching pair (two cards with the same number) is removed from the pack and that these cards are not returned to the pack. Let $(m/n)$ be the chance that two randomly selected cards from the pack now form a matching pair, find the least value of $(m+n)$.

Answer 1

Answer: 758

The probability that next pair is of the same card type is $\left(\frac{2}{38}\cdot \frac{1}{37}\right)$ [which has been removed]
The probability that next pair is of a different card type is $\left(\frac{4}{38}\cdot \frac{3}{37}\right)$.
However, there are 9 ways this can happen, so this case contributes $9\left(\frac{4}{38}\cdot \frac{3}{37}\right)$
Hence probability is $\frac{2}{38}\cdot \frac{1}{37}+9\left(\frac{4}{38}\cdot \frac{3}{37}\right)=\frac{55}{703}=\frac{m}{n}\Rightarrow m+n=758$