DIE DESCRIPTION
There is a $12$-sided Die with each face numbered as integer from $1$ to $12$.
The die is loaded! $40\%$ of rolls are of face $12$. While $60\%$ of rolls are of any other integer from $1$ to $11$, each integer with equal probability.
GUESSING GAME
Player $\mathcal{A}$ and $\mathcal{B}$ guess the number on a $12$-sided die. Player $\mathcal{A}$ guesses first. Player $\mathcal{B}$ guesses next! The closest guess wins the face value of the die.
Now, if Player $\mathcal{A}$ guesses 10.
What should player $\mathcal{B}$ guess?
I think it should be $9$, since $10$, $11$ and $12$ are together roughly equally probable to the numbers $1-9$ occurring, and if $\mathcal{B}$ guesses $9$, he will be win if it’s any number $1-8$.Is it more beneficial to guess first or second?
I believe it’s second, since you can make your choice based on acquired information (namely, what was the guess of First person). If you go first, second guesser can pick the number/range in better probability.When is it beneficial to pick $12$?
I believe it’s never, since $1-11$ have $60\%$ chance of winning, so you’ll always lose
Can anyone confirm if I am thinking correctly or not?
I think you miss the rule "The closest guess wins the face value of the die",if you select 9 ,maybe you will win the game but you can only get less value of the die, for example if the value of the die is 1, you only get one point,but if the value of the die is 12 he get 12 point.so if you Calculate the mathematical expectation of each value of the die, the sum of 1-11 is only 3.6 ,and 12 is 4.8,so i think guess 12 will win the game.