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If 13 players are each dealt four cards from a 52-card deck, what is the probability that each player gets one card of each suit?

So I chose to do the situation where we have no repetition (i.e the values are all distinct as well as suits). I ended up with an answer of $$\frac{{13 \choose 1}{12 \choose 1}{11 \choose 1}{10 \choose 1}{4 \choose 1}^4}{52 \choose 4}.$$ My reasoning: First off, we know there are ${52 \choose 4}$ ways to get 4 cards in a card. We also know there are ${13 \choose 1}$ ways to get a card in a suit and ${4 \choose 1}$ ways to get a card from 4 different suits but same value. Once chosen, to avoid repetition of suit and value, we remove all cards of similiar value and move onto the next suit. Then, choosing our next card, there are ${12 \choose 1}$ ways to do it and the pattern continues a few more times. In the case that we can have repeated values but still different suits, I believe the answer would be $$\frac{{13 \choose 1}^4{4 \choose 1}^4}{{52 \choose 4}}$$

My question: was this thinking correct or am I just plain wrong?

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The number you obtain is bigger than $1$. – André Nicolas Jun 30 '14 at 17:38
up vote 1 down vote accepted

Let us imagine that $4$ cards are dealt to A, B, C, D in that order. First we find the probability that A gets $1$ of each suit. There are $13^4$ "favourable" choices, out of the $\binom{52}{4}$ equally likely choices. Now it is B's turn, $12^4$ good choices out of $\binom{48}{4}$ equally likely choices, and so on. We end up with a probability of $\frac{13^412^411^410^4}{\binom{52}{4}\binom{48}{4}\binom{44}{4}\binom{40}{4}}$.

Remark: Some parts of the analysis in the OP were correct. The multiplication by $4^4$ was not, and leads to a number greater than $1$.

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Thank you! This was a very concise, clear answer and I now understand my errors. Oh, you confuse me... :P – user137087 Jun 30 '14 at 18:04
You are welcome. It takes a while until the ideas get internalized. – André Nicolas Jun 30 '14 at 18:07

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