# Probability of a Specific 5 Card Poker Hand

1. Sep 23, 2009

### wooster

Hello everyone,

First post here so be kind :)

I am working on a problem, that should be easy, but I just want to make sure my logic is correct. The problem is to find the probability of being dealt event A in a 5 card poker hand, where

A = {2 pair and 1 card all of the same colour}

Now I took a counting approach to this problem, where I counted all the possible ways for A, then divided by 52 choose 5 (the total number of unordered five card poker hands). I tried two ways which seemed to make sense, but they give me different answers.

Method 1

First, choose two ranks: 13 choose 2
Then choose two cards from first rank: 2 choose 2
Then choose two cards from second rank: 2 choose 2
Choose last card in 26-4 ways: 22 choose 1

So P(A) = (13 choose 2)*(2 choose 2)*(2 choose 2)*(22 choose 1)/(52 choose 5) = 0.00066

Method 2

First, choose colour: 2 choose 1
Choose two ranks: 13 choose 2
Then choose two cards from first rank: 2 choose 2
Then choose two cards from second rank: 2 choose 2
Choose last card in 26-4 ways: 22 choose 1

So P(A) = (2 choose 1)*(13 choose 2)*(2 choose 2)*(2 choose 2)*(22 choose 1)/(52 choose 5) = 0.00132

The second probability is obviously two times the first due to the initial choosing of the colour, is this redundant? Which method, if any, is correct?

Thanks for the help.

2. Sep 23, 2009

### Office_Shredder

Staff Emeritus
In the first method, you "dealt" with the color issue by having the number of ways of picking a pair of a certain rank be 2 choose 2. That means you're only allowing the choice of black cards (alternatively: only allowing red cards). Which means that you didn't count all the ways of getting the hand with red cards (or black cards). So your first method is incorrect since you only counted half the possible hands

3. Sep 23, 2009

### wooster

Okay, that makes sense. So that means the second method is correct?