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I am trying to figure out the solution to the following - "A standard 52-card deck consists of one of each combination of: 13 different ranks (Ace, 2, 3,...,10, Jack, Queen, King) and 4 different suits (clubs, diamonds, hearts, spades), since $13 \cdot 4 = 52$. In how many ways a 52-card deck can be dealt to thirteen players, four to each, so that every player has one card of each suit?"

I came up with $(13!) ^ 4 \cdot $ (number of ways to pick the players).

My reasoning is as follows - let's take a suit (say Clubs) - its first card (out of 13) can be dealt to any of the players (say player 1), then to player 2, player 3 ... player 13. So, $13 \cdot 12 \cdot 11 \ldots 1$, i.e. $13!$. Same reasoning for the other $3$ suits. So, we get $(13!)^4$.

Is this correct ?

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  • $\begingroup$ @JMoravitz - Thanks for the reply. My reasoning is as follows - let's take a suit (say Clubs) - its first card (out of 13) can be dealt to any of the players (say player 1), then to player 2, player 3 ... player 13. So, 13 * 12 * 11 .... 1 i.e. 13!. Same reasoning for the other 3 suits. So, we get $(13!) ^ 4$. Additionally, we have 13 choices to pick player 1, then 12 for player 2 ... which makes it another 13!. Hence, the answer is $13! ^ 5$. $\endgroup$
    – batman08
    Oct 2, 2023 at 14:13
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    $\begingroup$ Thirteen players? Four card hands each? I read too quickly it seems and am far too used to games having only four players (as is more common in reality). Yes... this appears to be correct. For the first player, you pick which spade they get... then pick which heart they get, and so on noting that they get exactly one of each. The related question (which I mistakenly read this as at first) of if you had four players who each got thirteen cards and you ask that they all have at least one of each suit is far more complicated. $\endgroup$
    – JMoravitz
    Oct 2, 2023 at 14:13
  • $\begingroup$ @JMoravitz - However, the book that I'm referring gives the following 2 solutions - Solution 1: Let's assign each player one at a time. The first player has 13 choices for the club, 13 for the heart, 13 for the diamond, and 13 for the spade, for a total of $13^4$ ways. The second player has $12^4$ choices (since there are only 12 of each suit remaining). And so on, so the answer is $13^4 . 12^4 .  11^4 .... 2^4 ... 1^4$. $\endgroup$
    – batman08
    Oct 2, 2023 at 14:20
  • $\begingroup$ Solution 2 - Alternatively, we can assign each suit one at a time. For the clubs suit, there are 13! ways to distribute them to the 13 different players. Then, the diamonds suit can be assigned in 13! ways as well, and same for the other two suits. By the product rule, the total number of ways is $(13!)^4$. Shouldn't we also consider that there are 13 ways to pick first player, 12 for second player ... ? $\endgroup$
    – batman08
    Oct 2, 2023 at 14:20
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    $\begingroup$ If you cared about assigning seats to the players in addition to assigning the cards, then your answer would have been $(13!)^5$, and this would have been true for both the first and the second approaches to the problem. Traditionally, in problems like this... the players are already sitting around the table and are distinguishable (e.g. they have different names). We do not need to decide who goes first. $\endgroup$
    – JMoravitz
    Oct 2, 2023 at 14:23

1 Answer 1

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Collating comments:

Yes, your answer is correct for the question of distributing the 52 cards to thirteen players who each receive four cards apiece such that each receives exactly one card of each suit. (Note that the related question of distributing thirteen cards to four players such that each receives at least one of each suit is far far more complicated)

The mentioned approaches of "For each suit, iterate through the cards and assign which player receives the card. Repeat for each suit" giving $(13\times 12\times 11\times \cdots \times 2\times 1) \times (13\times 12\times \cdots \times 2\times 1)\times (13\times \cdots \times 1)\times (13\times \cdots \times 1) = (13!)^4$ and the approach of "For each player, iterate through the suits and assign a remaining card of that suit to the player. Repeat for each player" giving $(13\times 13\times 13\times 13)\times (12\times 12\times 12\times 12)\times \cdots \times (1\times 1\times 1\times 1) = (13^4)\times (12^4)\times \cdots \times (1^4)$ are both correct and both give the same final value.

These can be seen to be equal by considering that if these were expanded we would have a total of $52$ terms in the overall products of each, and in each we have exactly four $13$'s... exactly four $12$'s... exactly four $11$'s and so on albeit in a different order. But, thanks to the commutativity and associativity properties of multiplication this does not change the final resulting value. Order within a product does not matter.

The concern about player order is unfounded. When talking about "people" in a combinatorics or probability setting, it is generally assumed that each person is distinguishable. With distinguishable objects, we may always assume that there exists some "canonical order" for them. In the case of people, for example, we may assume that they all have different names and we can refer to them in alphabetical order... or for machines we may assume they all have different serial numbers and we can refer to them in order of their serial numbers etc... When distributing the cards to the people in terms of "the first player" vs "the second player" this could have been done in the canonical ordering of the players by name or it could have been done in the canonical ordering of the players by turn order... but the concept of turn order is unnecessary to exist here for the problem to make sense.

If desired, you may include in your problem the concept of turn order and if you say that the turn order of the players has not yet been defined and you wish to consider two ways to deal the cards to be distinct if the turn order was different, then this introduces an additional factor of $13!$, and this is true regardless which approach you had taken to see the original answer. Traditionally however, in problems like this it is assumed that the turn order is already predescribed and the players are already sitting in their respective seats.

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  • $\begingroup$ Thanks for the detailed reply! $\endgroup$
    – batman08
    Oct 2, 2023 at 14:57

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