What is the probability of drawing a straight flush in a hand of 5 cards from a standard deck?
Instruction: Assume every unordered five-card hand from a standard 52-card deck is equally likely. Count royal flushes as straight flushes, allow ace low or high without wrapping, and justify the favorable count and denominator.
Updated
Official answer
A straight flush has five consecutive ranks in one suit. Include the royal flush in this event. For each of 4 suits there are 10 sequences: ace-low A2345, then 23456 through 9TJQK, and ace-high TJQKA. Ace does not wrap around in a sequence such as QKA23.
There are 40 favorable hands out of C(52,5) = 2,598,960 equally likely unordered hands:
P = 40 / 2,598,960 = 1 / 64,974 ≈ 0.0000153908, or 0.00153908%.
If a classification reports royal flushes separately, exclude those 4 hands and use 36 in the numerator. For a uniformly shuffled five-card deal, an ordered calculation gives the same ratio: every distinct hand has exactly 5! orders, so that factor cancels.
Assessment: A complete solution states whether royal flushes count, handles ace endpoints, and uses the same ordered or unordered convention in both numerator and denominator.