From a shuffled 52-card pack, two cards are lifted out at one go. What is the probability that the pair consists of a king together with a queen?

Total = $C(52,2)=1326$. A favourable hand pairs any of the 4 kings with any of the 4 queens: $4\times 4=16$ such pairs. Probability = 16/1326 = 8/663. Trap 16/663 doubles the count 'for the two orders', but a pair drawn together has no order — the 16 already covers every hand. Trap 4/663 = 16/2652 divides the unordered 16 by the ordered total $52\times 51$. Trap $1/169=(4/52)\times (4/52)$ assumes the first card is put back.