 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  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%?

 See The Solution Submitted by Brian Smith Rating: 4.6667 (3 votes) Comments: ( Back to comment list | You must be logged in to post comments.) re: Triangles (solution) different simplification | Comment 4 of 5 | (In reply to Triangles (solution) by Jer)

(a/(a+b))*(b/(a+b-1))=1/4 also simplifes to

a+b = (a-b)^2

In other words, the total number of cards must be the square of the color card difference. Unfortunately, a + b = 49 does not work since a-b <= 52 - 49 = 3. The next square, a + b = 36 is a valid possibility and forces a - b = 6.  This reproduces your solution: 52 - 36 = 16 (with a = 21 and b =15).

 Posted by ajosin on 2007-03-07 00:53:08 Please log in:

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