# Probability question (choosing cards from a deck)

1. Sep 23, 2009

### jumbogala

1. The problem statement, all variables and given/known data
A special deck of 40 cards has 4 suits (hearts, diamonds, spades, and clubs) and 10 denominations (aces, ones, ... tens).

Four cards are randomly chosen from the deck. What is the probability that you get:
a) two pairs?
b) the third card selected is the first ace chosen?

2. Relevant equations
Note: when I say 40 c 4, I mean 40 choose 4 or (40!) / (4!36!)

3. The attempt at a solution
First I found the number of ways 4 cards can be selected from 40. It's 40 c 4, or 91390 ways.

a) So first you choose a denomination. There's 10 ways that can happen, so it's 10 x (4 c 2). Then choose a second denomination. There are 9 ways that can happen, so it's 9 x (4 c 2). All together 10 x (4 c 2) x 9 x (4 c 2) = 3240.

3240 / 91390 = 0.035... but that's not the answer. Where did I go wrong?

b) The first card can be any of 36 cards. The second card can be any of 35 cards. The third card can be any of 4 cards (only aces), and the fourth can be any of 37 cards.

Number of ways it can happen then: 36 x 35 x 4 x 37 = 186480. But that's way more than the sample space so that can't be right =\

2. Sep 23, 2009

### Dick

For a) I think you are forgetting the order of choosing the denominations doesn't matter. You are treating it like it does. For b), again, it's matter of order. You seem to think the sample space is only 91390. It's not. The number of ways to pull 4 cards in a specific order is 40*39*38*37=2193360. Right?

3. Sep 24, 2009

### jumbogala

For a) I'm not sure how to fix it so that the order doesn't matter. Within (4 c 2) the order is of no importance, so I need to fix the 10 and 9. The answer in the booklet shows to pick 2 denominations first, so (10 c 2) and then multiply by (4 c 2) twice. But I don't quite understand why... (and couldn't you have two pairs from the same denomination?)

And for b), that makes sense. I'm confused though, because it's only the third card that matters. The order of the first two doesn't make a difference, so do I need to account for that somewhere?

4. Sep 24, 2009

### Dick

You can't have two pairs from the same denomination. That would be four of a kind. (10 c 2) is exactly the number of ways to pick two denominations without regard for order. For b) you should divide your number of ways to select four cards IN ORDER with the first ace in the third place by ALL ways to select four cards IN ORDER. The fact the order of the other three cards doesn't really matter isn't important. You are just taking a ratio of possible ordered draws.

5. Sep 24, 2009

### jumbogala

Okay, a makes total sense now.

I still don't get b, but I don't know how else you could explain it. I don't get how the order of the cards comes in to play here. Sigh...

6. Sep 24, 2009

### Dick

I'll try this once more. If you count ALL of the ways to pick four cards in order, with the restriction that the 3rd card is an ace, and then you divide by ALL of the ways to pick four cards in order, then you will get the correct probability, right?

7. Sep 25, 2009

### jumbogala

But I thought the four cards weren't picked in any order (except the third card). Is there a way to compute the probability when the order doesn't matter?

8. Sep 25, 2009

### Dick

Now you are just confusing me. How can you pick only one out of the four cards 'in order'? You pick all four cards in order. Only the third one needs to be an ace.

9. Sep 25, 2009

### jumbogala

I'm confusing myself too! I was confused about whether the order of the first two cards mattered. I think I get it now though, so thanks!