Balls in an urn (Posted on 2008-11-11)

	Submitted by pcbouhid
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Firstly, we can rule out a trivial case. Clearly there must initially be more than one ball of each color, for otherwise the probability of drawing two balls of matching color would be zero before adding the new balls, and greater than zero afterwards. We now calculate the probability of drawing matching colors, before and after adding the extra balls. All drawings are understood to be without replacement. BEFORE: Let there initially be cn balls; comprised of c colors, with n > 1 balls of each color. The number of ways of drawing two balls is cn(cn − 1). (There are cn choices for the first ball; cn − 1 choices for the second.) The number of ways of drawing two balls of a particular color is n(n − 1). Summing over all colors, the number of ways of drawing matching colors is cn(n − 1). Hence the probability of drawing matching colors is cn(n-1)/cn(cn-1) = (n-1)/(cn-1). AFTER: Let k balls of a new color be added. (We will set k = 20 at an appropriate point.) The number of ways of drawing two balls is (cn + k)(cn + k − 1). The number of ways of drawing matching colors is cn(n − 1) + k(k − 1). Hence the probability of drawing matching colors is [cn(n-1)+k(k-1)] / [(cn+k)(cn+k-1)]. Equating the above before and after probabilities, we get (cn − 1)[cn(n − 1) + k(k − 1)] = (n − 1)(cn + k)(cn + k − 1) Expanding, we have: c2n3 − cn2 − c2n2 + cn + cnk2 − cnk − k2 + k = = c2n3 + 2cn2k + nk2 − cn2 − nk − c2n2 − 2cnk − k2 + cn + k. Simplifying, we find that most terms cancel, yielding cnk2 = 2cn2k + nk2 − nk − cnk. Dividing by nk (which is non-zero), and regrouping: c(k + 1 − 2n) = k − 1 Substituting k = 20, we get c(21 − 2n) = 19. The only solution with c > 1 is c = 19, n = 10. Hence there were initially 19 × 10 = 190 balls in the urn.

	Subject	Author	Date
	Solution	Praneeth	2008-11-12 07:02:32
	Agreed	ed bottemiller	2008-11-11 19:03:23
	Computer Search	ed bottemiller	2008-11-11 18:35:11
	computer trial and error	Charlie	2008-11-11 18:22:28