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There are two boxes. Box 1 has \$60 and Box 2 has \$40. Each of you and your friend picks one of the boxes. Each of you get paid by splitting what you open collectively. For example, if both of you pick Box 2, each gets paid \$20. If one picks Box 1 and the other picks Box 2, each gets paid \$50. You two cannot communicate in any form. You two cannot pass on any objects. Both of you are rational and will follow a same strategy.

What strategy would you use?

P.S. The deterministic method will end up in \$30 for each when both pick one. There probably is a solution involving random number for each getting paid \$50. But I cannot come up with one.

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  • $\begingroup$ Ask this on puzzling.stackexchange.com $\endgroup$ – user535339 Mar 13 '18 at 22:01
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    $\begingroup$ @idk this is a mathematics question tha is absolutely on topic here. Because a mathematical correct answer is required it does not fit to the puzzling site. $\endgroup$ – miracle173 Mar 13 '18 at 22:12
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If you each pick box 1 with probability $\frac35$ and box 2 with probability $\frac25$, then each of you gets paid $$ \$30 \cdot \frac{9}{25} + \$20 \cdot \frac{4}{25} + \$50 \cdot \frac{12}{25} = \$38 $$ in expectation.

This is the best we can do with a random strategy: the function $$f(p) = 30p^2 + 20(1-p)^2 + 50(2p(1-p))$$ is maximized at $p = \frac35$.

Aside from "outside-the-box" solutions that depend on being able to distinguish each other somehow in a canonical way so that you can deterministically make opposite decisions (e.g by using the alphabetical order of your names, or sharing an entangled pair of qubits), such a random strategy is the most general thing you can do, so you cannot win more than $\$38$ in expectation.

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