One common toy, two rarer ones

Question

A cereal brand hides one of 3 toys per box, but not equally: the common toy appears with probability $\frac{1}{2}$ , and each of the two rarer toys with probability $\frac{1}{4}$ , independently per box.

On average, how many boxes until you own all three toys?

Accepted Answer

With unequal odds, use the weighted collection formula, which comes from inclusion-exclusion on "type still missing after t boxes": E[T] = \sumi \frac{1}{pi} - \sum{i<j}\frac{1}{pi+pj} + \frac{1}{p1+p2+p3}. Plug in p = \left(	frac12,	frac14,	frac14ight): - Singles: 2 + 4 + 4 = 10. - Pairs: 	frac12{+}	frac14=	frac34,\ 	frac12{+}	frac14=	frac34,\ 	frac14{+}	frac14=	frac12, giving \frac{4}{3}+\frac{4}{3}+2 = \frac{14}{3}. - Triple: \frac{1}{1} = 1. E[T] = 10 - \frac{14}{3} + 1 = 11 - \frac{14}{3} = \frac{19}{3} \approx 6.33 	ext{ boxes.} Takeaway. The weighted-collection average i…

One common toy, two rarer ones

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