I have a wording question regarding the following question:

"Four players, named A, B, C, and D, are playing a card game. A standard, well-shuffled deck of cards is dealt to the players (so each player receives a 13-card hand).

How many possibilities are there for the hand that player $A$ will get? (Within a hand, the order in which cards were received doesn't matter.)"

$\require{cancel}$ From my understanding, the possibilities for player $A$'s hand is $\binom{52}{13}$ if the cards are distributed in groups of 13s. However, if the cards are distributed one after another to each of the four players, the possibilities will be $ \cancel{ \binom{52}{1}+\binom{52-4}{1}+\binom{52-8}{1}\cdots\binom{52-48}{1} }$ correct?

Sorry I'm not familiar with how most card games are distributed among players, hence, the question. And does "within a hand" simply means all the cards in hand during the game?

Kindly advise

Thanks to Henry & JMoravitz comments, I've changed my title and added the following:

$$ \binom{n}{k} = \binom{52}{13} = \frac{\prod_{i=0}^{n-1} \binom{52-i}{1}}{13!} $$

  • 2
    $\begingroup$ Since the player does not see the cards the other players get, you may want ${52 \choose 1} \times {52-1 \choose 2} \times \cdots \times {52-12 \choose 1}$, which is $13!$ times the first result as here order did matter. $\endgroup$
    – Henry
    Apr 25, 2022 at 12:10
  • 2
    $\begingroup$ As an aside, with a "well-shuffled deck" it makes no difference whether you give the cards out one at a time or whether you give them out in groups of 13 at a time. The reason why in practice we usually deal cards out one at a time is two-fold. First, it is difficult to actually shuffle a deck well in real life. You might accidentally have certain cards stick together in the shuffling process or similar. Second, it can be an anti-cheating tool. Savvy cheaters may be able to shuffle in a specific way to keep certain cards where they want them to be. It is harder if cards are separated. $\endgroup$
    – JMoravitz
    Apr 25, 2022 at 12:16
  • $\begingroup$ With regards to $(52-1)$ options versus $(52-4)$ options for the "second" card that player $A$ gets... while yes player $A$ can not get as their second card any of the cards that players B,C,D got for their first... given a first card for $A$ and if we don't keep track of what those were that the other players got it still remains the case that any of the $51$ cards are equally likely to be player A's second card. See this. $\endgroup$
    – JMoravitz
    Apr 25, 2022 at 12:20
  • $\begingroup$ @JMoravitz so the answer to (52-1) vs (52-4) depends on whether the cards are in player $A$'s hand. Hence, cards that are/were distributed to players B,C,D are considered "still" in the deck when calculating the possibilities. Therefore, the correct way to to view (n-k); is k = number of cards player $A$ has in hand, not the total number of cards distributed. $\endgroup$ Apr 25, 2022 at 13:05
  • 1
    $\begingroup$ @JMoravitz is correct. Working an example with 6 cards and 2 players (which I did but am too lazy to post) helps explain why. When you choose your 2nd card, you ARE choosing from a set of 48 cards, but WHICH set of 48 cards? It can be any of 51 choose 3 sets leftover after the other 3 players have chosen their cards. You have to consider all of these 51 choose 3 sets when computing the number of ways. $\endgroup$ Apr 25, 2022 at 13:29


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