# Thread: Game of bridge and probability

1. ## Game of bridge and probability

Hello Members,
There are 4 players in game of bridge and thirteen cards out of 52 cards are to be distributed.(A deck of bridge cards are arranged in a four suits of 13 each. The 4 suits are called spades, clubs, hearts, diamonds. The last 2 are red, the first 2 are black. Cards of the same face value are called cards of the same kind. There are 13 face values(2,3,...,10,jack,queen,king,ace)in each suit. Playing bridge means distributing the cards to 4 players namely North, South, East, West so that each player receives 13 cards.) Now what is the probability that some player has complete suit?
Answer: I am referring to various books on statistics to answer this question, if any member knows the answer, he may answer this question

2. ## Re: Game of bridge and probability

For simplicity I will assume that the player is the first dealt to. (It can be shown that it doesn't matter but that proof is more tedious.) The probability the player is dealt, say, a heart on the first deal is 13/52= 1/2. That leaves 51 cards, 39 of them non-spades. So the probability the other three players are dealt non-spades is (39/51)(38/50)(37/49). When we get back to the player, there are 48 cards in the deck, 12 of them spades so the probability this player gets a spade as the second card is 12/48= 1/4. Continue that until all cards have been dealt. That would be the probability that player received all spades. Multiply by 4 to get the probability that player get a hand of all cards from any one of the four suits. Finally, multiply by 4 to get the probability any one of the 4 players has such a hand.

3. ## Re: Game of bridge and probability

There are $\dbinom{52}{13}$ different possible hands.

There are $4$ hands consisting entirely of 1 suit

$p = \dfrac{4}{\dbinom{52}{13}} = \dfrac{1}{158753389900} \approx 6.3 \times 10^{-12}$

4. ## Re: Game of bridge and probability Originally Posted by romsek There are $\dbinom{52}{13}$ different possible hands.

There are $4$ hands consisting entirely of 1 suit

$p = \dfrac{4}{\dbinom{52}{13}} = \dfrac{1}{158753389900} \approx 6.3 \times 10^{-12}$
The question does not ask about one hand. It suggests all four hands are dealt, what is the probability that at least one hand is 13 cards of the same suit? Unless I am misreading...

5. ## Re: Game of bridge and probability

The number of ways to deal four hands in bridge is $\dfrac{52!}{(13!)^4}$.

There are $\dfrac{39!}{(13!)^3}$ ways for North to have all spades. Thus, there are $4^2\cdot \dfrac{39!}{(13!)^3}$ ways for one of the directions to have all of one suit.
There are $\dfrac{26!}{(13!)^2}$ ways for North and South to have Spades and Hearts respectively. Thus, there are $2\cdot 6^2 \dfrac{26!}{(13!)^2}$ ways for two of the directions to both have all of one suit.
There are $\dfrac{13!}{13!}$ ways for North, South, and West to have Spades, Hearts, and Clubs respectively. Thus, there are $4!$ ways for three of the direction to have all of one suit.
There is $1$ way for North, South, East, and West to have Spades, Hearts, Clubs, and Diamonds respectively. Thus there are $4!$ ways for all four of the directions to have all of one suit.

The probability for at least one direction to have all of one suit is given by:

$\dfrac{4^2\cdot \dfrac{39!}{(13!)^3}-2\cdot 6^2 \dfrac{26!}{(13!)^2}+4!-4!}{\dfrac{52!}{(13!)^4} \approx 2.52\times 10^{-11}$

6. ## Re: Game of bridge and probability

I have solved this question as follows.
In a game of bridge, let $A_i$ be the event "Player number i has a complete suit". Then $p_i=\frac{4}{\binom{52}{13}}$; the event that both players i and j have complete suits can occur in 4*3 ways and has probability $p_{ij}=\frac{12}{\binom{52}{13}\binom{39}{13}}$. Similarly we get $p_{ijk}=\frac{24}{\binom{52}{13}\binom{39}{13} \binom {26}{13}}$

Finally $p_{ijkl}=p_{ijk}$ since whenever the 3 players have the complete suit so does the fourth.
The probability that some player has a complete suit is therefore $P_1=4*p_i-6*p_{ij}+4*p_{ijk}-p_{ijkl}$.Hence the final answer $2.519631234522642 * 10^{-11}$

7. ## Re: Game of bridge and probability Originally Posted by Vinod I have solved this question as follows.
In a game of bridge, let $A_i$ be the event "Player number i has a complete suit". Then $p_i=\frac{4}{\binom{52}{13}}$; the event that both players i and j have complete suits can occur in 4*3 ways and has probability $p_{ij}=\frac{12}{\binom{52}{13}\binom{39}{13}}$. Similarly we get $p_{ijk}=\frac{24}{\binom{52}{13}\binom{39}{13} \binom {26}{13}}$

Finally $p_{ijkl}=p_{ijk}$ since whenever the 3 players have the complete suit so does the fourth.
The probability that some player has a complete suit is therefore $P_1=4*p_i-6*p_{ij}+4*p_{ijk}-p_{ijkl}$.Hence the final answer $2.519631234522642 * 10^{-11}$
Yup, that is exactly what I had in post #5.

8. ## Re: Game of bridge and probability

Hello,
If you put your last calculation in latex in good legible form, it would be better for readers to read and understand.

9. ## Re: Game of bridge and probability Originally Posted by Vinod Hello,
If you put your last calculation in latex in good legible form, it would be better for readers to read and understand.
$\dfrac{ 4^2 \cdot \dfrac{ 39! }{ (13!)^3 } - 2\cdot 6^2 \dfrac{ 26! }{ (13!)^2 } + 4! - 4! }{ \dfrac{ 52! }{ (13!)^4 }} \approx 2.52 \times 10^{-11}$

#### Search Tags

combinatorial problem, probabaility, selfstudy 