Waiting for the next new hero

Question

A game unlocks one random hero per pull, drawn uniformly from 12 heroes, independently each time. You already own 9 distinct heroes.

On average, how many pulls until you unlock a hero you do not already have?

Accepted Answer

You only care about the next brand-new hero, so ignore the past pulls. On any single pull, the hero is new whenever it is one of the 12 - 9 = 3 you still lack, which happens with probability p = \frac{3}{12} = \frac{1}{4}. Waiting for a repeated independent event of probability p takes on average \frac{1}{p} tries (a geometric distribution). So the expected number of pulls is \frac{1}{p} = \frac{12}{3} = 4. Takeaway. When you hold k of n equally likely types, the wait for the next new type is geometric with success probability \frac{n-k}{n}, so it averages \frac{n}{n-k} draws.

Waiting for the next new hero

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