STEP 2013 P1 Statistics and Probability Question

There is a natural pairing of the cards into $26$ pairs that sum to $53$: $1-52,2-51$ and so on. Indeed, these are the only ways to make $53$ from two cards in the set.

The number of hands with only two cards – one pair – summing to $53$ is the product of

• $26$ ways to select the pair in hand
• $\binom{25}5$ ways to select the pairs the other five picked cards reside in – each of these cards must reside in its own pair
• $2^5=32$ ways to select which card in the selected single-card pairs is actually in the hand

Thus the final probability is $$\frac{26×\binom{25}5×32}{\binom{52}7}$$

There are $52\cdot51\cdot50\cdots46$ ways to choose $7$ cards. There are $\binom72=21$ ways to choose which two cards will be paired, $26$ ways to choose the pair, and $2$ ways to distribute the pair to the chosen positions.

The first of the remaining $5$ cards can be chosen in $50$ ways, the next in $48$ ways, and so on, because once we choose a card we cannot choose it or its pair. This gives a probability of $$\frac{21\cdot52\cdot50\cdot48\cdot46\cdot44\cdot42}{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46}$$