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Consider a standard pack of $52$ playing cards. The cards are distributed into $4$ piles completely randomly by tossing a four-sided die for each card. How many possible arrangements are there for distributing the cards among the four piles?

My attempt goes as follows:

$$\frac{52!}{(52-13)!\cdot13!\cdot2!}\approx 3.18\times 10^{11}$$ My reasoning for calculating it this way is because we are choosing $13$ cards from a selection of $52$ cards but there are $2$ different colors so I divide by these repeats.

The correct answer is $2.03\times 10^{31}$ or in exact form, it is

$20282409603651670423947251286016$.

I am $20$ orders of magnitude away from the correct answer which shows that combinatorics is not my strong point.

Could anyone please give me any hints or tips on how to reach the correct answer?

Edit:

Answers below indicate there is ambiguity in the word "arrangements" shown in the quote (the ordering of each pile may or may not matter). In which case I can only apologize for this ambiguity but I am simply quoting the professors question word for word.

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    Although @trueblueanil's answer was removed for reasons, he did bring up yet another interesting point that so far all answers given have assumed that the four piles are labeled (e.g. pile1,pile2,... or north,south,east,west) and even this is not a safe assumption. This reiterates the point that any question in enumerative combinatorics must be very precisely worded to remove any possible ambiguity, or disclaimers should be present with any answer as to what assumptions were made. – 2017-01-04
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    What JMoravtiz said :) – 2017-01-04

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The answer is simply $4^{52} $!!

Take for suppose you have to divide the cards into four groups numbered $1$, $2$, $3$ and $4$ respectively. These same numbers are also on the die. Each card has four ways of going into any one of the groups so hence for $52$ cards, we have the required number of possibilities. Hope it helps.

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    This answer does not take into account arrangement within each pile – 2017-01-04
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    @JMoravitz I worried about this at first, too. However, the deck already comes with a prearranged order, so we don't have the freedom to arrange the cards within a pile. – 2017-01-04
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For each of the 52 cards there are four possible choices for what pile it is in. So there's 4 choices for the first card, times 4 choices for the second card, times 4 for the third and so on.

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Assuming arrangement within each pile matters (unlike how the other answerers assumed) imagine three additional cards which are used to indicate when to switch to the next pile.

There are then $\frac{55!}{3!}\approx 2.1\cdot 10^{72}$ such arrangements.

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EDIT: (This all depends on what is meant by 'arrangements'; I think that in this case the correct answer is probably not what I give below (which assumes the deck is also randomized); but instead the correct answer is $4^{52}$ (we don;t care about the ordering in each pile)).

Suppose we take the usual deck of 52 cards; and add three blank White cards to the deck.

Shuffle that deck and lay the cards out on the table, face up.

Take the cards to the left of the leftmost White card (possibly none), and place them in a pile. Take the cards between the leftmost White and the second leftmost White card and place them in a second pile. Take the cards between the second leftmost White card and the third leftmost White and place them in a third pile. Take the remaining cards (those to the right of all the White cards) and place them in a fourth pile.

This is the number of possibilities as in your problem statement; so there are $\frac{(52+3)!}{3!}$ possible arrangements.

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    I agree with you completely. But all I can do is quote the professors' question and give the solution which was calculated via a computer program he wrote. Thanks for your answer. – 2017-01-04
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    No worries; this just goes to show that often one must (and should!) get a bit picky about informal terms like 'arrangement' to agree upon an answer :). – 2017-01-04