Solved with stroller . The answer is yes. Yes as in a definitive, positive, and affirmative yes. Note that if Peter doesn't take $x_6x_7x_8x_9x_{10}$, then he is doomed to eternal death if Basil chooses that number and sets $x_1=x_2=\cdots=x_5=0$ (and of course $x_6=x_7=x_8=x_9=x_{10}=\pi^{1+\frac{2e}{\pi}}$). Now, suppose that Peter does take that number, and now in order to avoid dying in his own trap, Basil will choose $x_5x_7x_8x_9x_{10}$. Now, suppose Peter didn't choose $x_5x_6x_8x_9x_{10}$, which enables Basil to do so. This instantly will doom Peter, as Basil will move in to sweep up that number. Now Basil will set $x_1=x_2=x_3=x_4=0$, $x_5=x_6=x_7=e$, and $x_8=x_9=x_{10}=\pi^{2\cdot\pi^\pi}$. Noting that $10^{40}\geq x_8\geq 10^{36}$ while $x_7\leq 3$, which indicates that Basil has a product of at least \[x_5x_6x_8x_9x_{10}+x_5x_7x_8x_9x_10\geq 2\cdot e^2\cdot x_8^3\]and Peter has a product of at most \[x_5x_6x_7x_9x_{10}+x_5x_6x_7x_8x_{10}+x_5x_6x_7x_8x_9+x_6x_7x_8x_9x_{10}\leq e^2\cdot x_8^3+\pi\cdot e^3\cdot x_8^2\leq e^2x_8^3+4\cdot 1000\cdot 10^{50}\leq 1.5\cdot e^2\cdot x_8^3\] Thus, Peter is forced to take $x_5x_6x_8x_9x_{10}$. However, we can turn this on him very quickly by taking the $x_4x_7x_8x_9x_{10}$ card, and then setting $x_1=x_2=x_3=0$, $x_4=x_5=x_6=e$, $x_7=x_8=x_9=x_{10}=\pi^{2\cdot \pi^{\pi}}$. Then, Basil has a sum of at least \[x_5x_7x_8x_9x_{10}+x_4x_7x_8x_9x_{10}\geq 2e\cdot x_7^4\]while Peter has a sum of at most (there are only \[x_6x_7x_8x_9x_{10}+17378\cdot x_5x_6x_8x_9x_{10}\leq e\cdot x_7^4+10^6\cdot e^2\cdot x_8^3\leq e\cdot x_7^4+10^{140}\leq 1.5\cdot e\cdot x_7^4\] Thus, no matter what Basil does, he ends up beating Peter. Note to the kids: It's very crucial we chose the numbers as we did - it was important to choose $\pi^{2\cdot \pi^{\pi}}$ and $e$, as otherwise the solution completely fails and falls apart. @below After taking a look at the Riemann Hypothesis, it became apparent we needed to randomly throw in $\pi$s everywhere to make sense. And this turned out to work.

Wow, interesting solution! Can you explain where you got $\pi^{2\cdot \pi^{\pi}}$?