Question 1210213: The following cards are split into four piles at random, so that every pile contains the same number of cards. What is the probability that every pile contains an Ace?
The cards are:
Ace of clubs
Ace of diamonds
Ace of spades
Ace of hearts
King of clubs
King of diamonds
King of spades
King of hearts
Found 2 solutions by CPhill, greenestamps:
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Solution:
The set of cards consists of 4 Aces (A♣, A♦, A♠, A♥) and 4 Kings (K♣, K♦, K♠, K♥), totaling 8 cards.
These 8 cards are split into four piles at random, such that every pile contains the same number of cards. Since there are 8 cards and 4 piles, each pile must contain $8 / 4 = 2$ cards.
We want to find the probability that every pile contains an Ace. This means that the four Aces must be distributed such that each of the four piles has exactly one Ace.
First, let's determine the total number of ways to split the 8 cards into four piles of 2 cards each. We can do this by considering the ways to choose the cards for each pile sequentially.
The number of ways to choose 2 cards for the first pile is $\binom{8}{2}$.
The number of ways to choose 2 cards for the second pile from the remaining 6 is $\binom{6}{2}$.
The number of ways to choose 2 cards for the third pile from the remaining 4 is $\binom{4}{2}$.
The number of ways to choose 2 cards for the fourth pile from the remaining 2 is $\binom{2}{2}$.
The product of these binomial coefficients is $\binom{8}{2} \times \binom{6}{2} \times \binom{4}{2} \times \binom{2}{2} = 28 \times 15 \times 6 \times 1 = 2520$.
However, since the order of the piles does not matter, we must divide by the number of permutations of the 4 piles, which is $4! = 24$.
Total number of ways to split the cards into four unordered piles of 2 = $2520 / 24 = 105$.
Now, let's consider the number of ways to split the cards such that every pile contains an Ace. This means the four Aces are paired with one other card each. The other cards must be the four Kings.
We need to form four pairs, each containing one Ace and one King.
The possible pairings are (A♣, one King), (A♦, one of the remaining Kings), (A♠, one of the remaining Kings), (A♥, the last King).
Consider the pairings as forming the four piles. We can pair A♣ with any of the 4 Kings.
Once A♣ is paired, A♦ can be paired with any of the remaining 3 Kings.
Once A♦ is paired, A♠ can be paired with any of the remaining 2 Kings.
Once A♠ is paired, A♥ must be paired with the last King.
The number of ways to form these four Ace-King pairs is $4 \times 3 \times 2 \times 1 = 24$.
Each pair forms one of the four piles. The order of the piles does not matter, so this number directly corresponds to the number of favorable outcomes.
The probability that every pile contains an Ace is the ratio of the number of favorable outcomes to the total number of ways to split the cards into four unordered piles:
Probability = (Number of ways every pile contains an Ace) / (Total number of ways to split the cards)
Probability = $24 / 105$
Simplifying the fraction: $24 / 105 = (3 \times 8) / (3 \times 35) = 8 / 35$.
Final Answer: The final answer is $\boxed{8/35}$
Answer by greenestamps(13215)  (Show Source):
You can put this solution on YOUR website!
The AI solution from the other "tutor" involves finding the number of ways of splitting the 8 cards into 4 piles of 2 card each.
The way I read the problem, the 8 cards have already been split into 4 piles of 2 cards each; the question is the probability that each of those piles of 2 cards contains an ace.
So the problem is only asking the probability that the 4 aces get in separate piles.
The first ace can be in any of the 4 piles -- probability 4/4
The second ace can be in any of the 3 remaining piles -- probability 3/4
The third ace can be in any of the 2 remaining piles -- probability 2/4
The fourth ace can be in any of the 1 remaining piles -- probability 1/4
The probability that the 4 aces are in different piles is
It doesn't matter which king gets with which ace, so that is the final answer to the problem.
ANSWER: 3/32
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