Q:

Bridge probability question

Standard 52-card deck:
  suits: clubs (♣), diamonds (♦), hearts (♥) and spades (♠)
  each suit possible cards and their values being:A,2,3,4,5,6,7,8,9,10,J,Q,k

now 4 men pick the deck in turn. pick one card each time.
at last each one has 13 cards in hand.

The probability question is 
  <1> what is the probability for at least one men hold 13 cards of same suit.
  <2> what is the probability for only the first man hold 13 cards of same suit

I think the 1> in halfway and stuck there:
     4 *  13! * 39! / 52!, but this number is not the result, I think it still need to be divided by a number, I think the number is just the ways of allocating 13 same balls to 52 boxes.

A:

Here's the first one: Let $A_1$ be the event that player 1 has 13 cards of the same suit, and define $A_2$, $A_3$ and $A_4$ similarly.
Then the probability that at least one player has 13 cards of the same suit is \begin{align*}
P(A_1 \cup A_2 \cup A_3 \cup A_4) &= \sum\limits_{i = 1}^4 P(A_i) - \sum\limits_{i \neq j} P(A_i \cap A_j) + \sum\limits_{i \neq j \neq k} P(A_i \cap A_j \cap A_k) \\
&- P(A_1 \cap A_2 \cap A_3 \cap A_4).
\end{align*}
This is from the principle of inclusion-exclusion.  By the symmetry of the problem, we know that $P(A_1) = P(A_2) = P(A_3) = P(A_4)$, and similarly $P(A_1 \cap A_2) = P(A_1 \cap A_3) =...$ and so on.  This allows us to write $$P(A_1 \cup A_2 \cup A_3 \cup A_4) = 4 P(A_1) - 6 P(A_1 \cap A_2) + 4 P(A_1 \cap A_2 \cap A_3) - P(A_4).$$ Let's examine each term individually:

$P(A_1)$: Player $1$ must have all cards of the same suit, but there are four possible suits.  Thus there are four hands (without order) that give player 1 a hand of all one suit; since there are a total of $\binom{52}{13}$ hands player 1 may have, we get $$P(A_1) = \frac{4}{\binom{52}{13}} = \frac{4 \cdot 13! \cdot 39!}{52!}.$$
$P(A_1 \cup A_2)$:  We have $4$ choices of suit for player $1$'s hand, and $3$ choices of suit for player $2$'s hand after picking the suit for player $1$.  Player $1$ has $\binom{52}{13}$ total possible hands, and player $2$ has $\binom{39}{13}$ possible hands after choosing player $1$'s hand.  Thus, we get $$P(A_1 \cup A_2) = \frac{4\cdot 3}{\binom{52}{13}\cdot \binom{39}{13}} = \frac{4\cdot 3 \cdot 13! \cdot 13! \cdot 26!}{52!}.$$
$P(A_1 \cap A_2 \cap A_3)$:  By the same logic, we get $$P(A_1 \cup A_2 \cup A_3) = \frac{4\cdot 3 \cdot 2}{\binom{52}{13}\cdot \binom{39}{13}\cdot\binom{26}{13}} = \frac{4\cdot 3 \cdot 2 \cdot (13!)^4}{52!}.$$
$P(A_1 \cap A_2 \cap A_3 \cap A_4)$: Note that in order for 3 people to all have the same suit, all 4 people must.  Thus, this probability is the same as above: $$P(A_1 \cap A_2 \cap A_3 \cap A_4) = \frac{4\cdot 3 \cdot 2 \cdot (13!)^4}{52!}.$$

Putting this all together, we get $$P(A_1 \cup A_2 \cup A_3 \cup A_4) = \frac{16\cdot 13! \cdot 39! - 72\cdot (13!)^2\cdot (26!) + 72\cdot (13!)^4}{52!}.$$

