Exactly two rolls show a chosen face

Question

Six balls are tossed independently into 3 bins, each ball equally likely to land in any bin. What is the probability that bin 1 holds exactly 2 of the six balls?

Accepted Answer

Each ball lands in bin 1 with probability \frac{1}{3} or elsewhere with probability \frac{2}{3}, independently. So the count in bin 1 is binomial with 6 trials and success probability \frac{1}{3}. For exactly 2 balls in bin 1: P = \binom{6}{2}\left(\frac13\right)^2\left(\frac23\right)^4 = 15 \cdot \frac{1}{9}\cdot\frac{16}{81} = \frac{240}{729} = \frac{80}{243} \approx 0.3292. Takeaway. The load in one bin is binomial \binom{m}{j}\left(\frac1n\right)^j\left(\frac{n-1}{n}\right)^{m-j}, regardless of what the other bins do.

Exactly two rolls show a chosen face

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