Combinations problem involving a standard pack of $52$ playing cards and a $4$ sided die: Part 2

Question

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 $4^{52}$

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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 $13$.

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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.

Answer 1

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.