A bag contains 16 billiard balls, some white and some black. You draw two balls at the same time. It is equally likely that the two will be the same color as different colors. What is the proportion of colors within the bag?

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The obvious answer, that there are eight balls of each color, doesn’t quite work: Of the 120 possible pairs drawn from such a bag, 64 have one ball of each color (8 white balls × 8 black balls) and 56 have two of the same color (28 possible pairs of white balls + 28 possible pairs of black balls). That gives a probability of 64/120 = 0.53 of getting a mixed draw and 56/120 = 0.47 of getting two balls of the same color.
But if the bag contains 6 balls of one color and 10 of the other, then there are 60 ways of getting a mixed draw, 45 ways of drawing the more common color, and 15 ways of drawing the less common color. 60 = 45 + 15, so the two outcomes are equally likely.
