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so just a quick note. My question is very similar (but not the same) to the question at this link:

Probability of getting two pair in poker

I calculated the probability of getting two pair as

$P=$ $$\frac{{13\choose1}{4\choose2}{12\choose1}{4\choose2}{11\choose1}{4\choose1}}{52\choose5}$$

This was wrong, and the correct answer is:

$P=$ $$\frac{{13\choose2}{4\choose2}{4\choose2}{11\choose1}{4\choose1}}{52\choose5}$$

Upon looking at the correct answer, I now understand that:

$$\frac{{13\choose1}{4\choose2}{12\choose1}{4\choose2}}{52\choose5}$$

Introduces redundancy for the two pair. Ie it counts both a (6H, 6D, 3H, 3D, X) and a (3H, 3D, 6H, 6D, X). Which is why the correct usage is $13\choose2$, but there is another thing I don't understand.

My problem is I feel like by the same way my original answer introduces redundancy, the correct answer also has a redundancy with the remaining card. How does

$${13\choose2}{4\choose2}^2{11\choose1}{4\choose1}$$

not count (6H, 6D, 3H, 3D, X) and (X, 6H, 6D, 3H, 3D)?

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You pick the two (distinct) ranks that are to occur as pairs. That's $13\choose 2$.

You pick the suits for the higher of these two ranks. That's $4\choose 2$.

You pick the suits for the lower of these two ranks. That's $4\choose 2$.

You pick the rank of the additional card from the ranks not used for the pairs. That's $11\choose 1$.

You pick the suit for that card. That's $4\choose 1$.

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  • $\begingroup$ Ok, but I still don't understand how the calculation doesn't introduce redundancy for the last card. In my head, I would divide the entire probability by 2! $\endgroup$
    – Abid Rizvi
    Commented Feb 4, 2017 at 21:20
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    $\begingroup$ @Abid Keep in mind that the multiplication principle has you set up your steps in a particular order. In this case you pick the rank and suit for the singleton only after having picked the pairs. There exist no sequence of choices then with how this specific solution is phrased that end up with the singleton being picked before the others. Multiplication principle works because every outcome is counted via exactly one sequence of choices. The aforementioned mistake was because some were counted via two different sequences of choices. $\endgroup$
    – JMoravitz
    Commented Feb 4, 2017 at 21:23
  • $\begingroup$ @JMoravitz Ok I think I get it now. $\endgroup$
    – Abid Rizvi
    Commented Feb 4, 2017 at 21:27

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