Probability with cards

[GreyhoundHBD120102]

What is the probability of getting 5 pairs of cards in the first 13 cards drawn of a 52 card standard deck? Not talking about one after another, but just 5 sets of the same card in any placement in those 13 cards. I did some math and I don't think it is accurate for my specific hand. If D is diamonds, H is hearts, S is spades, and C is clubs, I got 10D, 6H, 4S, 4C, 7D, 7S, 8S, QH, 10H, QD, 2S, AS, 8H. So I guess I have 2 questions. The first being already stated, and the second being, what is the probability of 2 of them being one after the other?. For my hand, I did this math and got this answer.
((4/52)(4/50)(3/49)(4/48)(3/47)(4/46)(4/45)(3/44)(3/43)(3/40))/(1/4) = 9/407132976250

 

[R.M.]

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.

 

[GreyhoundHBD120102]

Choosing the cards won't change anything with the probability, though it seems the only thing that might is the order of which the pairs come in. There is always only 4 possiblities for each ranks and with there being pairs, then the second time there is only 3. That does not change depending on rank, nor does the probability of 4 change on the one-of-a-kind. It shouldn't matter what pairs nor what one-of-a-kind cards they are. I am not positive, but it seems that if you get a pair in order from 1st card and 2nd card, it will have a different probability than that of 1st card and 3rd card. And that being amplified by the pairing of 5 ranks and 10 cards total. The 3 one of a kind only have to defy the majority or minority odds of creating a 3-of-a-kind.

(2 hours and 13 min. of hand written math later well on a call being slightly distracted) The probability of my equation actually came out to be 162/29184262103125 so the calculator couldn't resolve the fractions correctly with the way I put them in.

With combinations alone of any 13 cards, there is 635,013,559,600 possibilities. That is not separating them into 4 suits of the same card and then deciding on dealing with probability change of repeating cards (ignoring suit assignment) in 5 pairs with 3 one-of-a-kind cards. This is using the 52C13 formula or 52!/13!*(52-13)!. Now that I have all the 13 card possibilities along with in my original post, and the 13 cards that I pulled, can you help me figure out the probability of my 13 cards specifically? The suit can be ignored, though it would be cool to see the jurassic difference in probability with the suit being a factor as well.

 

[GreyhoundHBD120102]

648/29184262103125* my bad. Forgot to multiply it by 4 for the 1/4 of card being taken out and being divided by 1/4.

 

[Dr.Peterson]

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?