Question

A pair of fair dice is rolled once. Suppose that you lose ​$ 9 9 if the dice sum to 10 10 and win ​$ 13 13 if the dice sum to 4 4 or 12 12. How much should you win or lose if any other number turns up in order for the game to be​ fair?

Answer 1

Number of combinations to get sum of 10 are (4, 6), (5, 5) , (6, 4)

Probability of getting a sum of 10 = 3/36 = 1/ 12

Number of combinations to get sum of 4 or 12 are (1, 3), (2, 3) , (3, 1), (6, 6)

Probability of getting a sum of 4 or 12 = 3/36 = 4/36 = 1/9

Let x be the amount that  you win or lose if any other number turns up.

Probability of any other outcome = 1 - (1/12 + 1/9) = 29/36

For fair game, the expected value of the game should be 0,

Expected value of the game = -9 * (1/12) + 13 * (1/9) + x * 29/36 = 0

=> -9 * (3/36) + 13 * (4/36) + x * 29/36 = 0

=> 29x = -25

=> x = -25/29 = -$0.86

Thus, if you lose $0.86 on any other number turns up then the game is​ fair.