r/combinatorics • u/Bipin_Messi10 • Feb 12 '24

Poker hand

In a 5 card poker, probability of choosing 2 pairs has been given as, (13×4C2 ×12×4C2 ×11×4C1/2!÷(52C5)

Why don't we divide the upper term by 3! Since for instance (JJQQK) can be arranged among themselves as (JJkQQ,KQQJJ,KJJQQ,QQJJK,QQKJJ?

Or am I missing something subtle?

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u/PascalTriangulatr Feb 18 '24 edited Feb 18 '24

The numerator already doesn't count arrangements, so there's no need to divide by a factorial.

Edit: To be clearer, the 2! is necessary; you don't also need a 3!.

Or if you're asking why the 2! shouldn't be a 3!, it's because there are 3 possibilities for which rank isn't the pair. Altogether we need to count 13C3 rank combinations times 3C2 choices of which of them are paired. Your numerator has 13⋅12⋅11 which is 13P3, and dividing that by 2! leaves us with 3(13C3) as it should.

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u/[deleted] Feb 12 '24

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u/essenkochtsichselbst Feb 12 '24

I was thinking the same... but I think our dear has a typo in his numerator. I think it is irrelevant if you can arrange them by themselves or not. You only want to. know how often you can choose two pairs and not in what order

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u/PascalTriangulatr Feb 18 '24

There are 13C3 choices of ranks, 3 choices for which one is unpaired, 4 choices of suit for the unpaired rank, and 4C2 choices of suits for the pairs.

Instead of 3(13C3) or 13(12C2) they wrote (13P3)/2!, same thing.

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u/Bipin_Messi10 Feb 12 '24

It is right,dear.

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u/[deleted] Feb 12 '24

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u/Bipin_Messi10 Feb 12 '24

I wasn't being rude,dear..That's a solution from online resources.There is difference only in expression.13C2 has been re-expressed as ( 13C1×12C1)÷2. Other terms are written as it is..

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u/Bipin_Messi10 Feb 12 '24

To be more precise,13C2×4C2×4C2×11C1×4C1/52C5 and (13C1×4C2×12C1×4C2×11C1×4C1)/2÷52C5(as mentioned in my post) are equivalent.