Probability of getting 2 pairs

[cse63146]

Homework Statement

You are dealt 4 cards (each card is numbered from 0 - 9). Each card is independed of the other card. What is the probability of getting 2 pairs?

Homework Equations

The Attempt at a Solution

P(2 pair) = *4C2P(1st = 2nd)P(3rd ≠ 2nd = 1st) P(4th = 3rd ≠ 2nd = 1st) = (6)(0.1)(0.9)(0.1)=0.054
Is this correct?

 

[Robert1986]

You need to tell us what deck the cards are being dealt from. It doesn't look like a regular deck, so it is kind of hard.

 

[cse63146]

It's not really a deck per say.

You are dealt a card from a stack of 10 cards (0-9), record the number, return the card back to the deck,shuffle, and draw again. Do this until you recorded 4 numbers.

The probability of getting any number is 0.1.

 

[HallsofIvy]

That is not at all what you said in your first post.

One way to get two pair (which does NOT include "four of a kind") is to get the first two the same (the first card can be anything, then the probability the second card matches is .1), the third card different from the first two (probability .9) and the fourth card matching that (probability .1).
But the order does not matter so multiply by 4!.

 

[cse63146]

P(2 pairs) = 4!(0.1)²(0.9). That makes sense. Thank you.

Just one more question.

Would the probability of getting 1 pair be:

₄C₂P(second card matches 1st)P(3rd card is different from 1st and 2nd)P(4th different from 3rd)

P(pair) = ₄C₂(0.1)(0.9)(0.8)?

 

[Dick]

That is not at all what you said in your first post.

One way to get two pair (which does NOT include "four of a kind") is to get the first two the same (the first card can be anything, then the probability the second card matches is .1), the third card different from the first two (probability .9) and the fourth card matching that (probability .1).
But the order does not matter so multiply by 4!.

4! isn't right. If you draw 1,2,3,4, then there are 4! ways to permute it. If you draw 1,1,2,2 there aren't 4! different arrangements of that.

 

[cse63146]

4! isn't right. If you draw 1,2,3,4, then there are 4! ways to permute it. If you draw 1,1,2,2 there aren't 4! different arrangements of that.

So would a combination fit more than a permutation (ie 4 Choose 2)?

 

[Dick]

So would a combination fit more than a permutation (ie 4 Choose 2)?

Sort of. Except now there's a subtle overcounting going on. If you pick 1 first, and then the other number 2, you are going to recount those cases when you pick 2 first and then the other number to be 1. Why don't you try counting how may different ways there are to pick two pairs and then divide by the total number of ways to pick 4 cards?

 

[cse63146]

That would mean I'm counting the pairs twice. So it would be:

(ie 4 Choose 2)/2 = 3 ways of choosing?

 

[Dick]

That would mean I'm counting the pairs twice. So it would be:

(ie 4 Choose 2)/2 = 3 ways of choosing?

Yes. Like I said, it's maybe more clear if you count the number of ways to choose two pair. First pick 2 numbers from the 10. Then multiply by the number of ways to distribute the first number, so C(10,2)*C(4,2), right? Then divide by the total number of ways to choose,10^4. Also if you start getting really confused, try and check by simplifying the problem. Suppose you have only 2 numbers in the deck. Now there's only 16 possible ways to choose and you can write them all down. See if your calculation also works in that case.

 

[cse63146]

I figured it out. Thank you all for your help.