Six balls into three boxes: no box left empty

Question

Six 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 3^6 = 729 equally likely outcomes. Count the "bad" outcomes (some box empty) with inclusion-exclusion. One specific box empty: 2^6 = 64 ways, times \binom{3}{1} = 3. Two specific boxes empty: 1^6 = 1 way, times \binom{3}{2} = 3. So \(\text{some box empty}) = 3 \cdot 2^6 - 3 \cdot 1^6 = 192 - 3 = 189. Favorable (no empty box) = 729 - 189 = 540, so P(\text{no empty box}) = \frac{540}{729} = \frac{20}{27} \approx 0.7407. Takeaway. More balls make empty boxes less likely, so with six balls the no-empty probability climbs to 20/27.

Six balls into three boxes: no box left empty

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