dealing cards probability

Question

If a standard deck of cards is deal to 4 players, 13 cards each, how many possibilities are there assuming that it matters which player gets but card order does not matter. Why is the answer not (52 choose 13)^4?

Answer 1

Once we know which $13$ cards the first player gets, there are not $52$ cards remaining that could have been dealt to the second player. There are only $39.$

After we have looked at the cards dealt to the second player, only $26$ remain that might be in the third player's hand. And once we have looked at those, we know exactly which cards the fourth player received.

So the number of possible deals is just

$$ \binom{52}{13} \binom{39}{13} \binom{26}{13} = \frac{52!}{13!\,13!\,13!\,13!}.$$

Answer 2

The answer is not $\binom{52}{13}^4$ because the various people do not get cards from a fresh deck: If South has the Ace of $\spadesuit$, then West cannot have it. Your formula would be correct if after dealing $13$ cards to South, we threw away the rest of the deck, and got a fresh deck for the deal to West, and continued in this way for the other two players.

We can find the right answer as follows. There are $\binom{52}{13}$ hands that South can get. For each of these, there are $\binom{39}{13}$ hands that West can get. And for each of these, there are $\binom{26}{13}$ hands that North can get, for a total of $\binom{52}{13}\binom{39}{13}\binom{26}{13}$.

The product simplifies to $\dfrac{52!}{(13!)^4}$.

Answer 3

It will be something like:

You can choose $13$ cards out of $52$ cards for one person, then $13$ for other from the remaining $39$ and so on:

$$\binom{52}{13}\binom{39}{13}\binom{26}{13}\binom{13}{13}=\frac{52}{(13!)^4}\approx5.36\times10^{28}$$

Your assumption assumes that there are always $52$ cards remaining:

$$\binom{51}{13}^4$$

Answer 4

There are $52!$ to sort the deck of cards. The sorted deck is divided into 4 equal-sized stacks (the players' 'hands'). However, each stack can be sorted in $13!$ ways, each of which is considered equivalent.

So there are in fact only $\dfrac{52!}{13!^4}$ distinct ways to deal the cards.

Alternatively, you can deal cards from the deck to player 1 in $\frac{52!}{13!\,39!}$ distinct ways, and from the remaining deck you can deal cards to player 2 in $\frac{39!}{13!\,26!}$ distinct ways, and so on.   It's a different binomial coefficient for each consecutive player as they are dealt cards from a sequentially reduced deck size.   You may notice already that there is going to be a lot of cancelling.

$$\begin{align}{51\choose 13}{39\choose 13}{26\choose 13}{13\choose 13} &= \frac{52!}{13!39!}\frac{39!}{13!26!}\frac{26!}{13!13!}\frac{13!}{13!0!} \\ &= \frac{52!}{13!\,13!\,13!\,13!} \\ &= {52\choose 13,13,13,13}\end{align}$$

We call this a multinomial coefficient.   It is extension of the binomial coefficient.