Two frequent flavors and a scarce one

Question

A dispenser gives one of 3 flavors per draw, with probabilities 0.4, 0.4, and 0.2, independently. On average, how many draws until you have tasted all three flavors?

Accepted Answer

With unequal odds, apply the weighted-collection formula from inclusion-exclusion: E[T] = \sumi \frac{1}{pi} - \sum{i<j}\frac{1}{pi+pj} + \frac{1}{p1+p2+p3}. Plug in p = (0.4,\,0.4,\,0.2): - Singles: \frac{1}{0.4}+\frac{1}{0.4}+\frac{1}{0.2} = 2.5+2.5+5 = 10. - Pairs: sums 0.8,\ 0.6,\ 0.6 give \frac{5}{4}+\frac{5}{3}+\frac{5}{3} = \frac{55}{12} \approx 4.5833. - Triple: \frac{1}{1} = 1. E[T] = 10 - \frac{55}{12} + 1 = 11 - \frac{55}{12} = \frac{77}{12} \approx 6.42 	ext{ draws.} The two common flavors arrive quickly; the scarce 0.2 flavor is the main reason the average sits above the equal-o…

Two frequent flavors and a scarce one

Your answer

More Probability questions