# Using Combinations: I feel stupid.

1. Oct 20, 2008

### Lukie

1. In Poker:

A) A flush is 5 cards of the same suit not in any order. How many flushes are possible?

B) A full house is 3 of a kind plus a pair. How many full houses are possible?

2. n C r (That's all I can think of right now)

3. For A)

(13 C 5) / (52 C 5) = 33 / 66640

For some reason I don't think I'm doing it right. We're doing combinations and I'm really lost on this question for some reason!

2. Oct 20, 2008

### Tedjn

Just some general advice: in each of these questions, you want to think in steps.

Read A again to find out for what it is asking. Then, consider what you are currently doing. What are you doing wrong? These ideas might help you out: how many flushes per suit? how many suits?

For B, consider choosing the 3 of a kind first. How many ways are there to choose 3 of a kind among 4 of the same card? How many different-numbered 3 of a kinds are there? How many different-numbered pairs are there after the 3 of a kind is picked? How many ways are there to choose those 2 of a kinds for each number?

3. Oct 21, 2008

### HallsofIvy

Take a card from a deck of 52. It can be any card at all.

Now take another card from the remaining 51. In order to get a flush, it has to be the same suit as the first. There are 11 of those remaining.
Now take another card. There are 10 of that original suit remaining.
The the fourth and fifth card have to be of the same suit. How many cards, of that same suit are left in the deck eacht time?

Remember the "fundamental counting principle": if A can happen in m ways and B can happen in n ways, then AB can happen in m*n ways.

4. Oct 21, 2008

Staff Emeritus
Excellent advice. The #1 problem students of probability run into is insisting on solving the problem in one bite. To be a successful student, one needs to overcome this, and learn to proceed piece by piece.

5. Oct 21, 2008

### Lukie

It's hard for me to digest in steps. I'm not exactly a math person. :P
For A) would it be...

13C5

Would that make sense at all?

For B...
To choose 3 of a kind from 4 cards... it would be 4 choose 3? That would equal 4 ways... For the pair, if you took 3 cards from the original deck, you would have 49. If you had 49, you would have 24 pairs to choose from, but there would be a lot of different ways.

Would you take 49 and choose 2? But then I'm thinking how do you know you're choosing 2 of the same cards? I'm not sure how to put it so that you're choosing 2 of the same card.

Last edited: Oct 21, 2008