Four balls into three bins, all used

Question

Four balls are dropped independently into 3 bins, each landing in a uniformly random bin. What is the probability that all three bins are occupied?

Accepted Answer

There are 3^4 = 81 equally likely assignments. Use inclusion-exclusion on the events "bin i is empty." - All assignments: 3^4 = 81. - Subtract those that skip a chosen bin: 3 \cdot 2^4 = 3 \cdot 16 = 48. - Add back those that skip a chosen pair: 3 \cdot 1^4 = 3. Assignments with no empty bin: 81 - 48 + 3 = 36. So P(	ext{all occupied}) = \frac{36}{81} = \frac{4}{9} \approx 0.4444. With only four balls, one bin is fairly likely to be missed, so the probability sits below one half. Takeaway. When balls barely exceed bins, the all-occupied probability is modest; the surjection formula \sum{k}(-1…

Four balls into three bins, all used

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