# Bridge Probabilities exactly 1 person with 1 suit?

1. May 23, 2010

### ti-84minus

1. The problem statement, all variables and given/known data
In the game of bridge, four players are dealt 13 cards each from a well-shuffled deck of 52 playing cards. What is the probability that one of the players holds a hand that is made up of only one suit?

2. Relevant equations
nCr = n!/(r!(n-r)!)

3. The attempt at a solution
I know that when 1 player is dealt all of one particular suit, it should be 4/52C13
But for some reason, the answer at the back of the book for this question is very strange:
([(42*39!)/(13!3*3!) - (62*26!)/(13!2*2!) + 24] / (52!/(13!4*4!)

and this isn't equal to 4/52C13...

Since I already have the answer, I would like a thorough explanation of why it is so...
Thanks a lot!
:)

2. May 23, 2010

### Hurkyl

Staff Emeritus
You have stated three different problems in your post!

1. What are the odds that exactly one person is dealt an entire suit
2. What are the odds that one person is dealt an entire suit
3. What are the odds that one particular person is dealt an entire suit

I imagine you will be able to solve things yourself once you can recognize how these are three different problems.

3. May 23, 2010

### ti-84minus

But I still do not recognize the difference between those three.
exactly one person,

one person,

one particular person.
Shouldn't they all equal to 4/52C13.

Last edited by a moderator: May 27, 2010
4. May 23, 2010

### happyg1

You have to account for the possibility that 2 or 3 or 4 are all dealt the same suit...that's where the exactly one, one particular comes in.

5. May 27, 2010

### HallsofIvy

exactly one of East, West, North, or South is dealt all the same suit

At least one but possibly more are dealt all the same suit.

North is dealt all the same suit.

(or "South is dealt all the same suit" or "East is dealt all the same suit" or "West is dealt all the same suit". Those four have the same probability.)