Chasing the one toy your kid actually wants

Question

Each cereal box contains one toy, chosen uniformly at random from $8$ different toys (independently across boxes). Your child only wants the gold dragon.

On average, how many boxes must you buy to get the gold dragon?

Accepted Answer

Every box is an independent trial that "succeeds" (contains the gold dragon) with probability p = \frac{1}{8}. The number of boxes until the first success is a geometric random variable, and the expected number of trials until the first success is 1/p: E[	ext{boxes}] = \frac{1}{p} = \frac{1}{1/8} = 8. So you should expect to buy 8 boxes on average. Be careful about the difference between expecting 8 and being guaranteed the dragon within 8: after 8 boxes the chance of still having no dragon is \left(	frac78ight)^{8} \approx 0.344, about a 34\% chance of striking out. The average is 8, but…

Chasing the one toy your kid actually wants

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