# Easy Probability

1. Nov 5, 2005

### Cyrus

I have a quick one.

An example in a book im reading says

(a) what is the probability of being delt four of a kind, that is, for aces, four kings, and so forth?
(one standard 52 card deck)
And they say "there are only 48 possible hands containing 4 aces; another 48 containing 4 kings etc. Thus there are 13*48 possible four-of-a-kind hands"

There are 13 different cards, A-2-10-J-Q-K, doesnt that mean there are 13 possible four of a kind hands? What am I not seeing here......

Thank you.

2. Nov 5, 2005

### Cyrus

ooooooooooooooooooooooooooooooooooooo........... its a 5 card hand, so theres a remainer of 1 card.................how stupid of me.........sighhhhhhhhhhhhhhhh......... delete this thread please!

3. Nov 6, 2005

### Cyrus

Aha, I have a real question now.

The number of possible outcomes is given by the binomial coefficient, defined as:

$$(n,k) = \frac {n!} {k!(n-k!)}$$

I read a nice article online about how this is proved using binomial expansion and pascals triangle. Proof makes sense. I see how it works for pascals triangle, but I dont see the connection to the total number of outcomes possible. It goes for the problem of how many 5 card hands can you have with a 52 deck of cards. I dont see how this definition is applicable. (Please dont tell me plug in 52 for n and 5 for k, I know that, im asking WHY does this work). And I know im gonna get "because its a definition." Show me how its APPLICABLE. :-)

Cheers

4. Nov 6, 2005

### eok20

Say, you have n objects and you want to find out how many ways you can order k of them. You'll have n options for the first one, n-1 for the next...n-k+1 for the last, so by the multiplicative principle there will be n*(n-1)*...*(n-k+1) possibilities, which can be written as n!/(n-k)! Now you know that the number of combinations (where order does not matter) times the amount of ways to order them will be the same as k!/(k-n)! With k objects you have k options for the first item, k-1 for the second, and so on. so there are k! ways to order them. so nCk*k! = n!/(n-k)!. Dividing by n! yields what youre looking for.

EDIT: mixed some k's with n's

Last edited: Nov 6, 2005
5. Nov 6, 2005

### Cyrus

Say, you have n objects and you want to find out how many ways you can order k of them. You'll have n options for the first one, n-1 for the next...n-k+1 for the last, so by the multiplicative principle there will be n*(n-1)*...*(n-k+1) possibilities, which can be written as n!/(n-k)!

How come this alone is not the anwser? You said that you have n, or 52 cards, and you found out how many ways I can order k, or 5 of them. Im not following why you then multiplied this by k!

6. Nov 6, 2005

### eok20

n!/(n-k)! is caleed number of permutations of k elements. the reason that this is not the answer is because this counts 1,2,3 and 1,3,2 as two different permutations but for what you are looking for order doesn't matter and so 1,2,3 and 1,3,2 should not be counted twice (it doesn't matter in what order you get the cards). The reason that you have to divide by k! is because k! is how many ways you can order k elements.

7. Nov 7, 2005

### robert Ihnot

How to solve that lottery type of problem is: 13*4C4*48C1/52C5.

Here we divided the deck into two parts, one which contains all four cards of one suit, and the other which contains the remaining 48 cards, of which we draw one. Thus the numenator is: 13*1*48 = 624. This figure then is divided by 2598960, giving us .024%, which makes sense since it is very unlikely you will hold such a hand.

Why the 13? Well that comes about because THEY ARE 13 CASES HERE. If you'd rather we could just add the cases: for the Aces, for the duces, for the 3s, etc. Also, there is clearly only one way we can choose four from four, and 48 ways we can chose one from 48.

Last edited: Nov 7, 2005