+87 You take the four Aces, four 2's, and four 3's from a standard deck of 52 cards, forming a set of 12 cards. You then deal all 12 cards at random to four players, so that each player gets three cards. What is the probability that each player gets an Ace, a 2, and a 3?
+658 These numbers represent your cards if an ace = 1
1 1 1 1 2 2 2 2 3 3 3 3
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Plan:
Probability = Wanted Cases divided by Total Cases
Solve:
Total cases =
\({12 \choose 4}\) because it does not matter how you distribute the cards.
12!/(4!*8!) = 495
Wanted cases (order does not matter):
Only once case
So \(\frac{1}{495}\)? I'm not so sure about this one...
