52-card deck probability…

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?

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

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 $1$.
• My method: Consider a four-block grid for each player, assigned diamond, spade, club and heart for first, second and so on. There are $13!$ ways to fill the first blocks. Same for second, third and fourth blocks. So, favorable outcomes are $(13!)^{4}$. The total number of ways of distributing 4 cards to 13 players are $52!/(4!)^{13}$. So, the probability is $(13!)^{4}\times(4!)^{13}/52!$. Am I correct? – M. Kumar Apr 6 '19 at 16:27