Question

suppose that you lose $10 if the dice sum to 7 and win $11 if the dice sum to 3 or 2. How much should you win or lose if any other number turns up in order for the game to be fair?

Answer 1

A fair game is when the expected value/gain is zero, and is calculated by
E[X] = ∑ P(x)*x.

We assume we don't need to pay to play the game.

When throwing two dice, the probabilitiy of throwing a 7 is
P(7)=6/36  (pays x=$10)
(6 because there are six possible outcomes, {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)}.
Similarly, 
P(2)=1/36 (pays x=$11)
P(3)=2/36 (pays x=$11)
We assume the payout is x=k for the other outcomes, with total probability of (36-6-1-2)/36=27/36
Therefore the expected gain is
E[X]=P(7)*10+P(2)*11+P(3)*11+P(others)*k

For the game to be fair, we have E[X]=0, or
P(7)*10+P(2)*11+P(3)*11+P(others)*k=0
=>
k=-(P(7)*10+P(2)*11+P(3)*11)/P(others)
=-(6*10+2*11+3*11)/(27)
= -115/27
=-4.26 approximately.