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The probability of getting a hand with two pairs in poker is $C^{13}_2 \cdot C^4_2 \cdot C^4_2 \cdot 11 \cdot C^4_1.$

When I first started calculating the probability, I thought it was:

$$\binom{13}{1} \times \binom{4}{2} \times \binom{12}{1} \times \binom{4}{2} \times \binom{11}{1} \times \binom{4}{1}$$

Would someone please explain why my first thought is wrong? Since in the second part, you have already picked a pair, and there are only 12 ranks left to choose from to make the second pair.

Stefan4024
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Quaxton Hale
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    you might have picked : (1 1) (2 2) 3, and (2 2) (1 1) 3. In your calculation, those are counted as different ones. However, those are indistinguishable, so you divide by 2. – Sungjin Kim Oct 24 '13 at 19:28
  • OH, I see. Thank you! – Quaxton Hale Oct 24 '13 at 19:35
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    just thought I'd share that I thought the exact same thing and @SungjinKim you cleared it up for me thanks :) Great minds think alike ey Quaxton Hale – Kam Feb 19 '20 at 10:27

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