Bridge probability question

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.

Answer 1

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

Answer 2

Let me proceed step by step, giving fuller expressions than necessary to show the logic.

• P(any particular player has a perfect hand) $=\frac{4\binom{13}{13}}{\binom{52}{13}} = X$

• P(any particular two players have a perfect hand) $=\frac{4\binom{13}{13}3\binom{13}{13}}{\binom{52}{13}\binom{39}{13}}= Y$

• P(any particular three players have a perfect hand) $=\frac{4\binom{13}{13}3\binom{13}{13}2\binom{13}{13}}{\binom{52}{13}\binom{39}{13}\binom{26}{13}}= Z$

• P(all four players have a perfect hand) $=Z$

  We shall now name the $4$ players $A,B,C,D$, and (of course !) you are $A$

Part one

Now if $A$ has a perfect hand, we are also counting perfect hands of doubles like $AB$ twice, and triples like $ABC$ thrice, etc, so by PIE,

P(at least one person has a perfect hand)
$=\binom41X -\binom42Y + \binom43Z - \binom44 Z$ which yields
$\frac{18772910672458601}{745065802298455456100520000} = 2.519631234...e^-{11}$

Part two

Firstly, let us work out the probability that exactly one hand is perfect.

Again, we use PIE, but there is a difference !
We need to correct for double counting $2$ perfect hands, but when we do that, we have counted $3$ perfect hands $\binom31 - 2\binom32 = -3$ times, which we need to correct, and so on.

Thus the appropriate expression now is $4X - \binom21 6Y + \binom31 4Z - \binom41 Z$ which yields
$=\frac{242753155112819}{9634471581445544690955000} = 2.519631233...e^{-11} $

Note that the results tally with those quoted by wolfram alpha

But we are not done yet. The above probability is that of exactly any one hand is perfect. The probability that only your hand is perfect will be $\frac14$ of the above.