Brain Teaser on Simple Probability question 
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. 
 A: 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.
