Combinations of Poker HandsDate: 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 Check out our web site! http://mathforum.org/dr.math/ |
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