Expected number of rolling a pair of dice to generate all possible sums

Solution 1:

This is the coupon collectors problem with unequal probabilities. There is a treatment of this problem in Example 5.17 of the 10th edition of Introduction to Probability Models by Sheldon Ross (page 322). He solves the problem by embedding it into a Poisson process. Anyway, the answer is

$$ E[T]=\int_0^\infty \left(1-\prod_{j=1}^m (1-e^{-p_j t})\right) dt, $$

when there are $m$ events with probability $p_j, j=1, \dots, m$ of occurring.

In your particular problem with two fair dice, my calculations give

$$E[T] = {769767316159\over 12574325400} = 61.2173.$$

Solution 2:

Suppose the probabilities are $p_i$. For each set $S$, the probability that in the first $n$ terms we have only seen throws from $S$ is $$ p_S^n, \quad p_S \triangleq \sum_{i \in S} p_i. $$ The probability that we have not seen all outcomes by the $n$th throw is $$ r_n = \sum_{S \neq \emptyset} (-1)^{|S|+1} p_{\overline{S}}^n. $$ The required expectation is $$ E = \sum_{n \geq 1} r_n. $$ Changing the order of summation, the summand corresponding to $S$ contributes to the sum $$ (-1)^{|S|+1} \sum_{n \geq 1} p_{\overline{S}}^n = \frac{(-1)^{|S|+1} p_{\overline{S}}}{p_S}. $$ So we get the formula $$ E = \sum_{S \neq \emptyset} (-1)^{|S|+1} \left(\frac{1}{p_S} - 1 \right) = \sum_{S \neq \emptyset} \frac{(-1)^{|S|+1}}{p_S} - 1. $$ You can now in principle plug in the values of the dice and get the result. Since there are $11$ outcomes, you need to sum $2^{11} = 2048$ reciprocals (including the final $-1$).