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%?

(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