The 52 cards from a standard deck are distributed to four distinct people, 13 to each. Find the probability that each person gets at least three cards from each suit.
My professor gave us two different answers, both of which I don't understand. I would like to figure out if which one is correct and why OR if there is another answer. 
First attempt:
$$\frac{\left[4\binom{13}{4,3,3,3}\right]^{4}}{\binom{52}{13,13,13,13}}$$
Second attempt:
$$\frac{4!\binom{13}{4,3,3,3}^{4}}{\binom{52}{13,13,13,13}}$$
What in the world is going on with this problem. Please help and thank you!
 A: Given that each person gets at least three cards from each suit, that implies that every player will have three suits with $3$ cards each and one suit with $4$ cards and furthermore that those $4$ card suits must be different between each player.  Approach with multiplication principle.


*

*Step 1a: Pick who the "spadeheavy" player is.  


*

*Step 1b: Pick who the "clubheavy" player is.

*Step 1c: Pick who the "diamondheavy" player is.

*Step 1d: Pick who the "heartheavy" player is.


*Step 2a: For the spadeheavy player, pick which four spades he receives.


*

*Step 2b: For the spadeheavy player, pick which three clubs he receives.

*Step 2c: For the spadeheavy player, pick which three diamonds he receives.

*Step 2d: For the spadeheavy player, pick which three hearts he receives.


*Step 3a: For the clubheavy player, pick which four clubs he receives.


*

*$\vdots$


*$\vdots$

*Step 5: All remaining cards go to the heartheavy player
Note that as we move through step1a-1d the number of options available decreases at each step.  Further, as we move from step2 to step3 and beyond, the number of available cards decreases at each step along the way.

 Steps 1a-d can be accomplished in $4\cdot 3\cdot 2\cdot 1=4!$ ways.  Steps 2a-d can be accomplished in $\binom{13}{4}\cdot\binom{13}{3}\cdot\binom{13}{3}\cdot\binom{13}{3}$ ways.  Steps 3a-d can be accomplished in $\binom{10}{4}\binom{9}{3}\binom{10}{3}\binom{10}{3}$ ways, and so on.

$~$

 This gives a final total of $4!\binom{13}{4}\binom{13}{3}\binom{13}{3}\binom{13}{3}\binom{10}{4}\binom{9}{3}\binom{10}{3}\binom{10}{3}\binom{7}{4}\binom{6}{3}\binom{6}{3}\binom{7}{3}$ number of arrangements.  Dividing by $\binom{52}{13,13,13,13}$ will give the corresponding probability.


Alternatively:


*

*Step1a-1d: same as before

*Step 2a: Choose which four spades go to the spadeheavy player


*

*Step 2b: Choose which 3 spades from those remaining go to the player to the left of the spadeheavy player

*Step 2c: Choose which 3 spades from those remaining go to the player across from the spadeheavy player

*Step 2d: Choose which 3 spades from those remaining go to the player to the right of the spadeheavy player


*Step 3a: Choose which four clubs go to the clubheavy player


*

*$\vdots$


*$\vdots$

 For each of the steps $2$ through $5$, we can describe this either directly as $\binom{13}{4}\binom{9}{3}\binom{6}{3}\binom{3}{3}$, or we can describe as a sequence of $4$ H's (the positions of which representing which cards from that suit go to the corresponding suit-heavy player), $3$ L's (corresponding to player left of the suit-heavy player), $3$ R's (to the right) and $3$ A's (across), giving an equivalent count of $\binom{13}{4,3,3,3}$


Once you have gotten a count of how many hand distributions there are, divide by the total number of hands possible to get the corresponding probability.

 Once all is said and done, you will find your second answer to be the correct one and this equal to the answer arrived via the first method: $\frac{4!\binom{13}{4,3,3,3}^4}{\binom{52}{13,13,13,13}}$

