Five cards are drawn from an ordinary deck of 52 playing cards. Find the probability of getting two pairs?

For example, the hand could be A-A-5-5-Q or 4-4-7-7-K.








The answer in the back of the book is 0.0475.


But, I would like to know HOW you would get that answer.

Five cards are drawn from an ordinary deck of 52 playing cards. Find the probability of getting two pairs?
well if the condition is to get ONLY two pairs so





probability is the ratio of successful combs and possible combs...





Successful combs:





groups of numbers 13 x 4 = 52


C(13,2)xC(4,2)xC(4,2)xC(11,1)xC(4,1)


C(13,2) because we don't wanna them repeated


C(4,2) combs of each of C(13,2) combinations


C(11,1) because we don't need of a full house, so this eliminate the possibility.





Possible combs:


C(52,5)





[C(13,2)xC(4,2)xC(4,2)xC(11,1)xC(4,1)]...


(13*12*4*3*4*3*11*4/2*2*2)/(52*51*50*4...


.... too lazy to solve :P but im sure that is the same answer
Reply:52/52 x 3/51 x 50/50 x 3/49 x 48/48 = 1 x 1/17 x 1 x 3/49 x 1 and there you have it, just divide 3 by the denominator.
Reply:Let's calculate the probability of drawing the hand Q-Q-K-K-X, where X is a non-queen, non-king card.





If we draw five cards, the probability of getting exactly two queens is HYPGEOMDIST(2,5,4,52) = 0.0399298. We are requiring getting exactly two queens as our result, so there are now 50 cards left in the deck.





Assuming that two of the cards we've drawn are queens, it is as if we drew two queens, then drew three other cards. The probability of getting exactly two kings when drawing three cards is HYPGEOMDIST(2,3,4,50) = 0.0140816. We are requiring getting exactly two kings as our result, so there are now 48 cards left in the deck.





It is as if we drew two queens, then two kings, then drew one card. We simply require that it is not a king or a queen. There are two kings and two queens left in the deck, so the probability of drawing a non-king, non-queen card is just 44/48 = 0.916667





Multiplying these three probabilities together gives us the probability of drawing five cards and getting exactly two queens, two kings, and one card which isn't a king or a queen. The probability is 0.0399298 * 0.0140816 * 0.916667 = 0.000515421.





Now, that is only one possible pair of pairs. Other possible pairs of pairs would be jacks and tens, aces and sixes, fives and threes, etc. The number of unique pairs of pairs is 1/2*13*12 = 78. Getting any of these pairs of pairs is equally as likely as getting the Q-Q-K-K-X hand, so we multiply our probability by 78:





P = 0.000515421 * 78


P = 0.0402028





I realize this is a different answer than your book gives, but I'll stick by mine until I find a mistake in my reasoning.





Note:





In MS Excel, the formula


HYPGEOMDIST(A,B,C,D)


has four arguments which mean the following:


A = the number of the desired card drawn


B = the number of cards drawn


C = the number of the desired card in the deck


D = the number of cards in the deck
Reply:You are probably studying the combinatorial function right now... First, the number of ways of choosing 5 cards from 52 is C(52,5) and this is your denominator. Your numerator is 13*C(4,2)*12*C(4,2)*C(44,1)*(1/2) which is the number of ways of choosing 2 pair.

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