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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.

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# Probability of a Specific 5 Card Poker Hand

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