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 Fair Pairs (Posted on 2007-03-05)
The probability of drawing a pair of cards from a standard deck of 52 and having those cards be different colors is 26/51, slightly over 50%.

What is the smallest number of cards that need to be removed from the deck so that the probability of drawing a mismatched pair is exactly 50%?

 Submitted by Brian Smith Rating: 4.6667 (3 votes) Solution: (Hide) If there are R red cards and B black cards, then the number of ways to draw a matching pair is (R^2-R)/2 + (B^2-B)/2 and the number of ways to draw a pair of mismatched cards is R*B. For the probability of a mismatched pair to be 50%, then the two expressions must be equal. Then (R^2-R)/2 + (B^2-B)/2 = R*B This equation can be rearranged as (R-B)^2 = R+B. If N = R-B then N^2 = R+B. Then R = (N^2+N)/2 and B = (N^2-N)/2. The largest value of N which has R and B not exceeding 26 is N=6. Then R=21 and B=15, so the smallest number of cards that need to be removed is (26-21) + (26-15) = 16.

 Subject Author Date Answer K Sengupta 2008-10-03 05:57:21 re: Triangles (solution) different simplification ajosin 2007-03-07 00:53:08 Solution hoodat 2007-03-06 13:28:51 re: Triangles (solution)===> a referal Ady TZIDON 2007-03-05 13:42:43 Triangles (solution) Jer 2007-03-05 12:35:26

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