Probability All Players Receive a Spade 52 cards were distributed among 4 players, 13 cards for each. Calculate the probability that all players have received at least one spade. 
I do not know what to do with this question, i have the answer in my textbook however i do not understand where does it come from. How to solve this question?
EDIT: I am still looking for a good explanation. 
 A: Consider 4 compartments, each having $13$ slots, and although cards are not distributed in this manner, it will not matter if we first distribute all the spades.
If one compartment is barred, all the spades have to go into the remaining $39$ slots, if two compartments are barred, they have to go into the remaining 26 slots, and so on.
Then applying the principle of inclusion and exclusion, $Pr = \frac{\dbinom{52}{13} - \dbinom41\dbinom{39}{13} + \dbinom42\dbinom{26}{13} - \dbinom43\dbinom{13}{13}}{\dbinom{52}{13}} \approx 0.9489$ 
A: Let $A_i$ be the set of all deals in which player $i$ has no spades. The event that every player receives at least one spade is $(A_1 \cup A_2 \cup A_3 \cup A_4)^c$, so the desired probability is $1 - \operatorname{Pr}(A_1 \cup A_2 \cup A_3 \cup A_4)$. This probability can be computed with the principle of inclusion and exclusion. For ease of typesetting, let's first compute $| A_1 \cup A_2 \cup A_3 \cup A_4|$.
Taking advantage of the symmetry of the events, we can write 
$$
|A_1 \cup A_2 \cup A_3 \cup A_4| = \binom{4}{1}|A_1| - \binom{4}{2}|A_1 \cap A_2| + \binom{4}{3}|A_1 \cap A_2 \cap A_3| - \binom{4}{4}|A_1 \cap A_2 \cap A_3 \cap A_4|.
$$
For $|A_1|$, we have to ensure that player 1 does not receive any spades. We can do this by selecting his thirteen cards first from among the non-spades in $\binom{39}{13}$ ways, then choose all the other players cards in $\binom{39}{13,13,13}$ ways. (Note that we do not care whether the other players receive spades or not.) Thus, $|A_1| = \binom{39}{13}\binom{39}{13,13,13}$, where the latter factor is a multinomial coefficient.
Similarly, we have
\begin{align*}
|A_1 \cap A_2| &= \binom{39}{13}\binom{26}{13}\binom{26}{13,13}\\
|A_1 \cap A_2 \cap A_3| &= \binom{39}{13}\binom{26}{13}\binom{13}{13}\binom{13}{13}\\
|A_1 \cap A_2 \cap A_3 \cap A_4| &= 0 \text{ (it is impossible for all players to receive no spades)}
\end{align*}
Use these values to compute $|A_1 \cup A_2 \cup A_3 \cup A_4|$ (it's rather tedious to type it all out again). The total number of possible deals is $\binom{52}{13,13,13,13}$, so we finally obtain
$$
\operatorname{Pr}(A_1 \cup A_2 \cup A_3 \cup A_4) = \frac{|A_1 \cup A_2 \cup A_3 \cup A_4|}{\binom{52}{13,13,13,13}} \approx 0.051.
$$
Finally, $\operatorname{Pr}((A_1 \cup A_2 \cup A_3 \cup A_4)^c) \approx 1 - 0.051 = 0.949$.
A: I would say
Favorable event: Each player has one spade + 12 additional cards of, selected from the remaining 48 cards.
The number of favorable events: $\quad \omega ={48\choose12}$ 
the number of possible events: $\quad \Omega ={52\choose13}$
$\displaystyle P = \frac{\omega}{\Omega}=\frac{{48\choose12}}{{52\choose13}}=\frac{9139}{83300} \doteq 0.110$
Edit: $\color{green}{ \text{Apparently the wrong solution.}}$ georg
