An unfair dice question Unfair die of 12 faces. The number 11 has a chance of  40% while other
  faces are equally likely. If someone and you are playing a game and
  whoever is closer to the answer will win, what is your strategy? Also, do
  you start first or second?


So I believe the best strategy is to pick $8$ because the expected value of this dice is given by $8.05454546$

Also, I think it's best to go second because my opponent can just choose the bigger part of my numbers. For example, if I choose $11$, they can choose $10$, and they'll have $6/11$ chance to win. 

Is this right? I don't know if there's a better strategy.
The answer is 10

If you pick 10, and your opponent picks 9, they will win if the dice is 1-9. That means there is a 9/11 * 60% = 49% chance of winning. However, if your opponent picks 11, then there is a 1/11 * 60% + 40% = 45% chance of winning. Either way, you've won because your opponent is always under 50%
The answer is 10

If you pick 10, and your opponent picks 9, they will win if the dice is 1-9. That means there is a 9/11 * 60% = 49% chance of winning. However, if your opponent picks 11, then there is a 1/11 * 60% + 40% = 45% chance of winning. Either way, you've won because your opponent is always under 50%