Three coupon types appear with probabilities 1/2, 1/3 and 1/6 on each independent draw. What is the expected number of draws needed to collect all three?

Three coupon types appear with probabilities 1/2, 1/3 and 1/6 on each independent draw. What is the expected number of draws needed to collect all three?

Approach: The total wait is the maximum of three dependent waiting times, so apply inclusion-exclusion over the events that a given subset of coupons is still missing, using the tail formula for the expectation.

73/10. Write T as the maximum of waiting times for the three coupon types. Inclusion-exclusion on the events that a subset S is still missing gives E[T] = sum over non-empty S of (-1)^{|S|+1}/p_S, where p_S is the total probability of S. The singletons contribute 2 + 3 + 6 = 11, the pairs contribute -(6/5 + 3/2 + 2) = -47/10, and the triple contributes +1, so E[T] = 11 - 47/10 + 1 = 73/10 = 7.3. The equiprobable coupon collector would need 3(1 + 1/2 + 1/3) = 5.5 draws, so the imbalance costs 1.8 extra draws and nearly all of the excess is spent waiting for the 1/6 coupon.

Follow-up: How does the answer change if the rarest coupon's probability falls to 1/100 with the other two rescaled proportionally?

Key concepts: coupon collector, inclusion-exclusion, maximum of waiting times, imbalanced probabilities.