How to make a game with $2$ dice fair? You pick a number between $2$ and $12$. Then you roll $2$ dice. The result is the sum of the tosses.


*

*If your number is not the sum of the tosses then you lose a dollar.

*If your nmber is the sum of tosses, then you win $k$ dollars.


What is the best number to choose initially? What value of $k$ will make this game fair ?
Of course the most sum to appear is $7$.
But I want to know, are these question related or independent? Will I need to calculate value of $k$ dependent on what number I choose, or for all possible outcomes?
 A: If you want the game to be fair you want the expected gain/loss to be $0$, assuming you make the best choice. Put differently, you want to design the game that nobody can beat the host, the person that offers the game, but you also do not insist that the host always has an edge. 
Thus the two questions are related as you first need to know what is the optimal strategy.
As you said, it is picking $7$. When you pick the number $7$ your probability $p$ of winning is $6/36$, and that of loosing is $1-p= 30/36$. 
The expected gain/loss is 
$$(-1) \cdot \frac{30}{36} + k \ \frac{6}{36}.$$ 
You want this to be $0$. Solve for $k$.
A: The probability of the sum being $7$ is $\frac16$ and the probability of the sum not being $7$ is $\frac56$. To be fair you want to neither lose money or win money over the long run.
Pretend that playing the game $6$ times you lose $5$ of them and win once (this ratio does hold over the long run). Then you would lose five dollars and win $k$ dollars.


*

*To be fair $k=5$.

*If $k$ is less than $5$, you shouldn't play because over the long run you will lose.

*If $k$ is more than $5$, over the long run you will win.

