Seven darts, five sectors

Question

A board is divided into 5 equal sectors. You throw 7 darts, and each dart lands in a uniformly random sector, independently (assume every dart lands somewhere on the board).

On average, how many sectors receive no dart?

Accepted Answer

Linearity of expectation says the expected number of empty sectors is five times the probability that one chosen sector is empty. A single dart misses a given sector with probability \frac{4}{5}, so the sector stays empty across all 7 darts with probability \left(\frac{4}{5}ight)^{7} \approx 0.2097. Hence E[	ext{empty}] = 5 	imes \left(\frac{4}{5}ight)^{7} \approx 5 	imes 0.2097 = 1.049. Takeaway. The expected number of empty bins, n\left(\frac{n-1}{n}ight)^m, works for any balls-in-boxes setup, darts in sectors included.

Seven darts, five sectors

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