Rounds until one of two unequal switches trips

Question

Two independent switches are checked each round. Each round, switch one trips with probability $1/4$ and switch two trips with probability $2/5$ , independently. You wait until at least one has tripped.

On average, how many rounds does that take?

Accepted Answer

Get the per-round chance that at least one switch trips, then use the first-success wait. The chance neither trips in a round is \left(1 - 	frac14ight)\left(1 - 	frac25ight) = 	frac34 \cdot 	frac35 = 	frac{9}{20}, so the chance at least one trips is p = 1 - 	frac{9}{20} = 	frac{11}{20}. The expected wait is 1/p: E = \frac{1}{11/20} = \frac{20}{11} \approx 1.82 	ext{ rounds.} Takeaway. Combine unequal per-round odds via p = 1 - \prod(1 - pi) = 11/20, so the wait is 20/11 \approx 1.82.

Rounds until one of two unequal switches trips

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