The Probability of Getting At Least 3-of-a-Kind in a 3-Deck Card Game

[CanadianEh]

What is the probability of...

1. Ten cards are dealt from a deck of 156 cards (3 standard 52 card decks) What is the probability of getting at LEAST 3-of-a-kind (up to 10-of-a-kind).
My Attempt:
3. P(at least 3-of-a-kind)
= 1 - P(10 different ranks) - P(9 different ranks)
= 1 - [13C10 * (12C1)^10 + 13C9 * (9C1) * (12C2)^1 * (12C1)^8] / [156C10]
= 1 - [ 286 * 12^10 + 715 * 9 * 66 * 12^8 ] / 1752195368913990
= 1 - 200325892276224 / 1752195368913990
= 1 - 0.1143285137207
= 0.8856714862793This is a data management question. Please help! Thank you.

 

[larstuff]

1. Ten cards are dealt from a deck of 156 cards (3 standard 52 card decks) What is the probability of getting at LEAST 3-of-a-kind (up to 10-of-a-kind).



My Attempt:
3. P(at least 3-of-a-kind)
= 1 - P(10 different ranks) - P(9 different ranks)
= 1 - [13C10 * (12C1)^10 + 13C9 * (9C1) * (12C2)^1 * (12C1)^8] / [156C10]
= 1 - [ 286 * 12^10 + 715 * 9 * 66 * 12^8 ] / 1752195368913990
= 1 - 200325892276224 / 1752195368913990
= 1 - 0.1143285137207
= 0.8856714862793


This is a data management question. Please help! Thank you.

I'm not quite sure what the question is here. What does 3-of-a-kind mean in this situation? Does it mean 3 of the ace of Spades, 3 spades, 3 aces...
In other words, precicely what is the outcome we seek to get?

And - anyway, do not forget all possible combinations in your calculation.

 

[CanadianEh]

3-of-a-kind is like 3 Fours, 3 Fives, etc... I need help either finding the probability to getting 3-of-a-kind if 10 cards are dealt or AT LEAST 3-of-a-kind. I hope that clears it up.

 

[larstuff]

3-of-a-kind is like 3 Fours, 3 Fives, etc... I need help either finding the probability to getting 3-of-a-kind if 10 cards are dealt or AT LEAST 3-of-a-kind. I hope that clears it up.

Yes, it does. There are 3x4 of each kind then in the three decks, that's to say 12 Fours, 12 Fives and so on. This is not a simple task, as far as I can see. Try to rule out the "oposite", that's to say, find the probability of getting only 1 of each kind + 1 or 2 of each kind + 2 of each kind. If this is possible, the rest will be the probability of getting 3 or more of a kind.

 

[CanadianEh]

Exactly, there are 12 of each kind and 10 cards are being dealt.

 

[CompuChip]

I assume you want the total probability, so it suffices to calculate
1 - P
where P is the probability "there are at most 2 of every card".

Then because all cards are equivalent, I'd first calculate the probabilities of
2,3,4,5,6,7,8,9,10,J
2,3,4,5,6,7,8,9,10,10
2,3,4,5,6,7,8,8,9,9
2,3,4,5,6,6,7,7,8,8
2,3,4,4,5,5,6,6,7,7
2,2,3,3,4,4,5,5,6,6

Then you can rename the cards (for example, in the first one I have chosen 2 - J but instead of those 10 you can have any other, so you'd get a factor binom(13, 10) I think, similarly for the others).

Could this work or did I miss any important points?