Given a standard deck of 52 cards, what is the probability of drawing four cards in a row that form a two-pair poker hand?
Instruction: Analyze the structure of a two-pair hand and calculate accordingly.
Updated
Example Answer
Assuming a shuffled standard deck and four draws without replacement, I interpret this as exactly two distinct ranks, with two cards of each rank. The probability is C(13, 2) × C(4, 2)² / C(52, 4), which is 2,808/270,725, or about 1.037%.
I first choose the two ranks in C(13, 2) = 78 ways, then choose two of the four suits for each rank in six ways each. There are C(52, 4) = 270,725 equally likely four-card sets in total.
Why this works
Although the cards arrive in order, every four-card set has the same 4! possible orders, so counting unordered sets gives the same probability. Four cards of one rank are excluded. This is the four-card event requested, not the probability of a conventional five-card poker hand containing two pairs.