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Combinations of Poker Hands


Date: 05/15/98 at 20:34:12
From: Sam Bass
Subject: Poker hand combinations(different from similar ?)

I need to find how many combinations there are for each poker hand. I 
have the answers correct for: straight flush, 4 of a kind, full house, 
flush, and a straight. We were given the answers, but we have to 
explain the process. I have seen your explanation to 3 of a kind, 2 
pair, and 1 pair, and the answers you got are different from what we 
were given. Our answers are: 54,912, 123,552, and 1,098,240, 
respectively. I know these are right, because someone else in our 
class has gotten these answers.

Thanks.


Date: 05/18/98 at 12:47:20
From: Doctor Schwa
Subject: Re: Poker hand combinations(different from similar ?)

Sounds good so far. Of course, someone else getting the same answer is 
no guarantee that the answer is right. It has a lot more to do with 
the quality of the answer than the quantity. Still, your answers look 
correct to me. I'll work through my reasoning while I check your 
answers for myself, and maybe my thinking can help you do the required 
explaining of the process.

For 3 of a kind, first I choose which rank (2, 3, ..., J, Q, K, A)
it's going to be -- there are 13 ways to do that. Then, I have to 
choose 3 cards of that rank, and there are 4 ways to do that. Then I 
have to fill in the other two cards. They can't match, or the hand 
would be a full house. So I have 48 possible choices for the fourth 
card, and only 44 for the fifth card (the fifth card can't be from 
either of the previous ranks).

So my answer is 13 * 4 * 48 * 44 = 109,824.

But of course your answer is correct -- where did I go wrong? Yes, 
that's right, I counted each possible three of a kind twice. For 
instance, 3H 3C 3S 8D JS and 3H 3C 3S JS 8D were both counted under my 
scheme. So if I divide by 2 to eliminate that double counting, I get 
54,912 -- just like you did.

Two pair should be (13 choose 2) to choose which ranks the two pairs 
are, then (4 choose 2 * 4 choose 2) to choose the suits of the cards 
in the pairs, then 44 choices for the last card, for a total of 
123,552 -- so I agree with you again.

And finally for one pair, choose one rank to be the pair, 13 ways.
There are (4 choose 2) ways to pick the two suits. Then (12 choose 3) 
enumerates the ways to pick the ranks of the other 3 cards, since they 
can't match; and finally 4 * 4 * 4 ways to pick the suits for those 
three cards, for a total of 1,098,240.

My mistake on the three of a kind question is very interesting; when I 
saw the 109,824 up there being 10 times smaller than the number I just 
calculated, I noticed by chance that one pair is exactly 20 times more 
common than 3 of a kind. Why is that? There are two cards that could 
be the third of a kind, so where does the 40 come in to the one pair 
argument?

-Doctor Schwa, The Math Forum
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Associated Topics:
High School Permutations and Combinations

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