A pair of fair dice is rolled once. Suppose that you lose $8 if the dice sum to 10 and win $11 if the dice sum to 11 or 12. How much should you win or lose if any other number turns up in order for the game to be fair?

First find the possible dices rolls that give you 10, 11, and 12

Gives 10:

5,5

4,6

6,4

Gives 11 or 12:

5,6

6,5

6,6

The dice rolls have 6 * 6 = 36 possible combinations

The probability of getting a 10 is 3/32

The probability of getting an 11 or 12 is 3/32

multiplying those probabilities together with their reward/loss we have:

3/32 * - 8 = - 3/4

3/32 * 11 = 33/32

to make the game "fair" we need the probable amount of money you can win to be $0, so we can use the equation

-3/4 + 33/32 + (32/32 - 6/32) * x = 0

then you can just solve for x