Probability that every player is dealt a heart We've got a standard, 52-card deck. We're playing Bridge with 4 players, so every player is dealt 13 cards. There are $\frac{52!}{13!13!13!13!}$ ways to deal the cards to the four players. (Intuition for the computation: imagine we line the 52 cards up, and then divide them into four groups by splitting them after the $13$th, $26$th, and $39$th cards. Order doesn't matter, so then we divide by $13!$ once for every player.)
What we want to find is the probability that all of the four players are dealt at least one heart each, i.e. there is no player who is not dealt a heart. 
I know how to do this with just one player, but there's a dependence involved when dealing with multiple players, which makes things trickier. I'm thinking that the best way to do it would be to manipulate the order of $52!$ to include a heart in every partition of $13$, but I'm not sure how to do this. Surely there must be a straight-forward way?
 A: As per Rolf's hint, let $A_n$ represent the event that the North player was not dealt any hearts, $A_s$ that the south player wasn't dealt any hearts, etc.  Let $A_{e,w}$ represent that both the east and west players weren't dealt any hearts, etc.
With $S$, the sample space of all possible deals, and $A_\emptyset$ the event that every player has at least one heart, by inclusion-exclusion you have:
$|A_\emptyset| = |S| - |A_n| - |A_s| - |A_e| - |A_w| + |A_{n,s}| + |A_{n,e}|+\dots - |A_{n,e,s}| - |A_{n,e,w}| - \dots + |A_{n,e,s,w}|$
Noting further that there is a great deal of symmetry, this simplifies down to:
$|A_\emptyset| = |S| - 4|A_n| + 6|A_{n,s}| - 4|A_{n,s,e}| + |A_{n,e,s,w}|$
To calculate $|A_n|$, set it up as a multiplication principle question.  Choose the 13 non-heart cards for the north player.  Choose 13 of the remaining cards for each of the remaining players.  $|A_n| = \binom{39}{13}\binom{39}{13}\binom{26}{13}\binom{13}{13}$
To calculate $|A_{n,s}|$, choose 13 nonheart cards for north, choose 13 nonheart cards for south, choose 13 cards for each of the remaining players.  $|A_{n,s}| = \binom{39}{13}\binom{26}{13}\binom{26}{13}\binom{13}{13}$
Finally, for $|A_{n,e,s}|$ you choose the 13 nonheart cards for north, choose 13 more nonheart cards for east, choose 13 more nonheart cards for south (at this point they are the only nonheart cards left), and the only remaining cards in the deck are the hearts must go to west.  $|A_{n,e,s}|=\binom{39}{13}\binom{26}{13}\binom{13}{13}\binom{13}{13}$
Note finally that it is impossible for all players to not get a heart.
If you want the probability then, take the number of ways and divide by the size of the sample space.
A: What is the probability $A$ gets no heart? (call this event $a$)It is $\frac{39\cdot38\cdot37\dots 27}{52\cdot 51 \cdot 50 \dots 40}$
What is the probability $A$ and $B$ get no heart? (Call this event $ab$) It is $\frac{39\cdot38\dots 14}{52\cdot 51 \dots 27}$
What is the probability $A$ $B$ and $C$ get no heart? (Call this event $abc$) It is $\frac{39\cdot38\dots 1}{52\cdot 51 \cdot 50\dots 14}$
So we want:
$1-(P(a)+P(b)+P(c)+P(d))+(P(ab)+P(ac)+P(ad)+P(bc)+P(bd)+P(cd))-(P(abc)+P(acd)+P(abd)+P(bcd))$
So we want $1-4P(a)+6P(ab)-4P(abc)$, fortunately we can calculate each of the summands.
