Four balls into three boxes: no box left empty

Question

Four distinguishable balls are each dropped independently and uniformly at random into one of three distinguishable boxes. What is the probability that every box ends up with at least one ball?

Accepted Answer

There are 3^4 = 81 equally likely outcomes. Count the "bad" outcomes (some box empty) with inclusion-exclusion. One specific box empty: the four balls use two boxes, 2^4 = 16 ways, times \binom{3}{1} = 3. Two specific boxes empty: all balls in one box, 1^4 = 1 way, times \binom{3}{2} = 3. So \(\text{some box empty}) = 3 \cdot 2^4 - 3 \cdot 1^4 = 48 - 3 = 45. Favorable (no empty box) = 81 - 45 = 36, so P(\text{no empty box}) = \frac{36}{81} = \frac{4}{9} \approx 0.4444. Takeaway. With only four balls for three boxes, an empty box is more than half likely, so the no-empty probability drops to 4…

Four balls into three boxes: no box left empty

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