Eight monster figures from a gumball machine

Question

A gumball machine dispenses one plastic monster figure per turn, chosen uniformly at random from 8 different figures, independently each time.

On average, how many turns must you take to end up with all eight figures?

Accepted Answer

Measure progress by the number of distinct figures you own. Holding k different figures, the next turn is useful with probability \frac{8-k}{8}, so it takes on average \frac{8}{8-k} turns to gain one more distinct figure. Add the stage averages for k = 0 through 7: E = 8\left(1 + 	frac12 + 	frac13 + 	frac14 + 	frac15 + 	frac16 + 	frac17 + 	frac18ight). The bracket is H8 = \frac{761}{280}, so E = 8 	imes \frac{761}{280} = \frac{761}{35} \approx 21.74 	ext{ turns.} The last figure alone averages 8 turns, showing how the tail dominates as the set grows. Takeaway. The full-set average …

Eight monster figures from a gumball machine

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