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Bot GPT-4o To solve the problem of determining the probability that every pile contains an Ace when the given cards are split into three piles randomly, we need to proceed through a series of logical and combinatorial steps. The deck in question contains the following 6 cards:
- Ace of spades
- Ace of hearts
- Ace of diamonds
- Ace of clubs
- Two of spades
- Two of hearts
### Total Arrangements
Firstly, we calculate the total number of ways to distribute these 6 cards into 3 piles with each pile having 2 cards.
\[ \text{Number of ways to choose 2 cards out of 6 for the first pile} = \binom{6}{2} \]
\[ \text{Number of ways to choose 2 cards out of remaining 4 for the second pile} = \binom{4}{2} \]
\[ \text{Remaining 2 cards go to the third pile automatically} \]
Using the binomial coefficient formula:
\[ \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 \]
\[ \binom{4}{2} = \frac{4!}{2!2!} = \frac{4 \times 3}{2 \times 1} = 6 \]
\[ \binom{2}{2} = 1 \]
Since the order in which we choose the piles doesn't matter, we must divide by the number of permutations of the 3 piles to avoid double-counting:
\[ \frac{15 \times 6 \times 1}{3!} = \frac{15 \times 6}{6} = 15 \]
Thus, there are 15 different ways to distribute the 6 cards into 3 piles each containing 2 cards.
### Favorable Cases
Now, we need to count the number of favorable cases where each of the 3 piles has exactly one Ace. Since there are 4 Aces and 2 non-Ace cards, at least one pile must contain both non-Ace cards while the other two contain one Ace each along with a non-Ace.
To accomplish this:
1. Select one pile to contain the two non-Aces.
- There is 1 way to place the two non-Aces: choosing the two from the two non-Aces is trivial.
2. Distribute the 4 Aces into the remaining two piles such that each pile gets exactly two Aces.
We can count this as follows:
- Choose 2 out of 4 Aces for one pile (\(\binom{4}{2}\)).
Calculating:
\[ \binom{4}{2} = \frac{4!}{2!2!} = 6 \]
Thus, there are 6 favorable ways to ensure each pile ends up with exactly one Ace.
### Probability Calculation
Finally, the probability is the ratio of favorable outcomes to total possible outcomes:
\[ P(\text{Each pile contains an Ace}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{6}{15} = \frac{2}{5} \]
Therefore, the probability that every pile contains an Ace is \(\boxed{\frac{2}{5}}\).