To determine the probability that each of the 4 hands contains exactly one ace when dividing a standard deck of 52 cards into 4 hands of 13 cards each, let’s break down the problem.
First, we note that there are a total of 4 aces in the deck, and we want to distribute these 4 aces uniformly across the 4 hands. We can think of it as each hand must receive one of the 4 aces.
Next, let’s figure out the total number of ways to select the cards for the hands. We are using combinations since the order of the cards in each hand does not matter. The total number of ways to form 4 hands of 13 cards from 52 cards is given by:
C(52, 13) * C(39, 13) * C(26, 13) * C(13, 13)Now, let’s calculate how many favorable outcomes we have where each hand gets one ace. We can choose 1 ace to go into each hand in 4! (4 factorial) ways since there are 4 aces and each hand must receive one.
After placing the aces, we have 48 cards left (52 total – 4 aces), and we need to fill the remaining slots in the hands (12 cards per hand). The number of ways to do this is:
C(48, 12) * C(36, 12) * C(24, 12) * C(12, 12)Putting this all together, the probability that each hand has exactly one ace is given by the ratio of the number of favorable outcomes to the total outcomes:
P(Each hand has an ace) = (4! * C(48, 12) * C(36, 12) * C(24, 12) * C(12, 12)) / (C(52, 13) * C(39, 13) * C(26, 13) * C(13, 13))This calculation gives us the probability that each hand of 13 cards contains exactly one ace. Doing the arithmetic will give you the exact fraction or decimal value for the probability.