Coupon Collector "Paradox"

Question

Example Scenario:

• Imagine a Player has an unfair die with sides: $\{A,B,C,D,E,F\}$

• The probability of each side $p_i = \{1/12,1/6,1/4,1/12,1/6,1/4\}$

• The player's goal is to obtain at least one of each $S = \{A,B,C\}$

Questions:

The Player wishes to answer 3 questions about the game.

1. What is the Expected Number of die rolls until obtaining the 1st successful roll?
2. What is the Expected Number of desired sides seen at least once in $n$ rolls?
3. What is the Expected Number of die rolls to complete the collection?

My Solutions to the Questions:

1. Using a Geometric Distribution approach we have that the expected waiting time until the 1st success is $$E[x] = 1/p$$ where $$p = \sum_{i\in S} p_i$$ therefor $$p = 1/12 + 1/6 + 1/4 = 1/2$$ and $$E[x] = 1/(1/2) = 2$$

2. I believe this solution can be obtained from the the inclusion-exclusion principle: $$E[x] = \sum_{i\in S} 1-(1-p_i)^n$$ thus for $$n = 2$$ $$E[x] = 1-(1-1/12)^2 + 1-(1-1/6)^2 + 1-(1-1/4)^2 \approx 0.902777778$$

3. This answer comes from Ross; Introduction to Probability Models as the solution to the Unequal Coupon Collector Problem: $$\int_0^\infty1-(1-e^\frac{-x}{12})(1-e^\frac{-x}{6})(1-e^\frac{-x}{4})dx = 14.6$$

My "Paradox":

I am not 100% confident in the solution to question 2. I do not know how to prove that formula is correct, but feels correct to me. However, the confusing thing to me is that under the $n = 2$ scenario why does the result not equal 1.0? It feels to me like the result for that question should either equal 1.0 to match the solution to question 1 since they feel like the same question just from two different view points. Or if the result was slightly higher than 1.0 I could also reason to my self that the excess is attributable to the scenarios where you obtain 2 successes in 2 rolls. But with the solution of $\approx 0.9$ I do not have an intuitive understanding of why that is true.

So I am asking if my solution to question 2 is correct. And if it is, is there an intuitive explanation for why the solution to question 2 for when n = solution from question 1 the result is not 1.0?

Answer 1

"Expectation is linear", but linearly, if you roll twice and get AA, BB or CC, then you would count that as 2 successes.

But for the purposes of question 2, getting the same desired face twice counts as only 1 success.

Therefore the result of question 2 should not be $1$ as your intuition almost-correctly thinks it should be, but instead $1 - ((\frac 1 {12})^2 + (\frac 1 {6})^2 + (\frac 1 {4})^2) = 1 - 7/72 \approx 0.902777778$.