How many spins to see all four prizes?

Question

A prize wheel lands on one of 4 different prizes, each equally likely, independently on every spin. You keep spinning until you have seen all 4 prizes at least once.

On average, how many spins does that take?

Accepted Answer

Split the wait into 4 stages by how many distinct prizes you have already seen, and add the stage lengths with linearity of expectation. When you have already seen k - 1 prizes, a fresh prize comes up on the next spin with probability pk = \frac{4 - (k-1)}{4}. The number of spins in that stage is geometric with mean 	frac{1}{pk} = 	frac{4}{4-(k-1)}. Adding the four stage means, E[	ext{spins}] = 4\left(\frac11 + \frac12 + \frac13 + \frac14ight) = 4 	imes \frac{25}{12} = \frac{100}{12} = \frac{25}{3} \approx 8.333. Takeaway. The coupon-collector total for n equally likely outcomes is n\le…

How many spins to see all four prizes?

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