Four balls into two boxes: neither box left empty

Question

Four distinguishable balls are each dropped independently and uniformly at random into one of two distinguishable boxes. What is the probability that both boxes end up with at least one ball?

Accepted Answer

There are 2^4 = 16 equally likely outcomes. The only "bad" outcomes put every ball in a single box. Count them with inclusion-exclusion. One specific box empty forces all four balls into the other box: 1^4 = 1 way, and there are \binom{2}{1} = 2 choices of which box is empty. (Both boxes empty is impossible.) So \(\text{some box empty}) = 2 \cdot 1^4 = 2, namely "all in box 1" or "all in box 2." Favorable = 16 - 2 = 14, so P(\text{both nonempty}) = \frac{14}{16} = \frac{7}{8} = 0.875. Takeaway. With two boxes the bad set is just the two all-in-one outcomes, so no-empty is simply 1 - 2/2^4 = 7…

Four balls into two boxes: neither box left empty

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