What is the probability of drawing exactly two aces in four card draws without replacement from a standard deck?
Instruction: Calculate the probability considering the order of draws does not matter.
Context: This question tests the candidate's understanding of combinations and probability theory in a card game context.
Updated
Official answer
Assume a uniformly shuffled standard 52-card deck and four draws without replacement. Choose exactly two of the four aces and two of the 48 non-aces:
P = C(4,2) C(48,2) / C(52,4) = 6 × 1128 / 270725 = 6768/270725, approximately 2.50%.
The non-ace choice enforces “exactly two”: selecting the other cards from all remaining cards would also include three- or four-ace hands. The numerator and denominator both count unordered four-card hands, so no extra ordering factor is needed.