Tests to reach the single defective gadget

Question

A batch of 8 gadgets contains exactly 1 defective unit. They are tested one at a time in a uniformly random order until the defective one is found. On average, how many tests does it take to reach the defective gadget (including that test)?

Accepted Answer

With only one special item, the gap argument collapses to pure symmetry: the single defective is equally likely to be in any of the 8 positions, 1 through 8. Its expected position is just the average of 1 through 8: \frac{1 + 8}{2} = 4.5 	ext{ tests.} This matches the general formula \frac{N+1}{m+1} with N = 8, m = 1: \frac{9}{2} = 4.5. A lone special item lands, on average, right in the middle. Takeaway. One special item among N sits at expected position \frac{N+1}{2}, the midpoint; the lone defective among 8 averages 4.5.

Tests to reach the single defective gadget

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