At least one of three kids grabs their own backpack

Question

Three children each grab one of the three identical-looking backpacks at random, so all 3! = 6 assignments are equally likely. What is the probability that at least one child grabs their own backpack?

Accepted Answer

Use the complement. A full mismatch is a derangement of 3 items, and D3 = 2 (the only no-match arrangements send 1,2,3 to 2,3,1 or to 3,1,2). So the chance nobody matches is 2/6, and P(	ext{at least one match}) = 1 - \frac{2}{6} = \frac{4}{6} = \frac{2}{3} \approx 0.6667. Takeaway. With only three items you can list the two derangements directly, giving a clean 2/3 for at least one match.

At least one of three kids grabs their own backpack

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