# Studying Probability, confused about Combinations.

1. May 9, 2012

### daleklama

1. The problem statement, all variables and given/known data

Suppose 5 cards are dealt at random from a well shuffled, 52 card deck. The hand can be thought of as a random sample size 5 from 52, with no replacement.

What is:

(a) P (4 aces)
(b) P (3 aces and 2 kings)
(c) P (4 of a kind)

2. Relevant equations

I know the formula for 'combinations' (number of equally likely unordered outcomes, no replacement) is

number of equally likely unordered outcomes = n CHOOSE k, where n is the total sample size and k is the amount you're sampling/taking out.

3. The attempt at a solution

I know all of the solutions, I got them from my teacher, I just can't understand them at all!

These are the solutions:

(a) P (4 aces) = (4 choose 4) x (48 choose 1) / (52 choose 5)

(b) P (3 aces and 2 kings) = (4 choose 3) x (4 choose 2) / (52 choose 5)

(c) P (4 of a kind) = (13 choose 1) x (4 choose 4) x (48 choose 1) / (52 choose 5)

(a) I thought I understood this. I see why you'd put (52 choose 5) on the bottom, since that's what you're doing overall. And I thought it was (4 choose 4) because you have a total of 4 aces and you want all 4 of them, and I thought it was (48 choose 1) because you have 48 cards left to choose from, and you only need 1.
Is this right?

(b) Again, I kind of understand this, applying my previous logic. You have 4 aces and need 3, you have 4 kings and need 2, and that's your total hand dealt, so you don't need any more.

(c) I have no idea where this comes from.

Any help would be greatly appreciated!

2. May 9, 2012

### Infinitum

Hi daleklama!

Your understanding of the first two is right. Now for the third one. how many 'types' of cards of are there? once you chose one type of a card, you need all possible cards belonging to that kind(just like the first case). That's 4. Then, there's one spare card you need to choose again.

3. May 9, 2012

### daleklama

Ohhh, I completely understand now! Thanks a million!

(13 choose 1) because you're picking a 'type' of card, like 7, or an ace, or a jack. This is what you'll collect all 4 of. There are 13 different 'types' of cards.

(4 choose 4) because once you've chosen the 'type', there are four different suits and you need all 4.

and then you've only got four cards left, and you need a fifth, so you choose that from the remainder.

Fantastic! :) Thank you!

4. May 9, 2012

### Ray Vickson

To put it another way: the answer here is 13 times the answer to part (a), and (13 choose 1) = 13.

RGV

5. May 10, 2012

### HallsofIvy

Staff Emeritus
Let me point out a different way of doing at least the first. There are 52 cards in the deck, of which 4 are aces. The probability that the first card drawn is 4/52= 1/13. Once that is done, there are 51 cards left and 3 of them are aces. The probabilitythat the second card drawn is also and ace is 3/51. Then the probability the third is an ace is 2/50= 1/25 and the probability the last is an ace is 1/49. Since there are no aces left, the fifth card can be anything so the probability of that is 1. The probability all four cards are aces is (4/52)(3/51)(2/50)(1/49)(1). Now we note that there are 5 different possible orders- writing "A" for an ace, "O" for anything Other than an ace, they are AAAAO, AAAOA, AAOAA, AOAAA, and OAAAA. The probability of four aces in five cards is 5(4/52)(3/51)(2/50)(1/49)(1) The numerator is clearly 5(4!). We can make the denominator into a factorial by including 48(47)(46)...(3)(2)(1) in both numerator and denominator:5 [4(3)(2)(1)/(52)(51)(50)(49)][48(47)(46)...(3)(2)(1)/49(47)(46)...(3)(2)(1)]= 5(4! 48!)/ 52!