Probability of drawing a kind from a deck of poker

[Phys12]

Homework Statement

Find the probability that a hand of five cards in poker contains four cards of one kind.

Homework Equations

The Attempt at a Solution

Solution given in the book:[/B]
By the product rule, the number of hands of five cards with four cards of one kind is the product of the number of ways to pick one kind, the number of ways to pick the four of this kind out of the four in the deck of this kind, and the number of ways to pick the fifth card.
This is:
C(13, 1)C(4, 4)C(48, 1).
By Example 11 in Section 6.3 there are C(52, 5) different hands of five cards. Hence, the probability that a hand contains four cards of one kind is
C(13, 1)C(4, 4)C(48, 1)
------------------------------
C(52, 5)

I understand the last part (C(48,1)), we have chosen our 4 cards, now there are 48 cards remaining in the deck and we need the 5th card (choosing only 1 card), hence, we get C(48, 1). However, I don't get the first two (I kind of get C(13, 1), but not C(4,4)).

If I imagine a deck placed in 13 different sections each section containing 4 of each kind, then I get why we will have C(13,1), since of those 13 sections, we need to pick one kind. Is that a correct way of thinking about it? And if yes, then why even mention C(4,4) since we're picking up all the 4 cards when we pick 1 kind?

 

[Orodruin]

And if yes, then why even mention C(4,4) since we're picking up all the 4 cards when we pick 1 kind?

That is exactly the point. You must choose all four and there are C(4,4) = 1 ways of doing that.

 

[Phys12]

That is exactly the point. You must choose all four and there are C(4,4) = 1 ways of doing that.

Okay and the way that I am picturing the problem, is that correct? (The one in which I stack 4 cards in each section, all 4 belonging to the same kind?)

 

[Orodruin]

Okay and the way that I am picturing the problem, is that correct? (The one in which I stack 4 cards in each section, all 4 belonging to the same kind?)

Yes. The C(13,1) is choosing one of the 13 denominations to make your 4-of-a-kind. You need to pick all four of that denomination (C(4,4)), and you need to choose an additional card out of the 48 remaining (C(48,1)).

 

[Ray Vickson]

Homework Statement

Find the probability that a hand of five cards in poker contains four cards of one kind.

Homework Equations

The Attempt at a Solution

Solution given in the book:[/B]
By the product rule, the number of hands of five cards with four cards of one kind is the product of the number of ways to pick one kind, the number of ways to pick the four of this kind out of the four in the deck of this kind, and the number of ways to pick the fifth card.
This is:
C(13, 1)C(4, 4)C(48, 1).
By Example 11 in Section 6.3 there are C(52, 5) different hands of five cards. Hence, the probability that a hand contains four cards of one kind is
C(13, 1)C(4, 4)C(48, 1)
------------------------------
C(52, 5)

I understand the last part (C(48,1)), we have chosen our 4 cards, now there are 48 cards remaining in the deck and we need the 5th card (choosing only 1 card), hence, we get C(48, 1). However, I don't get the first two (I kind of get C(13, 1), but not C(4,4)).

If I imagine a deck placed in 13 different sections each section containing 4 of each kind, then I get why we will have C(13,1), since of those 13 sections, we need to pick one kind. Is that a correct way of thinking about it? And if yes, then why even mention C(4,4) since we're picking up all the 4 cards when we pick 1 kind?

Start again: your answer is orders of magnitude off. Your answer evaluates as 0.0002401, while the true answer is about 0.04292.

Hint: apply the hypergeometric distribution.

 

[Phys12]

Start again: your answer is orders of magnitude off. Your answer evaluates as 0.0002401, while the true answer is about 0.04292.

Hint: apply the hypergeometric distribution.

That's the answer that's given in the book. What's wrong with the procedure that I am using?

 

[Orodruin]

Start again: your answer is orders of magnitude off. Your answer evaluates as 0.0002401, while the true answer is about 0.04292.

Hint: apply the hypergeometric distribution.

If you are saying that the probability of drawing 4-of-a-kind in the deal in 5 card draw is 4.3% I would like to play a few hands of poker ...

Edit: The chance of obtaining two pairs in the deal is about 5%. This happens much more often than 4-of-a-kind.

Edit 2: You seem to be computing the probability of getting 4 cards of the same suite, not 4-of-a-kind.

 

[Ray Vickson]

If you are saying that the probability of drawing 4-of-a-kind in the deal in 5 card draw is 4.3% I would like to play a few hands of poker ...

Edit: The chance of obtaining two pairs in the deal is about 5%. This happens much more often than 4-of-a-kind.

Edit 2: You seem to be computing the probability of getting 4 cards of the same suite, not 4-of-a-kind.

Yes, you are right!

Back to the drawing board.

 

[Ray Vickson]

That's the answer that's given in the book. What's wrong with the procedure that I am using?

Nothing wrong: I computed the wrong probability. Your answer is OK.