A pair of fair dice is rolled once. suppose that you lose ​$ 8 if the dice sum to 3 and win ​$ 13 if the dice sum to 10 or 12. how much should you win or lose if any other number turns up in order for the game to be​ fair?

• MsEHolt: You should lose $1.20 for any other outcome to make it a fair game.
  
  There are 2 ways to get a sum of 3 out of 36 outcomes; this probability is 2/36.
  There are 4 ways to get a sum of 10 or 12 out of 36 outcomes; this probability is 4/36.
  There are 36-(2+4) = 36-6 = 30 other outcomes out of 36; this probability is 30/36.
  
  We lose $8 if we get a sum of 3; this expected value is -8(2/36) = -16/36.
  We win $13 if we get a sum of 10 or 12; this expected value is 13(4/36) = 52/36.
  Using x as the amount we win or lose for the other outcomes, the expected value is 30x/36.
  
  In order to be a fair game, the amount we win and/or lose should come outto( $0 total.
  
  Together we have the equation
  -16/36+52/36+30x/36 = 0
  
  Adding the fractions, we have
  (-16+52+30x)/36 = 0
  (30x+36)/36 = 0
  
  Multiplying both sides by 36,
  30x+36=0*36
  30x+36=0
  
  Subtracting 36 from both sides,
  30x+36-36=0-36
  30x=-36
  
  Dividing both sides by 30,
  30x/30 = -36/30
  x=-36/30 = -1.2
  
  Thus we should lose $1.20 for any other roll.