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a asymmetrical dice with 12 faces, 40% to be 12, others are equivalent. two people, choose a number. The man whose number is closer to the result will win. which number should you choose.

Attempt:

Calculate the expected values: E[X] = 0.4 * 12 + 0.6/11*(1+2+3…. +11) = 0.4 *12 + 0.6/11 * (11/2 *12) = 8.4 Choose 8.

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    I think you should actually choose the *median*, not the mean. – 2017-02-19
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    why is the median, instead of the expected value? – 2017-02-19
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    The number $m$ that minimizes $E[|X-m|]$ is the median. The number $\mu$ that minimizes $E[(X-\mu)^2]$ is the mean. Since you are measuring "closer to the result" using the usual distance (not square distance), you should use the median. – 2017-02-19

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I assume you are picking first and your opponent is picking second?

It matters cause, say you're picking second and your opponent picks $12.$ Then you should pick $11$ so that you maximize your probability of being closer. In fact, the optimal strategy for picking second is to pick one number to either side of the number your opponent picks, whichever side has the higher total probability. This is then your probability of victory.

So, then, if you're picking first, you want to minimize the probability on either side of your number. So your optimal choice is near the median. If you pick $12$ then there's $60\%$ to the left of you. If you pick $11$ there's $40\%$ to the right of you and $10*.6/11 \approx 54\% $ to the left of you. If you pick $10$, there is $.4+.6/11 \approx 45\%$ to the right of you and $9*.6/11\approx 49\%$ to the left of you. If you pick $9$ there's $.4+2*.6/11 \approx 51\%$ to the right of you and $8*.6/11 \approx 44\%$ to the left of you. So it's clear that the probability to the right is going to keep going up, so you want to pick $10.$ Notice that picking 10 gives you a $51\%$ chance of winning if your opponent chooses optimally and plays $9$ so the first player has the advantage.

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    Yes. I agree with you that the ordering matters a lot. But what if we are allowed to pick duplicate number? – 2017-02-19
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    @wrek Then I think both players should pick 10 and push every time and the game would be rather boring. – 2017-02-19
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    Thanks for your reply. What is the reasoning for picking 10? – 2017-02-19
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    @wrek The first player should pick 10 cause it maximizes their chances of winning assuming player 2 plays their best response. If the first player picks 10, player 2 should pick 9, giving player 2 a $49\%$ chance of winning, which is the best they can do. If the second player is allowed to pick 10, then they have a $100\%$ of tying, which is better, in most peoples' books, than a $49\%$ chance of winning, so they should pick $10.$ – 2017-02-19
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    I see. Thanks, that clear things up a lot! – 2017-02-19