Why does calculating probability of 2 pairs seem so different from for a full house?

Question

For 5 card poker without discards/draws, the probability of being dealt a 2 pair hand is about 0.047539, and for a full house is about 0.001441. 1

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I first learned about calculating for a 2 pair hand. I incorrectly came up with:

$${ {{13\choose 1}{4\choose 2}{12\choose 1}{4\choose 2}{44\choose 1}} \over{52\choose 5}}={{(13)(6)(12)(6)(44)}\over2598960}\approx0.095078$$

Notably, my answer is exactly double the correct answer.

I learned that the correct way to calculate this is to choose the 2 ranks together that will be the pairs:

$${ {{13\choose 2}{4\choose 2}{4\choose 2}{44\choose 1}} \over{52\choose 5}}={{(78)(6)(6)(44)}\over2598960}\approx0.047539$$

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So, later, I moved on to calculating for a full house. I proudly remembered to choose the ranks together and incorrectly came up with:

$${ {{13\choose 2}{4\choose 3}{4\choose 2}} \over{52\choose 5}}={{(78)(4)(6)}\over2598960}\approx0.000720$$

I learned that the correct way to calculate this is to choose the 2 ranks separately:

$${ {{13\choose 1}{4\choose 3}{12\choose 1}{4\choose 2}} \over{52\choose 5}}={{(13)(4)(12)(6)}\over2598960}\approx0.001441$$

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Q1: For a 2 pair hand, why does choosing the ranks separately double count the possibilities? Why must we choose the 2 ranks together (or divide by 2?)

Q2: When we move onto a full house, why does this change? Why does choosing the ranks together under count the possibilities? Why must we choose the 2 ranks separately (or multiply by 2?)

Q3: I'm having a surprisingly hard time grasping how to come up with these on my own. Are there any tips or resources to help polish up these issues?

Answer 1

You double count in the first case, because the hand 4, 4, 7, 7, 9 is the same as 7, 7, 4, 4, 9. You have to fix the pair of which you will draw a second card each (like 4, 7 in this case).

Otoh the full house 5, 5, 5, 2, 2 is different from 2, 2, 2, 5, 5. You first fix the triple and then the pair.

Hope this makes sense.

Answer 2

Your mistake in the first case was you calculated as though assigning a value to one pair was distinct and different than assigning the value to the other pair. As the pairs are both pairs it doesn't matter which of the two get each value.

Your mistake in the second case was you calculated as though assigning a value to the pair was the same and indistinguishable from assigning a value to the triple. As the pair and the triple are distinct and different things is does matter which of the two get each value.

${13\choose 2}$ is the way to choice two values from $13$ (and it doesn't matter which gets which value. ${13 \choose 1}$ is the way to choice one value to give the the first person and ${12\choose 1}$ is the way to choice another value to the second where it does matter which gets which.

For two-pair it doesn't matter which pair gets which as they are both the same thing: pairs. For full house it does matter what the triple gets and what the pair get as they are two distinct things: a triple and a pair.