+948 The following cards are split into three piles at random, so that every pile contains the same number of cards. What is the probability that every pile contains an Ace?
Ace of spades
Ace of hearts
Ace of diamonds
Ace of clubs
Two of spades
Two of hearts
+81 Combinatorics is not my strongest asset, but I'll give it a shot...
I think we could do this question using complementary counting. In other words, we could find all of the combinations of sets that do not contain an ace, and subtract that from the total number of sets. If the piles are indistinguishable, then one pile would have the two of spades and the two of hearts, and you would do \({ 4 \choose 2 }\), to find the number of ways to find the other sets, which evens out to 6. Now, we need to find the total number of sets, which I think would be \(\frac{{6 \choose2} \cdot {4 \choose2} \cdot 1 }{6} = 15\), since you would first choose two numbers out of the six to be in the first set, then 2 numbers out of the four remaining for the second set, and since the third set is already decided, you would just multiply by \({2 \choose 2} = 1\). However, I divided by 6, since if the sets were indistinguishable, you would need to divide by \(3!\) in order to get rid of the repeats. (I am pretty skeptical about this step, if anyone could verify / help me with this, that would be greatly appreciated!)
Finally, since there are 15 total cases, and there are 6 that do not work, there would be 9 cases that do work, so the final answer would be \(\boxed{\frac{3}{5}}\)
However, take whatever I am saying with a grain of salt, since as I said before, I am not the best at combinatorics... 
