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Find the probability that a hand of five cards in poker contains four cards of one kind.
asked in Mathematical Logic by Active (2k points)   | 36 views

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Case 1: 4 Ace, 1 card can any out of remaining 48

Case 2: 4 cards of face '2', 1 card can any out of remaining 48

And so on for each 13 kinds.

So, probablity = $\frac{\binom{48}{1}*13}{\binom{52}{5}}$ = $\frac{1}{4165}$
answered by Veteran (14.9k points)  

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