Number of ways to distribute 13 (out of 52) cards to 4 indistinguishable players.

Question

Typically, I would approach this problem using combinatorics, which would ask the number of ways to distribute to 1st player, 2, 3rd, 4th, giving $$ \binom{52}{13}\binom{39}{13}\binom{26}{13} $$

but I want to understand the permutations approach, which is $$ \frac{52!}{13!13!13!13!} $$ This is equivalent to the above, but I am wondering what is the name of this formula for the permutations approach?

It looks like the formula for permutations with duplications. If that's the case, then what are the duplications here? the 13 cards in each pile?

I understand 13! is the number of permutations in each pile, but I don't understand why dividing by it is correct.

Answer 1

First, if the players are indistinguishable, then both answers should be divided by $4!$.

Second, the second approach assumes the following: you shuffle the deck, give the first player the first 13 cards, the second player the next 13 cards, and so on. The duplications are the internal order the cards are given to each player. If you take the same deck and shuffle only the first 13 cards, all players will receive the same cards as before, but you count it as a new deck (there are $52!$ of those). So you divide by these repetitions.

Example Suppose you have 4 cards to give to 2 players. Cards are labeled ABCD. The number of ways is $\tfrac{4!}{2!2!}=6$ because

AB|CD = BA|CD = BA | DC = AB |CD (these are the $4=2!2!$ reptitions of the event that player 1 has in the end the cards $A$ and $B$ and player 2 the other two.

If, players are not distinguishable than you have only 3 options, because you have only 3 ways to split the 4 cards into two decks and it doesn't matter which gets which (you can't tell them apart anyway... )

Answer 2

You can imagine dealing $13$ cards to each of $4$ players as laying out the $52$ cards in a straight line, with player $1$ getting cards $1$ to $13$, player $2$ getting cards $14$ to $26$, and so on.

So you have $52!$ ways to lay out all the cards. But for each of those arrangements, there are many duplicate arrangements in which cards $1$ to $13$ are the same, but in a different order. Similarly for cards $14$ to $26$, and so on. For each group of $13$, there are $13!$ ways to rearrange them and not change the cards in each player's hands. That's why we see four factors of $13!$ in the denominator.

I am not aware of a specific name for this technique.

Answer 3

If the players are indistinguishable, the following approach can be taken. For convenience, let f(n, k) denote the number of ways of distributing n cards among k indistinguishable players. We want to find f(52, 4). Note that in general f(n, 1)=1.

One of the players will receive card x for some specified x. In how many ways can card x be distributed? There are 12 accompanying cards included with x that will be selected among the 51 remaining cards, so we can include card x in C(51, 12) ways. Once we distribute these 13 cards including card x to any one of the players, we have 3*13=39 cards to distribute among 3 players. f(52,4) = C(51,12)*f(39,3).

We can now proceed as before to compute f(39,3). For each of the groups of cards including x, let y be come card not included with card x. In how many ways can card y be distributed? Card y must be accompanied by 12 of the 38 remaining cards, so f(39,13)= C(38,12)*f(26,2) and f(52,4) = C(51,12)*C(38,12)*f(26, 2). Proceeding in the same way we get f(26, 2) = C(25, 12)*f(13,1), and f(13,1)=1.

So after all that explanation, we get f(52, 4) = C(51, 12)*C(38, 12)*C(25, 12).