Four balls, three boxes, an empty box

Question

You toss 4 distinguishable balls into 3 boxes, each ball landing in a uniformly random box, independently. What is the probability that at least one box ends up empty?

Accepted Answer

It is cleaner to find the complement first: the probability that every box gets at least one ball, then subtract from 1. There are 3^4 = 81 equally likely ways to place the four balls. Count the placements that fill all three boxes using inclusion-exclusion over the boxes that could be missed: \sum{k=0}^{3} (-1)^k \binom{3}{k}(3-k)^4 = 3^4 - \binom{3}{1}2^4 + \binom{3}{2}1^4 = 81 - 48 + 3 = 36. So the probability all boxes are filled is 	frac{36}{81} = 	frac49, and therefore P(	ext{some box empty}) = 1 - \frac{4}{9} = \frac{5}{9} \approx 0.556. With only one more ball than boxes, an empty …

Four balls, three boxes, an empty box

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