4 cards randomly dealt to every player out of 13 players While preparing for my midterm examination, I came across this review problem and I'm stuck.
4 cards are randomly dealt to every player, and there are a total of 13 players.

What is the probability of A = {each player has one card from each suit}
What is the probability of B = {each player has all four cards of the same value}
Compute $P_B(A)$ and $P_A(B)$. Are A and B independent?
What is the probability that each player has exactly three cards of the same value?

 A: There are 52! ways to deal out the cards. However, the order of cards within a hand doesn't matter, so we can divide by $$4!^{13}$$, giving a denominator of $$52! / 4!^{13}$$.
For P(A), there are 13 cards of each suit for player 1, 12 cards of each suit for player 2, and so on. Thus, the total is $$13^4 * 12^4 * 11^4 * \dots = 13!^4$$
For P(B), there are 13 possible hands for player 1, 12 possible hands for player 2, and so on. Thus, the total is $$13!$$
Note that in order for a player to have four of the same value, they must be different suits. Thus, $$P_B(A) = 1$$ and $$P_A(B) = P(B) / P(A) = 1/13!^3$$
It is also clear to see that the events are not independent.
A: A start: Line up the players in order of student number, and call them Player 1, Player 2, and so on up to Player 13. 
There are $\binom{52}{4}$ ways to choose the cards that Player 1 gets, and for each of these ways there are $\binom{48}{4}$ ways to choose the cards that Player 2 gets, and so on, for a total of $\binom{52}{4}\binom{48}{4}\binom{44}{4}\cdots \binom{4}{4}$ ways. 
This number simplifies to $\dfrac{52!}{(4!)^{13}}$. All these ways are equally likely. 
For event $A$, we count the "favourables." The spades that the various players get can be chosen in $13!$ ways. For each of these ways, the hearts they get can be chosen in $13!$ ways, and so on, for a total of $(13!)^4$.
Next we count the "favourables" for event $B$. The kinds the various players get can be chosen in $13!$ ways. 
Remark: The exactly $3$ will be tricky. It is not difficult to give an expression for the number of ways to distribute the "threes." However, for determining the distributions of the "odd" cards, we will need to ensure that everybody's odd card is of a different kind than the main $3$. To count, we need to know something about derangements. 
