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+b1))=1/4 also simplifes to
a+b = (ab)^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 ab <= 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 20070307 00:53:08 