Two dice sums

For any subset $S$ of $\{2,3,4,5,6,7,8,9,10,11,12\}$ , let $E_S$ be the expected number of dice throws needed to complete the set of all possible outcomes, given that outcomes in the set $S$ have already been obtained. We want to know the value of $E_\varnothing$ . On the other hand, we know that $E_{\{2,3,4,5,6,7,8,9,10,11,12\}} = 0$ and, for any proper subset $S$ of $\{2,3,4,5,6,7,8,9,10,11,12\}$ , $E_S \; = \; 1 + \sum_{j=2}^{12} \frac{6 - |j-7|}{36}E_{S \cup \{j\}}$ so that $\left(1 - \sum_{j \in S} \frac{6 - |j-7|}{36}\right)E_S \; = \; 1 + \sum_{j \not\in S} \frac{6 - |j-7|}{36} E_{S \cup \{j\}}$ and hence $E_{S} \; = \; \frac{36 + \sum_{j \not\in S}(6 - |j-7|)E_{S \cup \{j\}}}{36 - \sum_{j \in S}(6- |j-7|)}$ The rest is a matter of computation. We can show that $E_\varnothing \; = \; \frac{769767316159}{12574325400}$ and so the desired answer is $769767316159 + 12574325400 = \boxed{782341641559}$ .

Technic of this article give simple answer

$\int_0^{\infty} 1-(1-e^{\frac{-6x}{36}})(1-e^{\frac{-x}{36}})^2(1-e^{\frac{-2x}{36}})^2(1-e^{\frac{-3x}{36}})^2(1-e^{\frac{-4x}{36}})^2(1-e^{\frac{-5x}{36}})^2 dx = \frac {769767316159}{12574325400}$

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