Choosing the cards won't change anything with the probability
I think you are misunderstanding what
@R.M. said.
The idea is that the denominator is the number of ways to get a hand of 13 cards, and the numerator is the number of ways to get a hand that satisfies the requirements. We think of that as the
number of ways to choose such a set of cards. He did not say you have to actually choose some cards yourself.
Here is what he said:
The denominator is the number of possible 13 card hands: 52C13. I see this as a combination problem for the numerator. Then you need to choose which 5 cards from the 13 ranks will have pairs, then choose the remaining 3 one-of-a-kind cards from the remaining 8 ranks.
Suppose you want to intentionally build a hand with 5 pairs. How can you do so? You might first choose 5 "ranks" (numbers or faces) in which to make pairs; you can do that in 13 choose 5 ways, right? Then, for each of those, you have to pick 2 of the 4 suits -- how many ways can you do that? After you've done that, you've picked 10 cards for the hand; you have to pick 3 more cards, none of which are from the same rank. How many ways can you do that? Then multiply them all together. Put this numerator over the denominator you already know, and you have your probability.
What you tried to do, multiplying probabilities, doesn't allow for the fact that order doesn't matter. This approach is more natural.
(By the way, all this assumes you want
exactly 5 pairs, not
at least 5 pairs, and not allowing them to really be three or four of a kind.)
it would be cool to see the jurassic difference in probability with the suit being a factor as well.
What does
that mean?
But it appears that you are asking, ultimately, how rare a
specific hand you got is. That is really a fruitless question. Every hand is equally unlikely! See this page,
Should Rare Events Surprise Us?