# Thread: Cards & Probability: P(2 pairs) if dealt 4 cards

1. ## Cards & Probability: P(2 pairs) if dealt 4 cards

I just solved a multiple choice question but didn't get any of the answers. Let me know if I'm doing something wrong or if there is a mistake among the choices.

The Problem:

Given a standard playing card deck, what is the probability of being dealt 4 cards and getting exactly two differently ranked pairs?

My solution:

There are 52C4 ways to be dealt 4 cards from a deck.
There are 13C2 ways to select two different ranks from which to make pairs.
In each of those ranks there are 4C2 ways to select pairs.

P(2 pairs) = [(13C2)(4C2)(4C2)]/[52C4] = 0.0104, rounded to 4 decimal places

The "nearest" choice was 0.021, which is, after rounding, exactly double my answer. I suspect this resulted from a similar looking calculation:

P(2 pairs) = [(13x12)(4C2)(4C2)]/[52C4] = 0.0207

The reasoning behind this would be: "There are 13 ways to select your first rank and 12 ways to select your second rank."
However, doesn't that calculation double count the ways we can select two different ranks?

The other answers were all greater than 0.04.

Thanks for any help!

2. ## Re: Cards & Probability: P(2 pairs) if dealt 4 cards

Hello, kablooey!

It must be their error ... of the type you suspected.

3. ## Re: Cards & Probability: P(2 pairs) if dealt 4 cards

Well, by reasoning; after 1st 2 cards:
got pair : prob 3/51 [1]
not pair: prob 48/51 [2]

[1]:
3rd card: 48/50 (only 2 you don't want : matches to the pair)
4th card: 3/49 (to match your 3rd)

[2]:
3rd card: 6/50 (to get match with one of 1st two)
4th card: 3/49 (to match your 3rd)

SO:
[1] 3/51 * 48/50 * 3/49 + [2] 48/51 * 6/50 * 3/49 = .010372148...

4. ## Re: Cards & Probability: P(2 pairs) if dealt 4 cards

wiki "Poker Probability", you will get all the calculations

Thanks guys!