A player deals himself three cards off the top of a shuffled 52-card pack. What is the probability that all three of them fall in one and the same suit?

Total = $C(52,3)=22100$. A suit has 13 cards, so one suit supplies $C(13,3)=286$ triples; with four suits the favourable count is $4\times 286=1144$. Probability = 1144/22100 = 22/425. Trap 11/850 = 286/22100 counts one suit only and forgets to multiply by 4. Trap 11/1275 = 1144/132600 pairs an UNORDERED numerator with the ORDERED total $52\times 51\times 50$ — the two must be counted the same way. Trap $1/64=(13/52{)}^3$ assumes each card is replaced before the next draw.