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This is a subsequent question to this previous question.

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. Calculate how many microstates there are corresponding to having $13$ cards in each pile, i.e. the most-likely configuration, $\Omega_0$.

This question comes from statistical mechanics, but it is purely regarding the mathematics of finding how many microstates or combinations of the cards.

The correct answer to the previous question was $4^{52}$

My attempt and reasoning goes as follows:

$$\frac{4^{52}\cdot(13-4)!\cdot 4!}{13!}\approx 2.84\times 10^{28}$$ Where I have simply divided the previous answer by $13\choose 4$ or ${}^{13}C_4$ since by my logic we are now choosing $4$ piles from $13$.

But the correct answer is $5.36\times 10^{28}$ or in exact form: $53644737765488792839237440000$.

I am in the correct order of magnitude, but I cannot figure out how to get the correct answer.

If someone would please explain and/or give hints/tips to calculating the correct answer I would be most grateful. Thanks.

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    As an aside, for the previous part of the question, to get the intended given answer it was intended that order within the piles is irrelevant and the piles are distinct. The answer for this with those same assumptions will be what @trueblueanil's answer accidentally was before it was deleted for the previous question, $\binom{52}{13,13,13,13}=\frac{52!}{(13!)^4} \approx 5.36\times 10^{28}$. Here, we pick which thirteen of the cards goes to the first pile, which of the remaining thirteen cards goes to the second, etc... – 2017-01-04
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    In this and in the previous question, your personal attempts you seem very confused as to how to use binomial coefficients correctly. Make sure that the number on the top and the number on the bottom make sense together. Here, $\binom{52}{13}$ is for how many ways you can pick thirteen cards out of the fifty two cards, but $\binom{13}{4}$ you have thirteen ranks and four piles... we aren't picking four piles out of the thirteen ranks., those are completely different objects that shouldn't be compared. – 2017-01-04
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    @JMoravitz Yes, you're right, my personal attempts are flooded with confusion. As I said before; combinatorics is my weakness. You have answered this and my previous question very well. Would you please consider making an answer to this question as I am still confused why $\dfrac{52!}{(13!)^4}$ gives the correct answer. I am confused as to why $(13!)^4$ is required in the denominator and why is $52!$ being used in the numerator as opposed to $4^{52}$?. I'm sorry this is so easy for you, but it's not clear to me. I just can't figure out how you knew exactly what to do to get the correct answers – 2017-01-04
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    @JMoravitz I think I understand it now. Just one question: Does the notation $\binom{52}{13,13,13,13}=\dfrac{52!}{(52-13-13-13-13)!\cdot 13!\cdot 13!\cdot 13!\cdot 13!}=\dfrac{52!}{(13!)^4}$? As I am using the formula $\binom{n}{r}=\dfrac{n!}{(n-r)!\cdot r!}$. – 2017-01-04
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    $\binom{52}{13,13,13,13}$ is the [multinomial coefficient](https://en.wikipedia.org/wiki/Multinomial_theorem#Multinomial_coefficients), the coefficient of $x^{13}y^{13}z^{13}w^{13}$ in the expansion of $(x+y+z+w)^{52}$ is the number of ways of splitting a collection of $52$ distinct objects into four distinct groups each of size $13$ where order within each group doesn't matter and happens to also equal $\binom{52}{13}\binom{39}{13}\binom{26}{13}\binom{13}{13}$ and also equal $\frac{52!}{(13!)^4}$ – 2017-01-04

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Summarizing my comments above: from the earlier problem asked and given answer, we are able to understand that the question author is intending for order within each pile to not matter and for the piles to be labeled. Due to the nature of this resembling common card games, I will hereby refer to these piles as "hands" to hopefully avoid confusion and the pile labels to be $N,E,S,W$ (the cardinal directions north east south and west).

We are tasked with finding how many ways in which these cards may be distributed to the four distinct players' hands such that each player has 13 cards in their hand (where order within the hand is irrelevant).

We approach via multiplication principle:

  • Pick which $13$ cards out of the $52$ cards are given to $N$. There are $\binom{52}{13}$ ways to select the thirteen cards used for $N$ to hold where order of selection doesn't matter.
  • Pick which $13$ cards out of the remaining $39$ cards are given to $E$. There are $\binom{39}{13}$ ways to select the thirteen cards for $E$ (noting that depending on the selection made for $N$ the available selections will change for this step, but not the total number of selections)
  • Pick which $13$ cards out of the remaining $26$ cards are given to $S$. There are $\binom{26}{13}$ ways to select these
  • The final remaining $13$ cards in the deck will be given to $W$. There is only one option available for this step.

Multiplying, we get $\binom{52}{13}\binom{39}{13}\binom{26}{13}=\frac{52!}{13!\color{blue}{39!}}\cdot\frac{\color{blue}{39!}}{13!\color{green}{26!}}\cdot\frac{\color{green}{26!}}{13!13!}=\frac{52!}{(13!)^4}$ different arrangements.

(This can easily be searched using the phrase "number of bridge hands" as bridge is one of the more common card games utilizing thirteen card hands for each of four players from a 52 card deck)

More directly, we could have described this using multinomial coefficients giving the direct answer of $\binom{52}{13,13,13,13}$ which equals the same as above. The multinomial coefficient $\binom{52}{13,13,13,13}$ could be used to describe either the number of arrangements of $13\cdot N's,~13\cdot E's,~13\cdot S's,~13\cdot W's$, and we recognize that each arrangement corresponds to a specific distribution of the cards, or we may more directly recognize the multinomial coefficient as the number of ways of partitioning a set of $52$ distinct objects into four labeled subsets each of size 13.

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    Excellent answer, thank you very much for your time and effort. – 2017-01-04
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    @BLAZE as an aside, if order within the hands is considered relevant, the answer can quickly be seen to be $52!$ as the set of permutations of the $52$ cards are in bijection with the available results of dealing the cards (*e.g. the first thirteen cards go to $N$ in the order that they appeared in the deck and the next thirteen to $E$ etc...*) – 2017-01-04