Bridge Probabilities exactly 1 person with 1 suit?

[ti-84minus]

Homework Statement

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?

Homework Equations

_nC_r = n!/(r!(n-r)!)

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!
:)

 

[Hurkyl]

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.

 

[ti-84minus]

Thanks for the reply!
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.

 

[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.

 

[HallsofIvy]

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

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

one person,

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

one particular person.

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.)

Shouldn't they all equal to 4/₅₂C₁₃.