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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

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

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

RobPratt
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ManOnTheMoon
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    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. – Henry Apr 25 '22 at 12:10
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    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. – JMoravitz Apr 25 '22 at 12:16
  • 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. – JMoravitz Apr 25 '22 at 12:20
  • @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. – ManOnTheMoon Apr 25 '22 at 13:05
  • As @Henry notes, you should be multiplying, not adding. Also, divide by 13! at the end because the order in which you receive the cards shouldn't matter. Unfortunately, this still doesn't make the two numbers the same – Barry Carter Apr 25 '22 at 13:20
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    @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. – Barry Carter Apr 25 '22 at 13:29
  • @barrycarter, the whole idea of not counting cards that were dealt to the players when calculating the possibilities is honestly mind bending. And need time to adjust the institution of looking it in a micro level of the player $A$ perspective instead of the macro level of all players. Unless I keep doing problems that are in similar light, I'll probably fall back to the macro perspective of counting all cards dealt... – ManOnTheMoon Apr 25 '22 at 13:50
  • With regards to the latest update... you still have a summation symbol there. To emphasize, we are not adding. We are multiplying these together. Also, your iteration should have begun with $i=0$, not with $i=1$. If you insisted on writing it like this, it would have been $\dfrac{\prod\limits_{i=0}^{12}\binom{52-i}{1}}{13!}=\binom{52}{13}$... but at that point, why would you bother writing a binomial coefficient with $1$ as the bottom... rather than writing $\binom{50}{1}$ for instance, why not just write $50$? – JMoravitz Apr 25 '22 at 13:59
  • At that point, why bother writing with product symbol? Just write with falling factorial notation... that is, instead of writing $52\cdot 51\cdot 50\cdots 40$ write $52\frac{13}{~}$. At that point, the equality is hardly surprising because $\dfrac{52\frac{13}{~}}{13!}$ is precisely the definition of $\binom{52}{13}$ (or at least has been shown to be equivalent to $\binom{52}{13}$ several times over at this point in the course). Make sure that you know the difference between adding and multiplying and when each is to be used in counting problems. – JMoravitz Apr 25 '22 at 14:00
  • @JMoravitz, thanks for pointing out the product symbol (corrected). As for the use of the product symbol and binomial coefficient, I'm actually very new(or long forgotten) these concepts and symbols, hence, having them laid out is easier for me and i can verify the numbers with python when they are in pieces. fyi, python is also relatively new to me... lol – ManOnTheMoon Apr 25 '22 at 14:15

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