The question is,
S wants to wants to broadcast messages to A and B until both A and B receive at least K messages. Everytime S broadcast a message, A has a probability of $p_1$ to receive it and B has a probability of $p_2$ to receive it. Links S-A and S-B are assumed to be independent. Then what is the expectation of number of broadcast S needs to make sure both A and B receive at least K messages.
In my derivation, it is computed as the following, $ \begin{equation} \begin{array}{lcl} E & = & \sum_{x=K}^{\infty} x \cdot p(x)\\ \\ & = & \sum_{x=K}^{\infty} x \cdot\{ {{x-1}\choose{K-1}} p_1^{K-1}(1-p_1)^{x-K}p_1{{x} \choose {K}} p_2^{K}\\ \\ & & + {{x-1}\choose{K-1}} p_2^{K-1}(1-p_2)^{x-K}p_2{{x} \choose {K}} p_1^{K}\\ \\ & & - {{x-1}\choose{K-1}} p_1^{K-1}(1-p_1)^{x-K}p_1 {{x-1}\choose{K-1}} p_2^{K-1}(1-p_2)^{x-K}p_2\}\\ \\ & = & \sum_{x=K}^{\infty} \{ {{x-1}\choose{K-1}} x {{x} \choose {K}} p_1^K p_2^K (1-p_1)^{x-K}\\ \\ & & + {{x-1}\choose{K-1}} x {{x} \choose {K}} p_1^K p_2^K (1-p_2)^{x-K}\\ \\ & & - {{x-1}\choose{K-1}} x {{x-1} \choose {K-1}} p_1^K p_2^K (1-p_1)^{x-K} (1-p_2)^{x-K}\}\\ \\ & = & \sum_{x=K}^{\infty} \{ K{{x}\choose{K}} {{x} \choose {K}} p_1^K p_2^K (1-p_1)^{x-K}\\ \\ & & + K{{x}\choose{K}} {{x} \choose {K}} p_1^K p_2^K (1-p_2)^{x-K}\\ \\ & & - K{{x}\choose{K}} {{x-1} \choose {K-1}} p_1^K p_2^K (1-p_1)^{x-K} (1-p_2)^{K-k}\}\\ \\ & = & K p_1^K p_2^K \sum_{x=K}^{\infty} \{ {{x}\choose{K}} {{x} \choose {K}} (1-p_1)^{x-K} + {{x}\choose{K}} {{x} \choose {K}} (1-p_2)^{x-K}\\ \\ & & - {{x}\choose{K}} {{x-1} \choose {K-1}} (1-p_1)^{x-K} (1-p_2)^{x-K}\}\\ \end{array} \end{equation} $
How can I make this formula simpler? Thanks a lot.
Sorry about the misunderstanding. My goal is to compute the expectation of minimal number of broadcasts such that both A and B receive at least K messages.