One charm away from a milestone

Question

A machine dispenses one charm per token, uniformly from 8 charm types, independently. You currently own 5 different charms.

On average, how many charms must you get before one of them is a type you do not yet have?

Accepted Answer

Only the next new charm matters, so past draws are irrelevant. A single draw is new when it is one of the 8 - 5 = 3 missing types, with probability p = \frac{3}{8}. The wait for this event is geometric, averaging \frac{1}{p}: \frac{8}{3} \approx 2.67 	ext{ charms.} Takeaway. The next-new-type wait averages \frac{n}{n-k}; owning 5 of 8 makes it \frac{8}{3}.

One charm away from a milestone

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